# PEP 2017 Assignment 12

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1 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 of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.).17. In household wiring, copper wire.0 mm in diameter is model, graph the potential rises and drops in this circuit..33. When switch S in Fig. E.33 is open, the voltmeter V of the battery reads 3.08 V. When the switch is closed, the voltmeter reading drops to.97 V, Figure E.33 V and the ammeter A reads 1.6 A. Find the emf, the internal resistance of the battery, and the circuit resistance. r E Assume that the two meters are ideal, A so they don t affect the circuit. S.34. In the circuit of Fig. E.3, the.0-æ resistor is removed and replaced by a resistor of energy in the external resistor.... A typical small flashlight contains two batteries, each having an emf of 1. V, connected in series with a bulb having resistance 17 Æ. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for.0 h, what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.).3. A 40-W electric heater is designed to operate from 1 PEP 017 Assignment (a) What is the potential difference V ad in the circuit Figure P V 4.00 V of Fig. P.68? (b) What is the b c 9.00 V terminal voltage of the 4.00-V battery? (c) A battery with emf 6.00 V d V and internal resistance a 0.0 V 8.00 V 0.0 Æ is inserted in the circuit at d, with its negative 8.00 V terminal connected to the negative terminal of the 8.00-V battery. What is the difference of potential V bc between the terminals of the 4.00-V battery now?.69. The potential difference > across the terminals of a battery.74.. Acylindrical copper cable 1.0 km long is connected across a 0.0-V potential difference. (a) What should be its diameter so that it produces heat at a rate of 7.0 W? (b) What is the electric field inside the cable under these conditions?.7.. ANonideal Ammeter. Unlike the idealized amme- Calculate the emf E CP Consider the circuit shown in Fig. P.84. The battery has emf 60.0 V and negligible internal resistance. =.00 Æ, C 1 = 3.00 mf,and C = 6.00 mf. After the capacitors have attained their final charges, the charge on C 1 is Q 1 = 18.0 mc. (a) What is the final charge on C? (b) What is the resistance 1? Figure P.84 1 C E C In the circuit shown in Fig. E6. find (a) the current in resistor ; (b) the resistance ; (c) the unknown emf E. (d) If the circuit is broken at point x, what is the current in resistor? 6.6. Find the emfs E 1 and E in the circuit of Fig. E6.6, and find the potential difference of point b relative to point a. Figure E A a.00 A An Infinite Network. As shown in Fig. P6.91, a network of resistors of resistances 1 and extends to infinity toward the right. Prove that the total resistance T of the infinite network is equal to T = V 0.0 V 6.00 V 4.00 V 1.00 V E 1.00 V Figure P c 1 d 1 x (Hint: Since the network is infinite, the resistance of the network to the right of points c and d is also equal to T.) Suppose a resistor lies along Figure P6.9 a b.00 V E 1 Figure E V b E x 3.00 V A 6.00 V 6.00 A 6.4. A 1.4-mF capacitor is connected through a 0.89-MÆ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0,.0 s, 10.0 s, 0.0 s, and s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for t between 0 and 0 s CP In the circuit shown in and so on

2 = =.. Ω =..16. IDENTIFY: The geometry of the wire is changed, so its resistance will also change. ρl SET UP: =. Lnew = 3 L. The volume of the wire remains the same when it is stretched. A EXECUTE: Volume = LA so LA = L new A new. L A Anew = A=. L 3 new ρlnew ρ(3 L) ρl = = = 9 = 9. A A/3 A new EVALUATE: When the length increases the resistance increases and when the area decreases the resistance increases. ρl new

3 .33. IDENTIFY: The voltmeter reads the potential difference V ab between the terminals of the battery. SET UP: open circuit I = 0. The circuit is sketched in Figure.33a. EXECUTE: Vab ε = = 308. V Figure.33a SET UP: switch closed The circuit is sketched in Figure.33b. Figure.33b EXECUTE: Vab = Ir = 97. V ε 97. V r = I 308. V 97. V r = = Ω 16. A ε V 97 V And so ab. Vab = I = = = Ω. I 16. A EVALUATE: When current flows through the battery there is a voltage drop across its internal resistance and its terminal voltage V is less than its emf..34. I : The sum of the potential changes around the loop is zero.

