# Chapter 28 Solutions

Size: px
Start display at page:

Transcription

2 Chapter 8 Solutions (a) V term IR becomes 10.0 V I (5.60 Ω) so I 1.79 A (b) V term ε Ir becomes 10.0 V ε (1.79 A)(0.00 Ω) so ε 10.4 V 8.3 The total resistance is R 3.00 V A 5.00 Ω (a) (b) R lamp R r batteries 5.00 Ω Ω 4.59 Ω P batteries P total (0.408 Ω)I % (5.00 Ω)I 8.4 (a) Here ε I(R + r), so I ε R + r 1.6 V (5.00 Ω Ω).48 A Then, V IR (.48 A) ( 5.00 Ω) 1.4 V (b) Let I 1 and I be the currents flowing through the battery and the headlights, respectively. Then, I 1 I A, and ε I 1 r I R 0 so giving Thus, ε (I A)( Ω) + I (5.00 Ω) 1.6 V I 1.93 A V (1.93 A)(5.00 Ω) 9.65 V 8.5 V I 1 R 1 (.00 A)R 1 and V I (R 1 + R ) (1.60 A)(R Ω) Therefore, (.00 A)R 1 (1.60 A)(R Ω) or R Ω

3 136 Chapter 8 Solutions (a) R p ( Ω) Ω ( ) 4.1 Ω R s R 1 + R + R Ω (b) V IR 34.0 V I (17.1 Ω) I 1.99 A for 4.00 Ω, 9.00 Ω resistors Applying V IR, (1.99 A)(4.1 Ω) 8.18 V 8.18 V I (7.00 Ω) so I 1.17 A for 7.00 Ω resistor 8.18 V I (10.0 Ω) so I A for 10.0 Ω resistor *8.7 If all 3 resistors are placed in parallel, 1 R and R 100 Ω *8.8 For the bulb in use as intended, I P V 75.0 W V 0.65 A and R 10 V I 10 V 0.65 A 19 Ω Now, presuming the bulb resistance is unchanged, I 10 V Ω 0.60 A Across the bulb is V IR 19 Ω(0.60 A) 119 V so its power is P ( V)I 119 V(0.60 A) 73.8 W

4 Chapter 8 Solutions If we turn the given diagram on its side, we find that it is the same as Figure (a). The 0.0-Ω and 5.00-Ω resistors are in series, so the first reduction is as shown in (b). In addition, since the 10.0-Ω, 5.00-Ω, and 5.0-Ω resistors are then in parallel, we can solve for their equivalent resistance as: R eq 1 ( ).94 Ω Ω 5.00 Ω 5.0 Ω This is shown in Figure (c), which in turn reduces to the circuit shown in (d). Next, we work backwards through the diagrams applying I V/R and V IR. The 1.94-Ω resistor is connected across 5.0-V, so the current through the battery in every diagram is I V R 5.0 V 1.94 Ω 1.93 A (a) In Figure (c), this 1.93 A goes through the.94-ω equivalent resistor to give a potential difference of: V IR (1.93 A)(.94 Ω) 5.68 V From Figure (b), we see that this potential difference is the same across V ab, the 10-Ω resistor, and the 5.00-Ω resistor. (b) (b) Therefore, V ab 5.68 V (a) Since the current through the 0.0-Ω resistor is also the current through the 5.0-Ω line ab, I V ab R ab 5.68 V 5.0 Ω 0.7 A 7 ma (c) (d) ( 10 V) V IR eq I ρl + ρl + ρl + ρl, or Iρl A 1 A A 3 A A 1 A A 3 A 4 V Iρl A ( 10 V) 1 A V A 1 A A 3 A 4

5 138 Chapter 8 Solutions 8.11 (a) Since all the current flowing in the circuit must pass through the series 100-Ω resistor, P RI P max RI max so I max P R 5.0 W 100 Ω A R eq 100 Ω Ω 150 Ω V max R eq I max 75.0 V (b) P ( V)I (75.0 V)(0.500 A) 37.5 W total power P W P P 3 RI (100 Ω)(0.50 A) 6.5 W 8.1 Using.00-Ω, 3.00-Ω, 4.00-Ω resistors, there are 7 series, 4 parallel, and 6 mixed combinations: Series Parallel Mixed The resistors may be arranged in patterns:.00 Ω 6.00 Ω 3.00 Ω 7.00 Ω 4.00 Ω 9.00 Ω 5.00 Ω 0.93 Ω1.56 Ω 1.0 Ω.00 Ω 1.33 Ω. Ω 1.71 Ω 3.71 Ω 4.33 Ω 5.0 Ω 8.13 The potential difference is the same across either combination. V IR 3I 1 + R ( ) 1 R so R 1 R and R 1000 Ω 1.00 kω 8.14 If the switch is open, I ε /( R + R) and P ε R /( R + R) If the switch is closed, I ε /(R + R / ) and P ε ( R /)/(R + R /) Then, ε R ε ( R + R) R (R + R /) R +R R + R / R +R R + R The condition becomes R R / so R R (1.00 Ω) 1.41 Ω

6 Chapter 8 Solutions R p Ω R s ( ) Ω6.75 Ω I battery V R s 18.0 V 6.75 Ω.67 A P I R: P (.67 A) (.00 Ω) P 14. W in.00 Ω P 4 (.67 A) ( 4.00 Ω) 8.4 W in 4.00 Ω V (.67 A) (.00 Ω) 5.33 V, V 4 (.67 A) ( 4.00 Ω) V V p 18.0 V V V 4.00 V ( V 3 V 1 ) ( P 3 V 3) R 3 ( P 1 V 1) R 1 (.00 V) 1.33 W in 3.00 Ω 3.00 Ω (.00 V) 4.00 W in 1.00 Ω 1.00 Ω 8.16 Denoting the two resistors as x and y, x + y 690, and x + 1 y x x (690 x) + x x(690 x) x 690x + 103,500 0 x 690 ± (690) 414,000 x 470 Ω y 0 Ω

7 140 Chapter 8 Solutions 8.17 (a) V IR: 33.0 V I 1 ( 11.0 Ω) 33.0 V I (.0 Ω) I A I 1.50 A P I R: P 1 ( 3.00 A) ( 11.0 Ω) P ( 1.50 A) (.0 Ω) P W P 49.5 W The 11.0-Ω resistor uses more power. (b) P 1 + P 148 W P I( V) ( 4.50) ( 33.0) 148 W (c) R s R 1 + R 11.0 Ω+.0 Ω33.0 Ω V IR: 33.0 V I( 33.0 Ω), so I 1.00 A P I R: P 1 ( 1.00 A) ( 11.0 Ω) P ( 1.00 A) (.0 Ω) P W P.0 W The.0-Ω resistor uses more power. (d) P 1 + P I ( R 1 + R ) ( 1.00 A) ( 33.0 Ω) 33.0 W P I( V) ( 1.00 A) ( 33.0 V) 33.0 W (e) The parallel configuration uses more power (7.00)I 1 (.00)(5.00) I 1 so I A I 3 I 1 + I.00 A I.00 so I 1.9 A +ε.00(1.9) (5.00)(.00) 0 ε 1.6 V

8 Chapter 8 Solutions We name the currents I 1, I, and I 3 as shown. From Kirchhoff's current rule, I 3 I 1 + I Applying Kirchhoff's voltage rule to the loop containing I and I 3, 1.0 V (4.00)I 3 (6.00)I 4.00 V (4.00)I 3 + (6.00)I Applying Kirchhoff's voltage rule to the loop containing I 1 and I, (6.00)I 4.00 V + (8.00)I 1 0 (8.00)I (6.00)I Solving the above linear systems, I ma, I 46 ma, I A All currents flow in the directions indicated by the arrows in the circuit diagram. *8.0 The solution figure is shown to the right. *8.1 We use the results of Problem 19. (a) By the 4.00-V battery: U ( V)It 4.00 V( 0.46 A)10 s J By the 1.0-V battery: 1.0 V (1.31 A) 10 s 1.88 kj (b) By the 8.00 Ω resistor: I Rt (0.846 A) (8.00 Ω ) 10 s 687 J By the 5.00 Ω resistor: (0.46 A) (5.00 Ω ) 10 s 18 J By the 1.00 Ω resistor: (0.46 A) (1.00 Ω ) 10 s 5.6 J By the 3.00 Ω resistor: (1.31 A) (3.00 Ω ) 10 s 616 J By the 1.00 Ω r e s i s t o r : (1.31 A) (1.00 Ω ) 10 s 05 J (c) J kj 1.66 kj from chemical to electrical. 687 J + 18 J J J + 05 J 1.66 kj from electrical to heat.

