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

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1 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 6 and the total current /R = 1. This 1 will divide in the ratio of :1 through the 3 and 6 respectively so the 3 resistor receives /3 making the potential difference IR = (/3 )(3 ) = V. 3. dding resistors in parallel decreases the total circuit resistance, this increasing the total current in the circuit. 4. R = L/. Greatest resistance is the longest, narrowest resistor. 5. In parallel V 1 = V. Q 1 = 1 V 1 and Q = V so Q 1 /Q = 1 / = For steady power dissipation, the circuit must allow current to slow indefinitely. For the greatest power, the total resistance should be the smallest value. These criteria are met with the resistors in parallel. 7. To retain energy, there must be a capacitor that will not discharge through a resistor. apacitors in circuits and will discharge through the resistors in parallel with them. 8. P = I 9. W = Pt = I Rt 10. The resistance of the two resistors in parallel is 1. dded to the resistor in series with the pair gives R = L/. Least resistance is the widest, shortest resistor 1. The resistance of the two resistors in parallel is r/. The total circuit resistance is then 10 + ½ r, which is equivalent to /I = (10 V)/(0.5 ) = 0 = 10 + r/ 13. Resistance varies directly with temperature. Superconductors have a resistance that quickly goes to zero once the temperature lowers beyond a certain threshold. 14. The loop rule involves the potential and energy supplied by the battery and it s use around a circuit loop. 15. The capacitance of the 4 F and F in parallel is 6 F. ombined with the 3 F in series gives F for the right branch. dded to the 5 F in parallel gives a total of 7 F 16. Since the 5 F capacitor is in parallel with the battery, the potential difference across it is 100 V. Q = V 17. Total circuit resistance (including internal resistance) = 40 ; total current = 0.3. = IR 18. V XY = Ir where r is the internal resistance 19. P = I r 0. With more current drawn from the battery for the parallel connection, more power is dissipated in this connection. While the resistors in series share the voltage of the battery, the resistors in parallel have the full potential difference of the battery across them. 1. mperes = I (current); Volts = V (potential difference); Seconds = t (time): IVt = energy 147

2 . Resistance of the 1 and 3 in series = 4. This, in parallel with the resistor gives ( 4) /( + 4) = 8/6. lso notice the equivalent resistance must be less than (the resistor is in parallel and the total resistance in parallel is smaller than the smallest resistor) and there is only one choice smaller than. 3. The upper branch, with twice the resistance of the lower branch, will have ½ the current of the lower branch. 4. Power = IV = 480 W = 0.48 kw. nergy = Pt = (0.48 kw)( hours) = 0.96 kw-h 5. Total circuit resistance of the load = R/. Total circuit resistance including the internal resistance = r + R/. The current is then /(r + R/) and the total power dissipated in the load is P = I R load = ( R/)/(r + R/). Using calculus max/min methods or plotting this on a graph gives the value of R for which this equation is maximized of R = r. This max/min problem is not part of the curriculum but you should be able to set up the equation to be maximized. 6. The larger loop, with twice the radius, has twice the circumference (length) and R = L/ 7. y process of elimination, is the only possible true statement. 8. R = L/. If L, R and is r then 4 and R 4 making the net effect R 4 9. The motor uses P = IV = 60 W of power but only delivers P = Fv = mgv = 45 W of power. The efficiency is what you get what you are paying for = 45/ Resistance of the 000 and 6000 in parallel = 1500, adding the 500 in series gives a total circuit resistance of I total = I 1 = /R 31. I 1 is the main branch current and is the largest. It will split into I and I 3 and since I moves through the smaller resistor, it will be larger than I P = V /R 33. The current through R is found using the junction rule at the top junction, where 1 + enter giving I = 3. Now utilize Kirchhoff s loop rule through the left or right loops: (left side) + 16 V (1 )(4 ) (3 )R = 0 giving R = Utilizing Kirchhoff s loop rile with any loop including the lower branch gives 0 V since the resistance next to each battery drops the V of each battery leaving the lower branch with no current. You can also think of the junction rule where there is 0.04 going into each junction and 0.04 leaving to the other battery, with no current for the lower branch. 35. Summing the potential differences from left to right gives V T = 1 V ( )( ) = 16 V. It is possible for V T >. 36. urrent is greatest where resistance is least. The resistances are, in order, 1,, 4, and See above 38. Least power is for the greatest resistance (P = /R) 39. In series, the equivalent capacitance is calculated using reciprocals, like resistors in parallel. This results in an equivalent capacitance smaller than the smallest capacitor. 40. V T = Ir 41. Kirchhoff s junction rule applied at point X gives = I + 1, so the current in the middle wire is 1. Summing the potential differences through the middle wire from X to Y gives 10 V (1 )( ) = 1 V total 148