4 .. IDENTIFY: The power delivered to the bulb is I Energy = Pt.. SET UP: The circuit is sketched in Figure.. r is the combined internal resistance of both batteries. EXECUTE: (a) r total = 0. The sum of the potential changes around the circuit is zero, so 1. V 1. V I(17 Ω ) = 0. I = A. (3. 0 V)( A). (b) Energy = (0. 30 W)(. 0 h)(3600 s/h) = 940 J total P= I = ( A) (17 Ω ) = W. This is also 030. W (c) P 06 W. P 06. W = =. P= I so I = = = 01. A. 17 Ω The sum of the potential changes around the circuit is zero, so 1. V 1. V I Ir = V (01. A)(17 Ω) rtotal = = 70. Ω. 01. A EVALUATE: When the power to the bulb has decreased to half its initial value, the total internal resistance of the two batteries is nearly half the resistance of the bulb. Compared to a single battery, using two identical batteries in series doubles the emf but also doubles the total internal resistance. total Figure.

5 important..68. IDENTIFY: Consider the potential changes around the circuit. For a complete loop the sum of the potential changes is zero. SET UP: There is a potential drop of I when you pass through a resistor in the direction of the current. 80V. 40V. EXECUTE: (a) I = = 0167A.. Vd V I(0. 0Ω 800. Ω ) = Va, so 4. 0 Ω V = 800V. (0167A)(80.. Ω ) = 68V.. ad (b) The terminal voltage is V = V V. V V I(00. Ω ) = V and bc b c V = 400V. (0167A)(00.. Ω ) = 408V.. bc c (c) Adding another battery at point d in the opposite sense to the 8.0-V battery produces a counterclockwise V 8. 0 V 4. 0 V current with magnitude I = = 07A.. Then Vc V I(0. 0 Ω ) = Vb and 4. Ω V = 4.00 V (0.7 A) (0.0 Ω ) = 3.87 V. bc EVALUATE: When current enters the battery at its negative terminal, as in part (c), the terminal voltage is less than its emf. When current enters the battery at the positive terminal, as in part (b), the terminal voltage is greater than its emf..69. I : In each case write the terminal voltage in terms of ε, I and r. Since I is known, this gives two b

6 .74. (a) IDENTIFY: The rate of heating (power) in the cable depends on the potential difference across the cable and the resistance of the cable. SET UP: The power is P= V / and the resistance is = ρla /. The diameter D of the cable is twice its V V AV π r V radius. P = = = =. The electric field in the cable is equal to the potential ( ρla / ) ρl ρl difference across its ends divided by the length of the cable: E = VL /. EXECUTE: Solving for r and using the resistivity of copper gives 8 PρL (0. 0 W)( Ω m)(100 m) r = = = m. πv π(0. 0 V) (b) SET UP: E = VL / EXECUTE: E = (0 V)/(100 m) = V/m D= r = 0.184mm EVALUATE: This would be an extremely thin (and hence fragile) cable..7. I : The ammeter acts as a resistance in the circuit loop. Set the sum of the potential rises and

7 flowing though it..84. IDENTIFY: No current flows to the capacitors when they are fully charged. SET UP: V = I and V = Q/ C. Q µ C EXECUTE: (a) VC = = = V. V C 300µ F C 1 C C. C1 Q = C V = (6. 00 µ F)(6. 00 V) = µ C. = V = 600. V. (b) No current flows to the capacitors when they are fully charged, so ε = I 1 I. V V = V 600 V. C = V I = = = A. 00. Ω ε I V V = = = Ω. I 300. A EVALUATE: When a capacitor is fully charged, it acts like an open circuit and prevents any current from flowing though it..8. I and S U : Follow the steps specified in the problem.