9 14 Chapter 8 Solutions 8. We name the currents I 1, I, and I 3 as shown. [1] I (3.00 kω) I 1 (.00 kω) 0 [] 80.0 I 3 (4.00 kω) 60.0 I (3.00 kω) 0 [3] I I 1 + I 3 (a) Substituting for I and solving the resulting simultaneous equations yields I ma (through R 1 ) I 3.69 ma (through R 3 ) I 3.08 ma (through R ) (b) V cf 60.0 V (3.08 ma)(3.00 kω) 69. V Point c is at higher potential. 8.3 Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure. Apply Kirchhoff s loop rule to both loops in Figure (b) to obtain: (.71R)I 1 + ( 1.71R)I 50 and ( 1.71R)I 1 + ( 3.71R)I 500 (a) With R 1000 Ω, simultaneous solution of these equations yields: I ma and I ma From Figure (b), V c V a ( I 1 + I )( 1.71R) 40 V Thus, from Figure (a), I 4 V c V a 4R 40 V 60.0 ma 4000 Ω (b) Finally, applying Kirchhoff s point rule at point a in Figure (a) gives: I I 4 I ma 10.0 ma ma, or I 50.0 ma flowing from point a to point e.

10 Chapter 8 Solutions Name the currents as shown in the figure to the right. Then w + x + z y. Loop equations are 00w x x y y 70.0z Eliminate y by substitution. Eliminate x : x.50 w x 0.0w 0.0 z w 0.0x 90.0 z w 0.0 z w 90.0 z 0 Eliminate z w to obtain w w 0 w 70.0/ A upward in 00 Ω Now z 4.00 A upward in 70.0 Ω x 3.00 A upward in 80.0 Ω y 8.00 A downward in 0.0 Ω and for the 00 Ω, V IR (1.00 A)(00 Ω) 00 V 8.5 Using Kirchhoff s rules, 1.0 ( )I 1 ( )I ( 1.00)I ( )I 3 0 and I 1 I + I ( )I ( )I ( 1.00)I ( )I 3 0 Solving simultaneously, dead battery, I 0.83 A downward in the and I A downward in the starter.

11 144 Chapter 8 Solutions 8.6 V ab ( 1.00)I 1 + ( 1.00) ( I 1 I ) V ab ( 1.00)I 1 + ( 1.00)I + ( 5.00) ( I I 1 + I ) V ab ( 3.00) ( I I 1 )+ ( 5.00) ( I I 1 + I ) Let I 1.00 A, I 1 x, and I y Then, the three equations become: V ab.00 x y, or y.00x V ab V ab 4.00 x y and V ab x y Substituting the first into the last two gives: 7.00V ab 8.00 x and 6.00V ab.00 x Solving these simultaneously yields V ab 7 17 V Then, R ab V ab I 7 17 V 1.00 A or R ab 7 17 Ω 8.7 We name the currents I 1, I, and I 3 as shown. (a) I 1 I + I 3 Counterclockwise around the top loop, 1.0 V (.00 Ω)I 3 (4.00 Ω)I 1 0 Traversing the bottom loop, 8.00 V (6.00 Ω)I + (.00 Ω)I 3 0 I I 3 I I 3 and I ma (b) V a (0.909 A)(.00 Ω) V b V b V a 1.8 V

12 Chapter 8 Solutions We apply Kirchhoff's rules to the second diagram I 1.00I 0 (1) I I 0 () I 1 I + I 3 (3) Substitute (3) into (1), and solve for I 1, I, and I 3 I A; I 5.00 A; I A Then apply P I R to each resistor: (.00 Ω) 1 : P I 1 (.00 Ω) (0.0 A) (.00 Ω) 800 W (4.00 Ω) : P 5.00 A (4.00 Ω) 5.0 W (Half of I goes through each) (.00 Ω) 3 : P I 3 (.00 Ω) (15.0 A) (.00 Ω) 450 W 8.9 (a) RC ( Ω)( F) 5.00 s (b) Q Cε ( C)(30.0 V) 150 µc ε 30.0 (c) I(t) R e t/rc exp 10.0 ( )( ) 4.06 µa 8.30 (a) I(t) I 0 e t/rc I 0 Q RC C (1300 Ω)( F) 1.96 A I(t) (1.96 A) exp s (1300 Ω)( F) 61.6 ma (b) q(t) Qe t/rc (5.10 µc) exp s (1300 Ω)( F) 0.35 µc (c) The magnitude of the current is I A

13 146 Chapter 8 Solutions 8.31 U 1 C( V) and V QC Therefore, U Q C and when the charge decreases to half its original value, the stored energy is one-quarter its original value: U f 1 4 U (a) τ RC ( Ω)( F) 1.50 s (b) τ ( Ω)( F) 1.00 s (c) The battery carries current 10.0 V Ω 00 µa The 100 kω carries current of magnitude I I 0 e t/rc 10.0 V Ω e t/ 1.00 s So the switch carries downward current t / 1.00 s 00 µa + (100 µa)e 8.33 (a) Call the potential at the left junction V L and at the right V R. After a "long" time, the capacitor is fully charged. V L 8.00 V because of voltage divider: I L 10.0 V 5.00 Ω.00 A V L 10.0 V (.00 A)(1.00 Ω) 8.00 V Likewise, V R.00 Ω.00 Ω Ω 10.0 V.00 V or I R 10.0 V 10.0 Ω 1.00 A V R (10.0 V) (8.00 Ω)(1.00 A).00 V Therefore, V V L V R V 1 (b) Redraw the circuit R ( 1/ 9.00 Ω)+ 1/ 6.00 Ω RC s ( ) 3.60 Ω and e t/rc 1 10 so t RC ln µs

14 Chapter 8 Solutions (a) τ RC ( Ω) ( F) 1.0 s ε (b) I R e t/rc e t/1.0 s q Cε [ 1 e t/rc ] ( 1.0) [ 1 e t/1.0 ) q 36.0 µc[ 1 e t/1.0 ] I 3.00 µae t/ V 0 Q C Then, if q(t) Qe t/rc V(t) V 0 e t/rc V() t e t/rc V 0 Therefore 1 exp 4.00 R ( ) ln R R 1.60 MΩ ( ) 8.36 V 0 Q C Then, if q(t) Qe trc V(t) ( V 0 )e trc and V(t) RC e t ( V 0 ) When V(t) 1 ( V 0), then e trc 1 Thus, R t RC ln 1 ln t C ln ( )

15 148 Chapter 8 Solutions 8.37 q(t) Q[ 1 e t/rc ] so q(t) Q 1 e t/rc e 0.900/RC or e 0.900/RC ln( ) RC thus RC ln( ) 0.98 s 8.38 Applying Kirchhoff s loop rule, I g ( 75.0 Ω)+ ( I I g )R p 0 Therefore, if I 1.00 A when I g 1.50 ma, R p I g ( 75.0 Ω) ( I I g ) ( ) 75.0 Ω ( ) A 1.00 A A Ω 8.39 Series Resistor Voltmeter V IR: (R s ) Solving, R s 16.6 kω Figure for Goal Solution Goal Solution The galvanometer described in the preceding problem can be used to measure voltages. In this case a large resistor is wired in series with the galvanometer in a way similar to that shown in Figure P8.4b This arrangement, in effect, limits the current that flows through the galvanometer when large voltages are applied. Most of the potential drop occurs across the resistor placed in series. Calculate the value of the resistor that enables the galvanometer to measure an applied voltage of 5.0 V at full-scale deflection. G : O : The problem states that the value of the resistor must be large in order to limit the current through the galvanometer, so we should expect a resistance of kω to MΩ. The unknown resistance can be found by applying the definition of resistance to the portion of the circuit shown in Figure 8.4b. A : V ab 5.0 V; From Problem 38, I 1.50 ma and R g 75.0 Ω. For the two resistors in series, R eq R s + R g so the definition of resistance gives us: V ab I(R s + R g ) Therefore, R s V ab I 5.0 V R g Ω16.6 kω A L : The resistance is relatively large, as expected. It is important to note that some caution would be necessary if this arrangement were used to measure the voltage across a circuit with a comparable resistance. For example, if the circuit resistance was 17 kω, the voltmeter in this problem would cause a measurement inaccuracy of about 50%, because the meter would divert about half the current that normally would go through the resistor being measured. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.