3 4. When the switch is closed, the circuit behaves as if the capacitor were just a wire and all the potential of the battery is across the resistor. s the capacitor charges, the voltage changes over to the capacitor over time, eventually making the current (and the potential difference across the resistor) zero and the potential difference across the capacitor equal to the emf of the battery. 43. See above 44. See above 45. In series 1 = There are several ways to do this problem. We can find the total energy stored and divide it into the three capacitors: U = ½ V = ½ ( F)(6 V) = 36 J 3 = 1 J each 47. P = I R and R = L/ giving P L/d 48. P = I R 49. Since these resistors are in series, they must have the same current. 50. ach branch, with two capacitors in series, has an equivalent capacitance of F = 1 F. The three branches in parallel have an equivalent capacitance of 1 F + 1 F + 1 F = 3 F 51. For each capacitor to have 6, each branch will have 6 since the two capacitors in series in each branch has the same charge. The total charge for the three branches is then 18. Q = V gives 18 = (3 F)V 5. Utilizing Kirchhoff s loop rule starting at the upper left and moving clockwise: ( )(0.3 ) + 1 V 6 V ( )(0. ) ()(R) ()(1.5 ) = Summing the potential differences: 6 V ( )(0. ) ()(1 ) = 8.4 V 54. nergy = Pt = I Rt 55. When the switch is closed, the circuit behaves as if the capacitor were just a wire, shorting out the resistor on the right. 56. When the capacitor is fully charged, the branch with the capacitor is closed to current, effectively removing it from the circuit for current analysis. 57. Total resistance = /I = 5. Resistance of the 30 and 60 resistors in parallel = 0 adding the internal resistance in series with the external circuit gives R total = 0 + r = P = V /R and if V is constant P 1/R 59. For the ammeter to read zero means the junctions at the ends of the ammeter have the same potential. For this to be true, the potential drops across the 1 and the resistor must be equal, which means the current through the 1 resistor must be twice that of the resistor. This means the resistance of the upper branch (1 and 3 ) must be ½ that of the lower branch ( and R) giving = ½ ( + R) 60. Kirchhoff s loop rule (V = Q/ for a capacitor) 61. To dissipate 4 W means R = V /P = 6. The resistances, in order, are: 8, 4/3, 8/3, 1 and 6 6. imensional analysis: = /s /proton = protons/sec 10 9 protons/meter = 10 7 m/s 63. The equivalent capacitance of the two 3 F capacitors in parallel is 6 F, combined with the 3 F in series gives total = F 149

4 64. The equivalent capacitance between X and Y is twice the capacitance between Y and Z. This means the voltage between X and Y is ½ the voltage between Y and Z. For a total of 1 V, this gives 4 V between X and Y and 8 V between Y and Z. 65. losing the switch short circuits ulb causing no current to flow to it. Since the bulbs were originally in series, this decreases the total resistance and increases the total current, making bulb 1 brighter. 66. In series 1 = P = V /R 68. losing the switch reduces the resistance in the right side from 0 to 15, making the total circuit resistance decrease from 35 to 30, a slight decrease, causing a slight increase in current. For the current to double, the total resistance must be cut in half. 69. R = L/ L/d where d is the diameter. R x /R y = L x /d x L y /d y = (L y )d y /[L y (d y ) ] = ½ 70. Using all three in series = 3, all three in parallel = 1/3. One in parallel with two in series = /3, one in series with two in parallel = 3/ 71. Summing the potential differences from bottom to top: left circuit: (1 )r + = 10 V right circuit: + (1 )r + = 0 V, solve simultaneous equations 7. The equivalent resistance of the 0 and the 60 in parallel is 15, added to the 35 resistor in series gives = If you perform Kirchhoff s loop rule for the highlighted loop, you get a current of 0 through the 6 resistor. 74. N is in the main branch, with the most current. The current then divides into the two branches, with K receiving twice the current as L and M. The L/M branch has twice the resistance of the K branch. L and M in series have the same current. 75. See above. urrent is related to brightness (P = I R) 76. If K burns out, the circuit becomes a series circuit with the three resistors, N, M and L all in series, reducing the current through bulb N. 77. If M burns out, the circuit becomes a series circuit with the two resistors, N and K in series, with bulb L going out as well since it is in series with bulb M. 150