8 third loop. 6.. IDENTIFY: Apply Kirchhoff s point rule at point a to find the current through. Apply Kirchhoff s loop rule to loops (1) and () shown in Figure 6.a to calculate and ε. Travel around each loop in the direction shown. (a) SET UP: (d) IDENTIFY: If the circuit is broken at point x there can be no current in the Ω resistor. There is now only a single current path and we can apply the loop rule to this path. SET UP: The circuit is sketched in Figure 6.b. Figure 6.a EXECUTE: Apply Kirchhoff s point rule to point a: I = 0 so I A A = 0 I =.00 A (in the direction shown in the diagram). (b) Apply Kirchhoff s loop rule to loop (1): (6. 00 A)(3. 00 Ω) (. 00 A) 8. 0 V = V (. 00 Ω ) 8. 0 V = V V = = 00. Ω 00. A (c) Apply Kirchhoff s loop rule to loop (): (6. 00 A)(3. 00 Ω) (4. 00 A)(6. 00 Ω ) ε = 0 ε = V 4. 0 V = 4. 0 V EVALUATE: Can check that the loop rule is satisfied for loop (3), as a check of our work: 8. 0 V ε (4. 00 A)(6. 00 Ω) (. 00 A) = V 4. 0 V 4. 0 V (. 00 A)(. 00 Ω ) = 0. 0 V = 4. 0 V V. 0 V =. 0 V, so the loop rule is satisfied for this loop. (d) I : If the circuit is broken at point x there can be no current in the Ω resistor. There is Figure 6.b EXECUTE: 8. 0 V (3. 00 Ω) I (. 00 Ω ) I = V I = = 30. A 800. Ω EVALUATE: Breaking the circuit at x removes the 4.0-V emf from the circuit and the current through the Ω resistor is reduced. IDENTIFY: Apply the loop rule and junction rule.

9 . Ω 6.6. IDENTIFY: Apply the loop rule and junction rule. SET UP: The circuit diagram is given in Figure 6.6. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. EXECUTE: The loop rule applied to loop (1) gives: 0. 0V (1. 00 A)(1. 00 Ω ) (1. 00 A)(4. 00 Ω ) (1. 00 A)(1. 00 Ω) ε (1. 00 A)(6. 00 Ω ) = 0 ε 1 = 0. 0 V V V V V = V. The loop rule applied to loop () gives: 0. 0 V (1. 00 A)(1. 00 Ω) (. 00 A)(1. 00 Ω) ε (. 00 A)(. 00 Ω) (1. 00 A)(6. 00 Ω ) = 0 ε = 0. 0 V V. 00 V V V = 7. 0 V. Going from b to a along the lower branch, V (. 00 A)(. 00 Ω ) 7. 0 V (. 00 A)(1. 00 Ω ) = V V V = V; point b is at 13.0 V lower b a b a potential than point a. EVALUATE: We can also calculate Vb Va by going from b to a along the upper branch of the circuit. Vb (1. 00 A)(6. 00 Ω ) 0. 0 V (1. 00 A)(1. 00 Ω ) = Va and Vb Va = V. This agrees with Vb Va calculated along a different path between b and a. 1 Figure 6.6

10 6.4. t/ IDENTIFY: For a charging capacitor qt () Cε (1 e τ ε t/ τ = ) and it () = e. 6 6 SET UP: The time constant is C = ( Ω ) ( F) = 11.1s. tc / EXECUTE: (a) At t = 0 s: q= Cε (1 e ) = 0. tc / 6 (0. s)/(111. s) 4 At t = s: q= Cε (1 e ) = ( F)(60. 0 V)(1 e ) = C. tc / 6 (100. s)/(111. s) 4 At t = 10 s: q= Cε (1 e ) = ( F)(60. 0 V)(1 e ) = C. tc / 6 (00. s)/(111. s) 4 At t = 0 s : q= Cε (1 e ) = ( F)(60. 0 V)(1 e ) = C. tc / 6 (100 s)/(111. s) 4 At t = 100 s : q= Cε (1 e ) = ( F)(60. 0 V)(1 e ) = C. ε tc / (b) The current at time t is given by: i= e V 0/11 1 At t = 0 s : i = e. = A Ω V 111 / At t = s : i= e. = A Ω V 10/11 1 At t = 10 s : i = e. = A Ω V 0/11 1 At t = 0 s : i= e. = A Ω V 100/ At t = 100 s : i= e. = A Ω (c) The graphs of qt () and it () are given in Figure 6.4a and b. EVALUATE: The charge on the capacitor increases in time as the current decreases. Figure 6.4

11 too large, the current is not large enough IDENTIFY: Consider one segment of the network attached to the rest of the network. SET UP: We can re-draw the circuit as shown in Figure EXECUTE: T T T =. = T T = ±. > 0, so =. T T T = 0. 1 T 1 EVALUATE: Even though there are an infinite number of resistors, the equivalent resistance of the network is finite. Figure 6.91

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