16 Chapter 8 Solutions We will use the values required for the 1.00-V voltmeter to obtain the internal resistance of the galvanometer. V I g (R + r g ) Solve for r g : r g V I g R 1.00 V A 900 Ω 100 Ω We then obtain the series resistance required for the 50.0-V voltmeter: R V I g r g 50.0 V A 100 Ω 49.9 kω 8.41 V I g r g ( I I g )R p, or R p I g r g I I g ( ) ( ) I g 60.0 Ω ( I I g ) Therefore, to have I A 100 ma when I g ma: ma R p ( ) 60.0 Ω 99.5 ma ( ) 0.30 Ω Figure for Goal Solution Goal Solution Assume that a galvanometer has an internal resistance of 60.0 Ω and requires a current of ma to produce full-scale deflection. What resistance must be connected in parallel with the galvanometer if the combination is to serve as an ammeter that has a full-scale deflection for a current of A? G : O : A : An ammeter reads the flow of current in a portion of a circuit; therefore it must have a low resistance so that it does not significantly alter the current that would exist without the meter. Therefore, the resistance required is probably less than 1 Ω. From the values given for a full-scale reading, we can find the voltage across and the current through the shunt (parallel) resistor, and the resistance value can then be found from the definition of resistance. The voltage across the galvanometer must be the same as the voltage across the shunt resistor in parallel, so when the ammeter reads full scale, V ( ma) ( 60.0 Ω) 30.0 mv Through the shunt resistor, I 100 ma ma 99.5 ma Therefore, R V I 30.0 mv 99.5 ma 0.30 Ω L : The shunt resistance is less than 1 Ω as expected. It is important to note that some caution would be necessary if this meter were used in a circuit that had a low resistance. For example, if the circuit resistance was 3 Ω, adding the ammeter to the circuit would reduce the current by about 10%, so the current displayed by the meter would be lower than without the meter. Problems 46 and 59 address a similar concern about measurement error when using electrical meters.

17 150 Chapter 8 Solutions 8.4 R x R R 3 R 1 R R 3.50 R 1000 Ω Ω 8.43 Using Kirchhoff s rules with R g << 1, ( 1.0 Ω)I 1 + ( 14.0 Ω)I 0, so I 1 3 I I ( I 1 + I g ) 0, and I 7.00( I I g ) 0 The last two equations simplify to 1.0 Ω I 1 I 14.0 Ω + G I g 70.0 V I - I g 7.00 Ω I 1 + I g 7.00 Ω ( 3 I ) I g, and I I g Solving simultaneously yields: I g A 8.44 R ρl A and R i ρl i A i But, V AL A i L i, so R ρl V and R i ρl i V Therefore, R ρ ( L i + L) V [ ( )] ρl i 1 + LL i V R i [ 1 + α] where α L L This may be written as: R R i (1 + α + α ) 8.45 ε x R s ε s R s ; ε x ε sr x 48.0 Ω R s 36.0 Ω ( V) 1.36 V

18 Chapter 8 Solutions 151 *8.46 (a) In Figure (a), the emf sees an equivalent resistance of Ω. I V A Ω The terminal potential difference is V Ω V Ω Ω Ω Ω Ω (a) (b) (c) V IR ( A) ( Ω) V A V A 1 (b) In Figure (b), R eq Ω Ω Ω (c) The equivalent resistance across the emf is The ammeter reads and the voltmeter reads In Figure (c), Therefore, the emf sends current through The current through the battery is but not all of this goes through the ammeter. The voltmeter reads Ω Ω Ω Ω I ε R V Ω A V IR ( A) ( Ω) V Ω Ω Ω R tot Ω Ω Ω I V Ω A V IR ( A) ( Ω) V The ammeter measures current I V R V A Ω The connection shown in Figure (c) is better than that shown in Figure (b) for accurate readings (a) P I( V) So for the Heater, I P V 1500 W 10 V 1.5 A For the Toaster, And for the Grill, I 750 W 10 W 6.5 A I 1000 W 10 V 8.33 A (Grill) (b) A sufficient. The current draw is greater than 5.0 amps, so this would not be

19 15 Chapter 8 Solutions 8.48 (a) P I R I ρl A (1.00 A) ( Ω m)(16.0 ft)( m / ft) π( m) W (b) P I R 100(0.101 Ω) 10.1 W 8.49 I Al R Al I Cu R Cu so I Al R Cu I Cu R Al ρ Cu I Cu 1.70 ( ρ Al ) 0.776(0.0) 15.5 A *8.50 (a) Suppose that the insulation between either of your fingers and the conductor adjacent is a chunk of rubber with contact area 4 mm and thickness 1 mm. Its resistance is ( )( 10 3 m) R ρl A Ω m m Ω The current will be driven by 10 V through total resistance (series) Ω+10 4 Ω Ω Ω It is: I V R ~ 10 V Ω ~10 14 A (b) The resistors form a voltage divider, with the center of your hand at potential V h, where V h is the potential of the "hot" wire. The potential difference between your finger and thumb is V IR ~ ( A) ( 10 4 Ω)~10 10 V. So the points where the rubber meets your fingers are at potentials of ~ V h V and ~ V h V *8.51 The set of four batteries boosts the electric potential of each bit of charge that goes through them by V 6.00 V. The chemical energy they store is U q V (40 C)(6.00 J/C) 1440 J The radio draws current I V R 6.00 V 00 Ω A So, its power is P ( V)I (6.00 V)( A) W J/s Then for the time the energy lasts, we have P Et: t E P 1440 J J / s s We could also compute this from I Q/t: t Q I 40 C A s. h

20 Chapter 8 Solutions 153 *8.5 I ε R + r, so P I R ε R ( R + r) or R + r P R ( ) ε Let x ε P, then ( R + r) xr or R + ( r x)r r 0 With r 1.0 Ω, this becomes which has solutions of R + (.40 x)r , R (.40 x )± (.40 x ) ± 4.1 (a) With ε 90. V and P 1.8 W, x 6.61: R ( ) Ω or Ω (b) For ε ε 90. V and P 1. W, x P 3.99 R ± ( 1. 59) ± 3. The equation for the load resistance yields a complex number, so there is no resistance that will extract 1. W from this battery. The maximum power output occurs when R r 1.0 Ω, and that maximum is: P max ε 4r 17.6 W 8.53 Using Kirchhoff s loop rule for the closed loop, I 4.00 I 0, so I.00 A V b V a V (.00 A) ( 4.00 Ω) ( 0) ( 10.0 Ω) 4.00 V Thus, V ab 4.00 V and point a is at the higher potential The potential difference across the capacitor V() V t max [ 1 e trc ] Using 1 Farad 1 s Ω, 4.00 V 10.0 V ( ) 1 e 3.00 s Therefore, e ( Ω) R or e ( Ω) R ( ) ( ) R s Ω Taking the natural logarithm of both sides, Ω R ln( 0.600) and R Ω ln ( ) Ω 587 kω

21 154 Chapter 8 Solutions 8.55 Let the two resistances be x and y. x y Then, R s x + y P s I and R p xy x + y P p I 5 W 9.00 Ω y 9.00 Ω x 5.00 A ( ) 50.0 W 5.00 A ( ).00 Ω x y so ( ) ( ).00 Ω x 9.00 x x 9.00 Ω x x Ω x Factoring the second equation, ( x 6.00) ( x 3.00) 0 so x 6.00 Ω or x 3.00 Ω Then, y 9.00 Ω x gives y 3.00 Ω or y 6.00 Ω The two resistances are found to be 6.00 Ω and 3.00 Ω Let the two resistances be x and y. Then, R s x + y P s I and R p xy x + y P p I. From the first equation, y P s x, and the second I becomes x( P s I x) x + ( P s I x) P p I or x P s I x + P s P p I 4 0. Using the quadratic formula, x P s ± P s 4P s P p I. Then, y P s I x gives y P s > P s 4P s P p I. The two resistances are P s + P s 4P s P p I and P s P s 4P s P p I

22 Chapter 8 Solutions The current in the simple loop circuit will be I ε R + r (a) V ter ε Ir (b) I ε R + r εr R + r (c) P I R ε R (R + r) and and V ter ε as R I ε r as R 0 dp dr ε R( )(R + r) 3 + ε ε (R + r) R ε (R + r) 3 + (R + r) 0 Figure for Goal Solution Then R R + r and R r Goal Solution A battery has an emf ε and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. G : O : If we consider the limiting cases, we can imagine that the potential across the battery will be a maximum when R (open circuit), the current will be a maximum when R 0 (short circuit), and the power will be a maximum when R is somewhere between these two extremes, perhaps when R r. We can use the definition of resistance to find the voltage and current as functions of R, and the power equation can be differentiated with respect to R. A : (a) The battery has a voltage V terminal ε Ir εr R + r (b) The circuit's current is I ε R + r or as R, V terminal ε or as R 0, I ε r (c) The power delivered is P I R ε R (R + r) To maximize the power P as a function of R, we differentiate with respect to R, and require that dp /dr 0 Then R R + r and R r dp dr ε R( )(R + r) 3 + ε (R + r) ε R (R + r) 3 + ε (R + r) 0 L : The results agree with our predictions. Making load resistance equal to the source resistance to maximize power transfer is called impedance matching.

23 156 Chapter 8 Solutions 8.58 (a) ε I(ΣR) (ε 1 + ε ) V (4.00 A) [( R)Ω] ( ) V 0; so R 4.40 Ω (b) Inside the supply, Inside both batteries together, For the limiting resistor, P I R ( 4.00 A) (.00 Ω) 3.0 W P I R ( 4.00 A) ( Ω) 9.60 W P ( 4.00 A) ( 4.40 Ω) 70.4 W (c) P I(ε 1 + ε ) (4.00 A) [( )V] 48.0 W 8.59 Let R m measured value, R actual value, I R current through the resistor R (a) I current measured by the ammeter. (a) When using circuit (a), I R R V 0 000(I I R ) or R I 1 I R But since I V R m and I R V R, we have I I R R R m (b) Figure for Goal solution and R (R R m) R m (1) When R > R m, we require (R R m ) R Therefore, R m R ( ) and from (1) we find (b) When using circuit (b), R 1050 Ω I R R V I R (0.5 Ω). But since I R V R m, R m ( R) () When R m > R, we require (R m R) R From () we find R 10.0 Ω

24 Chapter 8 Solutions 157 Goal Solution The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in Figure P8.59. The ammeter has a resistance of Ω, and the voltmeter has a resistance of 0000 Ω. Within what range of actual values of R will the measured values be correct to within 5.00% if the measurement is made using (a) the circuit shown in Figure P8.59a (b) the circuit shown in Figure P8.59b? G : An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance, so that adding the meter does not alter the current or voltage of the existing circuit. For the non-ideal meters in this problem, a low values of R will give a large voltage measurement error in circuit (b), while a large value of R will give significant current measurement error in circuit (a). We could hope that these meters yield accurate measurements in either circuit for typical resistance values of 1 Ω to 1 MΩ. O : The definition of resistance can be applied to each circuit to find the minimum and maximum current and voltage allowed within the 5.00% tolerance range. A : (a) In Figure P8.59a, at least a little current goes through the voltmeter, so less current flows through the resistor than the ammeter reports, and the resistance computed by dividing the voltage by the inflated ammeter reading will be too small. Thus, we require that V/ I 0.950R where I is the current through the ammeter. Call I R the current through the resistor; then I I R is the current in the voltmeter. Since the resistor and the voltmeter are in parallel, the voltage across the meter equals the voltage across the resistor. Applying the definition of resistance: Our requirement is V I R R ( I I R )( 0000 Ω) so I I R (R Ω) I R R 0.95R I R (R Ω) 0000 Ω 0000 Ω Solving, 0000 Ω 0.95(R Ω)0.95R Ω and R 1000 Ω or R 1.05 kω 0.95 (b) If R is too small, the resistance of an ammeter in series will significantly reduce the current that would otherwise flow through R. In Figure 8.59b, the voltmeter reading is I( Ω)+ IR, at least a little larger than the voltage across the resistor. So the resistance computed by dividing the inflated voltmeter reading by the ammeter reading will be too large. We require V I 1.05R so that I (0.500 Ω) + IR 1.05R I Thus, Ω R and R 10.0 Ω L : The range of R values seems correct since the ammeter s resistance should be less than 5% of the smallest R value ( Ω 0.05R means that R should be greater than 10 Ω), and R should be less than 5% of the voltmeter s internal resistance ( R kω1 kω ). Only for the restricted range between 10 ohms and 1000 ohms can we indifferently use either of the connections (a) and (b) for a reasonably accurate resistance measurement. For low values of the resistance R, circuit (a) must be used. Only circuit (b) can accurately measure a large value of R.

25 158 Chapter 8 Solutions 8.60 The battery supplies energy at a changing rate Then the total energy put out by the battery is de dt P EI E E R e 1/RC de t0 ε R exp t RC dt ε de R ( RC) exp t 0 RC dt ε C exp t RC RC 0 ε C[0 1] ε C The heating power of the resistor is So the total heat is de ε dt P V R I I t R R exp R RC de 0 ε R exp t RC dt ε de R RC 0 exp t RC dt ε C t exp RC RC 0 ε C ε [0 1] C The energy finally stored in the capacitor is U 1 C ( V) 1 C ε. Thus, energy is conserved: ε C 1 ε C + 1 ε C and resistor and capacitor share equally in the energy from the battery (a) q C( V) [ 1 e trc ] q ( F) ( 10.0 V) 9.93 µc (b) I dq dt V R RC e t 10.0 V I Ω e A 33.7 na (c) du dt d 1 q dt C q dq C dt q C I du dt C CV A ( ) W 334 nw (d) P battery IE ( A) ( 10.0 V) W 337 nw

26 Chapter 8 Solutions Start at the point when the voltage has just reached V and the switch has just closed. The voltage is V 3 3 and is decaying towards 0 V with a time constant R B C. V C (t) 3 V e t/r B C We want to know when V C (t) will reach 1 3 V. Therefore, or t 1 R B C ln 1 3 V 3 V e t/rbc or e t/rbc 1 After the switch opens, the voltage is 1 V, increasing toward V with time constant ( R 3 A + R B)C: V C (t) V 3 V e t/(r A +R B )C When V C (t) 3 V, 3 V V 3 Ve t/(r A +R B )C or e t/(r A +R B )C 1 so t (R A + R B )C ln and T t 1 + t (R A + R B )C ln 8.63 (a) First determine the resistance of each light bulb: P ( V) R R ( V) P (10 V) 60.0 W 40 Ω We obtain the equivalent resistance R eq of the network of light bulbs by applying Equations 8.6 and 8.7: 1 R eq R Ω+10 Ω360 Ω ( 1/R )+ ( 1/R 3 ) The total power dissipated in the 360 Ω is P ( V) (10 V) R eq 360 Ω 40.0 W (b) The current through the network is given by P I R eq : I P R eq 40.0 W 360 Ω 1 3 A The potential difference across R 1 is V 1 IR A (40 Ω) 80.0 V The potential difference V 3 across the parallel combination of R and R 3 is V 3 IR A 1 ( 1/ 40 Ω)+ ( 1/ 40 Ω ) 40.0 V

27 160 Chapter 8 Solutions 8.64 V IR (a) 0.0 V ( A)(R Ω) R Ω kω (b) 50.0 V ( A)(R + R Ω) R 30.0 kω (c) 100 V ( A)(R 3 + R Ω) R kω 8.65 Consider the circuit diagram shown, realizing that I g 1.00 ma. For the 5.0 ma scale: ( 4.0 ma) ( R 1 + R + R 3 ) ( 1.00 ma) ( 5.0 Ω) or R 1 + R + R Ω (1) For the 50.0 ma scale: ( 49.0 ma) ( R 1 + R ) ( 1.00 ma) ( 5.0 Ω+R 3 ) or 49.0( R 1 + R ) 5.0 Ω+R 3 () For the 100 ma scale: ( 99.0 ma)r 1 ( 1.00 ma) ( 5.0 Ω+R + R 3 ) or 99.0R Ω+R + R 3 (3) Solving (1), (), and (3) simultaneously yields R Ω, R 0.61 Ω, R Ω 8.66 Ammeter: I g r ( A I g )( 0.0 Ω) or Voltmeter: I g ( r Ω) V (1).00 V I g ( r Ω) () Solve (1) and () simultaneously to find: I g ma and r 145 Ω

28 Chapter 8 Solutions (a) After steady-state conditions have been reached, there is no DC current through the capacitor. Thus, for R 3 : I R3 0 (steady-state) For the other two resistors, the steady-state current is simply determined by the 9.00-V emf across the 1-k Ω and 15-k Ω resistors in series: ε For R 1 and R : I (R1 +R ) 9.00 V 333 µa (steady-state) R 1 + R (1.0 kω+15.0 kω) (b) After the transient currents have ceased, the potential difference across C is the same as the potential difference across R ( IR ) because there is no voltage drop across R 3. Therefore, the charge Q on C is Q C ( V) R C (IR ) (10.0 µf)(333 µa)(15.0 k Ω) 50.0 µc (c) When the switch is opened, the branch containing R 1 is no longer part of the circuit. The capacitor discharges through (R + R 3 ) with a time constant of (R + R 3 )C (15.0 k Ω k Ω)(10.0 µf) s. The initial current I i in this discharge circuit is determined by the initial potential difference across the capacitor applied to (R + R 3 ) in series: I i ( V) C (R + R 3 ) IR (333 µa)(15.0 kω) 78 µa (R + R 3 ) (15.0 kω+3.00 kω) Thus, when the switch is opened, the current through R changes instantaneously from 333 µa (downward) to 78 µa (downward) as shown in the graph. Thereafter, it decays according to I R I i e t/(r +R 3 )C (78 µa)e t/(0.180 s) (for t > 0) (d) The charge q on the capacitor decays from Q i to Q i /5 according to (a) q Q i e t/(r +R 3 )C Q i 5 Q s) ie( t/ e t/0.180 s ln 5 t 180 ms t (0.180 s)(ln 5) 90 ms

29 16 Chapter 8 Solutions 8.68 V ε e trc ε so ln ε V 1 RC t A plot of ln V versus t should be a straight line with slope 1 RC. Using the given data values: t s () V (V) ln ( ε V ) (a) A least-square fit to this data yields the graph to the right. Σx i 8, Σx i , Σx i y i 44, Σy i 4.03, N 8 Slope N ( Σx i y i ) Σx i N( Σx i ) Σx i ( )( Σy i ) ( ) Intercept Σx i ( ) Σy i ( ) Σx i N( Σx i ) Σx i ε The equation of the best fit line is: ln V ( )t ( )( Σx i y i ) ( ) (b) Thus, the time constant is τ RC slope s and the capacitance is C τ R 84.7 s µf Ω

30 Chapter 8 Solutions r i/6 a r r r r r r r r r r r b a i/3 i/6 i/3 i/3 i/6 i/3 i/6 i/6 i/3 i/6 i/3 b 3 junctions at the same potential another set of 3 junctions at the same potential 8.70 (a) For the first measurement, the equivalent circuit is as shown in Figure 1. a c R 1 b R ab R 1 R y + R y R y Ry Rx Ry so R y 1 R 1 (1) Figure 1 For the second measurement, the equivalent circuit is shown in Figure. a R c Thus, R ac R 1 R y + R x () Ry Ry Rx Figure Substitute (1) into () to obtain: R 1 1 R 1 + R x, or R x R 1 4 R 1 (b) If R Ω and R 6.00 Ω, then R x.75 Ω The antenna is inadequately grounded since this exceeds the limit of.00 Ω Since the total current passes through R 3, that resistor will dissipate the most power. When that resistor is operating at its power limit of 3.0 W, the current through it is R 1 R 3 I total P R 3.0 W.00 Ω 16.0 A, or I total 4.00 A Half of this total current (.00 A) flows through each of the other two resistors, so the power dissipated in each of them is: R R 1 R R 3.00 Ω P ( 1 I total) R (.00 A) (.00 Ω) 8.00 W Thus, the total power dissipated in the entire circuit is:

31 164 Chapter 8 Solutions P total 3.0 W W W 48.0 W 8.7 The total resistance between points b and c is: ( 00. k Ω )( 300. kω ) R 10. kω 00. k Ω kω The total capacitance between points d and e is: C.00 µf µf 5.00 µf The potential difference between point d and e in this series RC circuit at any time is: [ ] ( 10.0 V) [ 1 e 1000t 6 ] V ε 1 e trc Therefore, the charge on each capacitor between points d and e is: q 1 C 1 ( V) (.00 µf) ( 10.0 V) [ 1 e 1000t 6 ] ( 40 µc ) [ 1 e 1000t 6 ] and q C ( V) ( 3.00 µf) ( 10.0 V) [ 1 e 1000t 6 ] ( 360 µc ) [ 1 e 1000t 6 ].00 kω.00 kω 1.00 µf 3.00 µf 0 V *8.73 (a) R eq 3R I ε 3R P series ε I ε 3R 1 (b) R eq ( 1/R)+ ( 1/R)+ 1/R ( ) R 3ε 3 I R P parallel ε 3ε I R (c) Nine times more power is converted in the parallel connection.

### Chapter 28. Direct Current Circuits

Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining

### A positive value is obtained, so the current is counterclockwise around the circuit.

Chapter 7. (a) Let i be the current in the circuit and take it to be positive if it is to the left in. We use Kirchhoff s loop rule: ε i i ε 0. We solve for i: i ε ε + 6. 0 050.. 4.0Ω+ 80. Ω positive value

### Version 001 CIRCUITS holland (1290) 1

Version CIRCUITS holland (9) This print-out should have questions Multiple-choice questions may continue on the next column or page find all choices before answering AP M 99 MC points The power dissipated

### Electromotive Force. The electromotive force (emf), ε, of a battery is the maximum possible voltage that the battery can provide between its terminals

Direct Current When the current in a circuit has a constant magnitude and direction, the current is called direct current Because the potential difference between the terminals of a battery is constant,

### Electric Currents. Resistors (Chapters 27-28)

Electric Currents. Resistors (Chapters 27-28) Electric current I Resistance R and resistors Relation between current and resistance: Ohm s Law Resistivity ρ Energy dissipated by current. Electric power

### Physics 115. General Physics II. Session 24 Circuits Series and parallel R Meters Kirchoff s Rules

Physics 115 General Physics II Session 24 Circuits Series and parallel R Meters Kirchoff s Rules R. J. Wilkes Email: phy115a@u.washington.edu Home page: http://courses.washington.edu/phy115a/ 5/15/14 Phys

### Chapter 26 & 27. Electric Current and Direct- Current Circuits

Chapter 26 & 27 Electric Current and Direct- Current Circuits Electric Current and Direct- Current Circuits Current and Motion of Charges Resistance and Ohm s Law Energy in Electric Circuits Combination

### Chapter 7 Direct-Current Circuits

Chapter 7 Direct-Current Circuits 7. Introduction... 7. Electromotive Force... 7.3 Resistors in Series and in Parallel... 4 7.4 Kirchhoff s Circuit Rules... 6 7.5 Voltage-Current Measurements... 8 7.6

### Chapter 26 Direct-Current Circuits

Chapter 26 Direct-Current Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing

### Capacitance, Resistance, DC Circuits

This test covers capacitance, electrical current, resistance, emf, electrical power, Ohm s Law, Kirchhoff s Rules, and RC Circuits, with some problems requiring a knowledge of basic calculus. Part I. Multiple

Superconductors A class of materials and compounds whose resistances fall to virtually zero below a certain temperature, T C T C is called the critical temperature The graph is the same as a normal metal

### AP Physics C. Electric Circuits III.C

AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the cross-sectional area of the conductor changes. If a conductor has no current,

### Chapter 26 Direct-Current Circuits

Chapter 26 Direct-Current Circuits 1 Resistors in Series and Parallel In this chapter we introduce the reduction of resistor networks into an equivalent resistor R eq. We also develop a method for analyzing

### M. C. Escher: Waterfall. 18/9/2015 [tsl425 1/29]

M. C. Escher: Waterfall 18/9/2015 [tsl425 1/29] Direct Current Circuit Consider a wire with resistance R = ρl/a connected to a battery. Resistor rule: In the direction of I across a resistor with resistance

### 3/17/2009 PHYS202 SPRING Lecture notes Electric Circuits

PHYS202 SPRING 2009 Lecture notes Electric Circuits 1 Batteries A battery is a device that provides a potential difference to two terminals. Different metals in an electrolyte will create a potential difference,

### Ch 28-DC Circuits! 1.) EMF & Terminal Voltage! 9.0 V 8.7 V 8.7 V. V =! " Ir. Terminal Open circuit internal! voltage voltage (emf) resistance" 2.

Ch 28-DC Circuits! 1.) EMF & Terminal Voltage! 9.0 V 8.7 V 8.7 V V =! " Ir Terminal Open circuit internal! voltage voltage (emf) resistance" 2.) Resistors in series! One of the bits of nastiness about

### The RC Time Constant

The RC Time Constant Objectives When a direct-current source of emf is suddenly placed in series with a capacitor and a resistor, there is current in the circuit for whatever time it takes to fully charge

### Chapter 26 Direct-Current and Circuits. - Resistors in Series and Parallel - Kirchhoff s Rules - Electric Measuring Instruments - R-C Circuits

Chapter 26 Direct-Current and Circuits - esistors in Series and Parallel - Kirchhoff s ules - Electric Measuring Instruments - -C Circuits . esistors in Series and Parallel esistors in Series: V ax I V

### Chapter 6 DIRECT CURRENT CIRCUITS. Recommended Problems: 6,9,11,13,14,15,16,19,20,21,24,25,26,28,29,30,31,33,37,68,71.

Chapter 6 DRECT CURRENT CRCUTS Recommended Problems: 6,9,,3,4,5,6,9,0,,4,5,6,8,9,30,3,33,37,68,7. RESSTORS N SERES AND N PARALLEL - N SERES When two resistors are connected together as shown we said that

### DC Circuits. Electromotive Force Resistor Circuits. Kirchoff s Rules. RC Circuits. Connections in parallel and series. Complex circuits made easy

DC Circuits Electromotive Force esistor Circuits Connections in parallel and series Kirchoff s ules Complex circuits made easy C Circuits Charging and discharging Electromotive Force (EMF) EMF, E, is the

### Resistivity and Temperature Coefficients (at 20 C)

Homework # 4 Resistivity and Temperature Coefficients (at 0 C) Substance Resistivity, Temperature ( m) Coefficient, (C ) - Conductors Silver.59 x 0-0.006 Copper.6 x 0-0.006 Aluminum.65 x 0-0.0049 Tungsten

### Exam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.

Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular

### Chapter 3: Electric Current And Direct-Current Circuits

Chapter 3: Electric Current And Direct-Current Circuits 3.1 Electric Conduction 3.1.1 Describe the microscopic model of current Mechanism of Electric Conduction in Metals Before applying electric field

### ELECTRIC CURRENT. Ions CHAPTER Electrons. ELECTRIC CURRENT and DIRECT-CURRENT CIRCUITS

LCTRC CURRNT CHAPTR 25 LCTRC CURRNT and DRCTCURRNT CRCUTS Current as the motion of charges The Ampère Resistance and Ohm s Law Ohmic and nonohmic materials lectrical energy and power ons lectrons nside

### Physics 1214 Chapter 19: Current, Resistance, and Direct-Current Circuits

Physics 1214 Chapter 19: Current, Resistance, and Direct-Current Circuits 1 Current current: (also called electric current) is an motion of charge from one region of a conductor to another. Current When

### physics 4/7/2016 Chapter 31 Lecture Chapter 31 Fundamentals of Circuits Chapter 31 Preview a strategic approach THIRD EDITION

Chapter 31 Lecture physics FOR SCIENTISTS AND ENGINEERS a strategic approach THIRD EDITION randall d. knight Chapter 31 Fundamentals of Circuits Chapter Goal: To understand the fundamental physical principles

### Chapter 27. Circuits

Chapter 27 Circuits 1 1. Pumping Chagres We need to establish a potential difference between the ends of a device to make charge carriers follow through the device. To generate a steady flow of charges,

### Chapter 19 Lecture Notes

Chapter 19 Lecture Notes Physics 2424 - Strauss Formulas: R S = R 1 + R 2 +... C P = C 1 + C 2 +... 1/R P = 1/R 1 + 1/R 2 +... 1/C S = 1/C 1 + 1/C 2 +... q = q 0 [1-e -t/(rc) ] q = q 0 e -t/(rc τ = RC

### Tactics Box 23.1 Using Kirchhoff's Loop Law

PH203 Chapter 23 solutions Tactics Box 231 Using Kirchhoff's Loop Law Description: Knight/Jones/Field Tactics Box 231 Using Kirchhoff s loop law is illustrated Learning Goal: To practice Tactics Box 231

### AC vs. DC Circuits. Constant voltage circuits. The voltage from an outlet is alternating voltage

Circuits AC vs. DC Circuits Constant voltage circuits Typically referred to as direct current or DC Computers, logic circuits, and battery operated devices are examples of DC circuits The voltage from

### On the axes of Fig. 4.1, carefully sketch a graph to show how the potential difference V across the capacitor varies with time t. Label this graph L.

1 (a) A charged capacitor is connected across the ends of a negative temperature coefficient (NTC) thermistor kept at a fixed temperature. The capacitor discharges through the thermistor. The potential

### PEP 2017 Assignment 12

of the filament?.16.. Aductile metal wire has resistance. What will be the resistance of this wire in terms of if it is stretched to three times its original length, assuming that the density and resistivity

### Direct-Current Circuits. Physics 231 Lecture 6-1

Direct-Current Circuits Physics 231 Lecture 6-1 esistors in Series and Parallel As with capacitors, resistors are often in series and parallel configurations in circuits Series Parallel The question then

### REVISED HIGHER PHYSICS REVISION BOOKLET ELECTRONS AND ENERGY

REVSED HGHER PHYSCS REVSON BOOKLET ELECTRONS AND ENERGY Kinross High School Monitoring and measuring a.c. Alternating current: Mains supply a.c.; batteries/cells supply d.c. Electrons moving back and forth,

### AP Physics C - E & M

AP Physics C - E & M Current and Circuits 2017-07-12 www.njctl.org Electric Current Resistance and Resistivity Electromotive Force (EMF) Energy and Power Resistors in Series and in Parallel Kirchoff's

### Chapter 2. Engr228 Circuit Analysis. Dr Curtis Nelson

Chapter 2 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 2 Objectives Understand symbols and behavior of the following circuit elements: Independent voltage and current sources; Dependent voltage and

### 5. In parallel V 1 = V 2. Q 1 = C 1 V 1 and Q 2 = C 2 V 2 so Q 1 /Q 2 = C 1 /C 2 = 1.5 D

NSWRS - P Physics Multiple hoice Practice ircuits Solution nswer 1. The resistances are as follows: I:, II: 4, III: 1, IV:. The total resistance of the 3 and 6 in parallel is making the total circuit resistance

### = e = e 3 = = 4.98%

PHYS 212 Exam 2 - Practice Test - Solutions 1E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0.

### Chapter 19. Electric Current, Resistance, and DC Circuit Analysis

Chapter 19 Electric Current, Resistance, and DC Circuit Analysis I = dq/dt Current is charge per time SI Units: Coulombs/Second = Amps Direction of Electron Flow _ + Direction of Conventional Current:

### ENERGY AND TIME CONSTANTS IN RC CIRCUITS By: Iwana Loveu Student No Lab Section: 0003 Date: February 8, 2004

ENERGY AND TIME CONSTANTS IN RC CIRCUITS By: Iwana Loveu Student No. 416 614 5543 Lab Section: 0003 Date: February 8, 2004 Abstract: Two charged conductors consisting of equal and opposite charges forms

### Phys 2025, First Test. September 20, minutes Name:

Phys 05, First Test. September 0, 011 50 minutes Name: Show all work for maximum credit. Each problem is worth 10 points. Work 10 of the 11 problems. k = 9.0 x 10 9 N m / C ε 0 = 8.85 x 10-1 C / N m e

### Chapter 3: Electric Current and Direct-Current Circuit

Chapter 3: Electric Current and Direct-Current Circuit n this chapter, we are going to discuss both the microscopic aspect and macroscopic aspect of electric current. Direct-current is current that flows

### Chapter 21 Electric Current and Direct- Current Circuits

Chapter 21 Electric Current and Direct- Current Circuits Units of Chapter 21 Electric Current Resistance and Ohm s Law Energy and Power in Electric Circuits Resistors in Series and Parallel Kirchhoff s

### Direct Current (DC) Circuits

Direct Current (DC) Circuits NOTE: There are short answer analysis questions in the Participation section the informal lab report. emember to include these answers in your lab notebook as they will be

### [1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1

1 (a) Define capacitance..... [1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. S 1 S 2 6.3 V 4.5 μf Fig. 1.1 Switch S 1 is closed and switch S 2 is left

### Question 3: How is the electric potential difference between the two points defined? State its S.I. unit.

EXERCISE (8 A) Question : Define the term current and state its S.I unit. Solution : Current is defined as the rate of flow of charge. I = Q/t Its S.I. unit is Ampere. Question 2: Define the term electric

### Chapter 21 Electric Current and Direct- Current Circuits

Chapter 21 Electric Current and Direct- Current Circuits 1 Overview of Chapter 21 Electric Current and Resistance Energy and Power in Electric Circuits Resistors in Series and Parallel Kirchhoff s Rules

### Capacitance. A different kind of capacitor: Work must be done to charge a capacitor. Capacitors in circuits. Capacitor connected to a battery

Capacitance The ratio C = Q/V is a conductor s self capacitance Units of capacitance: Coulomb/Volt = Farad A capacitor is made of two conductors with equal but opposite charge Capacitance depends on shape

### physics for you February 11 Page 68

urrent Electricity Passage 1 4. f the resistance of a 1 m length of a given wire t is observed that good conductors of heat are also is 8.13 10 3 W, and it carried a current 1, the good conductors of electricity.

### ELECTRIC CURRENT IN CONDUCTORS CHAPTER - 32

1. Q(t) t + Bt + c a) t Q Q 'T ' T t T b) Bt Q B Q 'T' t T c) C [Q] C T ELECTRIC CURRENT IN CONDUCTORS CHPTER - 3 1 1 d) Current t dq d t Bt C dt dt t + B 5 5 + 3 53.. No. of electrons per second 16 electrons

### PhysicsAndMathsTutor.com

Electricity May 02 1. The graphs show the variation with potential difference V of the current I for three circuit elements. PhysicsAndMathsTutor.com When the four lamps are connected as shown in diagram

### Fig. 1 Fig. 2. Calculate the total capacitance of the capacitors. (i) when connected as in Fig. 1. capacitance =... µf

1. Fig.1 shows two capacitors, A of capacitance 2µF, and B of capacitance 4µF, connected in parallel. Fig. 2 shows them connected in series. A two-way switch S can connect the capacitors either to a d.c.

### Circuits Practice Websheet 18.1

Circuits Practice Websheet 18.1 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. How much power is being dissipated by one of the 10-Ω resistors? a. 24

### The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:

PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.

### Circuits. PHY2054: Chapter 18 1

Circuits PHY2054: Chapter 18 1 What You Already Know Microscopic nature of current Drift speed and current Ohm s law Resistivity Calculating resistance from resistivity Power in electric circuits PHY2054:

### Clicker Session Currents, DC Circuits

Clicker Session Currents, DC Circuits Wires A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? 1) it decreases

### Chapter 28. Direct Current Circuits

Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining

### Electric Charge. Electric Charge ( q ) unbalanced charges positive and negative charges. n Units Coulombs (C)

Electric Charge Electric Charge ( q ) unbalanced charges positive and negative charges n Units Coulombs (C) Electric Charge How do objects become charged? Types of materials Conductors materials in which

### Application of Physics II for. Final Exam

Application of Physics II for Final Exam Question 1 Four resistors are connected as shown in Figure. (A)Find the equivalent resistance between points a and c. (B)What is the current in each resistor if

### Farr High School HIGHER PHYSICS. Unit 3 Electricity. Question Booklet

Farr High School HIGHER PHYSICS Unit 3 Electricity Question Booklet 1 MONITORING ND MESURING.C. 1. What is the peak voltage of the 230 V mains supply? The frequency of the mains supply is 50 Hz. How many

### Lecture Outline Chapter 21. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.

Lecture Outline Chapter 21 Physics, 4 th Edition James S. Walker Chapter 21 Electric Current and Direct- Current Circuits Units of Chapter 21 Electric Current Resistance and Ohm s Law Energy and Power

### Switch. R 5 V Capacitor. ower upply. Voltmete. Goals. Introduction

Switch Lab 9. Circuits ower upply Goals + + R 5 V Capacitor V To appreciate the capacitor as a charge storage device. To measure the voltage across a capacitor as it discharges through a resistor, and

### MTE 2: Ch :30-7pm on Mar 26

MTE 2: Ch 2103 5:30-7pm on Mar 26 Contact me and Prof. Rzchowski after this lecture for Alternate Exams (also by email asap!) 2:30-4pm 6:00-7:30pm on Mar 26 Office hrs change this week Wed morning 1 Contents

### Direct-Current Circuits

Direct-Current Circuits A'.3/.". 39 '- )232.-/ 32,+/" 7+3(5-.)232.-/ 7 3244)'03,.5B )*+,"- &'&./( 0-1*234 35-2567+- *7 2829*4-& )"< 35- )*+,"-= 9-4-- 3563 A0.5.C2/'-231).D')232.')2-1 < /633-">&@5-:836+-0"1464-625"-4*43"

### To receive full credit, you must show all your work (including steps taken, calculations, and formulas used).

Page 1 Score Problem 1: (35 pts) Problem 2: (25 pts) Problem 3: (25 pts) Problem 4: (25 pts) Problem 5: (15 pts) TOTAL: (125 pts) To receive full credit, you must show all your work (including steps taken,

### Chapter 26 Examples : DC Circuits Key concepts:

Chapter 26 Examples : DC Circuits Key concepts: Internal resistance : battery consists of some idealized source of voltage (called the electromotive force, or EMF, which uses the symbol ξ) and an effective

Physics 142 Steady Currents Page 1 Steady Currents If at first you don t succeed, try, try again. Then quit. No sense being a damn fool about it. W.C. Fields Electric current: the slow average drift of

### ELECTRICITY. Prepared by: M. S. KumarSwamy, TGT(Maths) Page

ELECTRICITY 1. Name a device that helps to maintain a potential difference across a conductor. Cell or battery 2. Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of

### PH 222-2C Fall Circuits. Lectures Chapter 27 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

PH 222-2C Fall 2012 Circuits Lectures 11-12 Chapter 27 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter 27 Circuits In this chapter we will cover the following topics: -Electromotive

### Direct Current Circuits. February 18, 2014 Physics for Scientists & Engineers 2, Chapter 26 1

Direct Current Circuits February 18, 2014 Physics for Scientists & Engineers 2, Chapter 26 1 Kirchhoff s Junction Rule! The sum of the currents entering a junction must equal the sum of the currents leaving

### 1 Written and composed by: Prof. Muhammad Ali Malik (M. Phil. Physics), Govt. Degree College, Naushera

CURRENT ELECTRICITY Q # 1. What do you know about electric current? Ans. Electric Current The amount of electric charge that flows through a cross section of a conductor per unit time is known as electric

### Use these circuit diagrams to answer question 1. A B C

II Circuit Basics Use these circuit diagrams to answer question 1. B C 1a. One of the four voltmeters will read 0. Put a checkmark beside it. b. One of the ammeters is improperly connected. Put a checkmark

### Circuits. David J. Starling Penn State Hazleton PHYS 212

Invention is the most important product of man s creative brain. The ultimate purpose is the complete mastery of mind over the material world, the harnessing of human nature to human needs. - Nikola Tesla

### Prof. Anyes Taffard. Physics 120/220. Voltage Divider Capacitor RC circuits

Prof. Anyes Taffard Physics 120/220 Voltage Divider Capacitor RC circuits Voltage Divider The figure is called a voltage divider. It s one of the most useful and important circuit elements we will encounter.

### A Review of Circuitry

1 A Review of Circuitry There is an attractive force between a positive and a negative charge. In order to separate these charges, a force at least equal to the attractive force must be applied to one

### NATIONAL QUALIFICATIONS CURRICULUM SUPPORT. Physics. Electricity. Questions and Solutions. James Page Arthur Baillie [HIGHER]

NTIONL QULIFICTIONS CURRICULUM SUPPORT Physics Electricity Questions and Solutions James Page rthur Baillie [HIGHER] The Scottish Qualifications uthority regularly reviews the arrangements for National

### UNIT II CURRENT ELECTRICITY

UNIT II CUENT ELECTICITY Weightage : 07 Marks Electric current; flow of electric charges in a metllic conductor, drift velocity, mobility and their relation with electric current. Ohm s law electrical

### Version 001 HW 20 Circuits C&J sizemore (21301jtsizemore) 1

Version 00 HW 20 Circuits C&J sizemore (230jtsizemore) This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Serway

### Physics Module Form 5 Chapter 2- Electricity GCKL 2011 CHARGE AND ELECTRIC CURRENT

2.1 CHARGE AND ELECTRIC CURRENT Van de Graaf 1. What is a Van de Graaff generator? Fill in each of the boxes the name of the part shown. A device that produces and store electric charges at high voltage

### POWER B. Terms: EMF, terminal voltage, internal resistance, load resistance. How to add up resistors in series and parallel: light bulb problems.

iclicker Quiz (1) I have completed at least 50% of the reading and study-guide assignments associated with the lecture, as indicated on the course schedule. A. True B. False Hint: this is a good time to

### EXPERIMENT 12 OHM S LAW

EXPERIMENT 12 OHM S LAW INTRODUCTION: We will study electricity as a flow of electric charge, sometimes making analogies to the flow of water through a pipe. In order for electric charge to flow a complete

### CURRENT ELECTRICITY. Q1. Plot a graph showing variation of current versus voltage for a material.

CURRENT ELECTRICITY QUESTION OF ONE MARK (VERY SHORT ANSWER) Q. Plot a graph showing variation of current versus voltage for a material. Ans. Q. The graph shown in the figure represents a plot of current

### PH 102 Exam I N N N N. 3. Which of the following is true for the electric force and not true for the gravitational force?

Name Date INSTRUCTIONS PH 102 Exam I 1. nswer all questions below. ll problems have equal weight. 2. Clearly mark the answer you choose by filling in the adjacent circle. 3. There will be no partial credit

### Chapter 28: DC and RC Circuits Kirchhoff s Rules

Chapter 28: DC and RC Circuits Kirchhoff s Rules Series Circuits The current is the same in each device. The equivalent resistance of the circuit is the sum of the individual resistances. Parallel Circuits

### Louisiana State University Physics 2102, Exam 2, March 5th, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

### ANNOUNCEMENT ANNOUNCEMENT

ANNOUNCEMENT Exam : Tuesday September 25, 208, 8 PM - 0 PM Location: Elliott Hall of Music (see seating chart) Covers all readings, lectures, homework from Chapters 2 through 23 Multiple choice (5-8 questions)

### Introduction to AC Circuits (Capacitors and Inductors)

Introduction to AC Circuits (Capacitors and Inductors) Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/

### Physics 7B-1 (A/B) Professor Cebra. Winter 2010 Lecture 2. Simple Circuits. Slide 1 of 20

Physics 7B-1 (A/B) Professor Cebra Winter 2010 Lecture 2 Simple Circuits Slide 1 of 20 Conservation of Energy Density In the First lecture, we started with energy conservation. We divided by volume (making

### Physics 212 Midterm 2 Form A

1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 10-19 C B. 6.4 10-19 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus

### ELECTRICITY & CIRCUITS

ELECTRICITY & CIRCUITS Reason and justice tell me there s more love for humanity in electricity and steam than in chastity and vegetarianism. Anton Chekhov LIGHTNING, PART 2 Electricity is really just

### Electricity & Optics

Physics 241 Electricity & Optics Lecture 12 Chapter 25 sec. 6, 26 sec. 1 Fall 217 Semester Professor Koltick Circuits With Capacitors C Q = C V V = Q C + V R C, Q Kirchhoff s Loop Rule: V I R V = V I R

### 18 - ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS ( Answers at the end of all questions ) Page 1

( Answers at the end of all questions ) Page ) The self inductance of the motor of an electric fan is 0 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of 8 µ F (

### Laboratory 7: Charging and Discharging a Capacitor Prelab

Phys 132L Fall 2018 Laboratory 7: Charging and Discharging a Capacitor Prelab Consider a capacitor with capacitance C connected in series to a resistor with resistance R as shown in Fig. 1. Theory predicts

### 1. How much charge is stored in a capacitor, whose capacitance C = 2µF, connected to a 12V battery?

IMP 113: 2 nd test (Union College: Spring 2010) Instructions: 1. Read all directions. 2. In keeping with the Union College policy on academic honesty, you should neither accept nor provide unauthorized

### EXPERIMENT 5A RC Circuits

EXPERIMENT 5A Circuits Objectives 1) Observe and qualitatively describe the charging and discharging (decay) of the voltage on a capacitor. 2) Graphically determine the time constant for the decay, τ =.

### Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

University of Rhode Island DigitalCommons@URI PHY 204: Elementary Physics II Physics Course Materials 2015 10. Resistors II Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License

### Circuits. 1. The Schematic

+ ircuits 1. The Schematic 2. Power in circuits 3. The Battery 1. eal Battery vs. Ideal Battery 4. Basic ircuit nalysis 1. oltage Drop 2. Kirchoff s Junction Law 3. Series & Parallel 5. Measurement Tools