5 78. Using Kirchhoff s loop rule around the circuit going through either V or R since they are in parallel and will have the same potential drop gives: V (1.00 m)(5 ) V (1.00 m)(975 ) = The equivalent resistance in parallel is smaller than the smallest resistance. 80. When the capacitor is fully charged, the branch on the right has no current, effectively making the circuit a series circuit with the 100 and 300 resistors. R total = 400, = 10 V = IR 81. In series, they all have the same current,. P 3 = I 3 V 3 8. P = /R. Total resistance of n resistors in series is nr making the power P = /nr = P/n 83. The current through bulb 3 is twice the current through 1 and since the branch with bulb 3 is half the resistance of the upper branch. The potential difference is the same across each branch, but bulbs 1 and must divide the potential difference between them. 84. by definition of a parallel circuit 85. R = L/ L/d where d is the diameter. R II /R I = L II /d II L I /d I = (L I )d I /[L I (d I ) ] = ½ 86. P = IV 87. If the current in the 6 resistor is 1, then by ratios, the currents in the and 3 resistor are 3 and respectively (since they have 1/3 and 1/ the resistance). This makes the total current 6 and the potential drop across the 4 resistor 4 V. Now use Kirchhoff s loop rule for any branch. 88. The voltage across the capacitor is 6 V (Q = V) and since the capacitor is in parallel with the 300 resistor, the voltage across the 300 resistor is also 6 V. The 00 resistor is not considered since the capacitor is charged and no current flows through that branch. The 100 resistor in series with the 300 resistor has 1/3 the voltage ( V) since it is 1/3 the resistance. Kirchhoff s loop rule for the left loop gives = 8 V. 89. P = V /R 90. For the currents in the branches to be equal, each branch must have the same resistance. 91. R L/ = L/d. If d, R 4 and if L, R making the net effect R 8 9. ulbs in the main branch have the most current through them and are the brightest. 93. In parallel, all the resistors have the same voltage ( V). P 3 = I 3 V If the resistances are equal, they will all draw the same current. 95. Resistor is in a branch by itself while resistors, and are in series, drawing less current than resistor. 96. ven though the wires have different resistances and currents, the potential drop across each is 1.56 V and will vary by the same gradient, dropping all 1.56 V along the same length. 97. ach computer draws I = P/V = computers will draw 16.7, while 5 will draw over The capacitance of the two capacitors in parallel is. ombined with a capacitor in series gives = + = P = IV = 1.56 kw. nergy = Pt = 1.56 kw 8 h = 1.48 kw-h 100. Resistance of bulbs & = 0 combined with in parallel gives 6.7 for the right side. ombined with & in series gives a total resistance of 6.7. = IR 151

6 P Physics Free Response Practice ircuits NSWRS a. VT = Ir = 6 V b. In parallel, each resistor gets 6 V and P = V /R gives R = 3 c. For the 3 resistor we have I = V/R = leaving 1 for the branch with R1. R = V/I = a. The two batteries are connected with opposing emfs so the total emf in the circuit is = 60 V 1 V = 48 V The resistance of the parallel combination of resistors is ( ¼ + ¼ + ½ ) = 1 combining with the rest of the resistors in series gives a total circuit resistance of 8. The total current is then /R = 6. The voltage across the parallel combination of resistors is V P = IR P = 6 V so the current through the resistor is I = V/R = 3 b. P = I R = 108 W c. The current is forced through battery from the positive to the negative terminal, charging the battery. This makes the equation for the terminal voltage V T = + Ir = 18 V 1980 a. The resistance of the device is found from R = V/I = 6. With a 4 volt source, to provide a current of requires a total resistance of 1. For the additional 6 resistance, place two 3 resistors in series with the device. b. Since the device requires, a resistor in parallel with the device must carry a current of 6 = 4. In parallel with the device, the resistor will have a potential difference of 1 V so must have a resistance of V/I = 3. Thus, a single 3 resistor in parallel will suffice. c. P = I R = 48 W 157

7 1984 a. Since the clock requires 15 V it must be directly connected between and. Since the radio requires less than 15 V, there must be a resistor in series with it. b. The current through the radio (and R) is 10 m. The voltage across the radio is 9 V, which leaves 6 V across the resistor giving R = V/I = 600 c. P = IV where V = 15 V and I = 10 m + 0 m = 30 m so P = 0.45 W and energy = Pt = 7 J a. The two batteries are connected with opposing emfs so the total emf in the circuit is = 0 V V = 18 V The equivalent resistance of the two parallel resistors is (6 1)/(6 + 1) = 4 and since R is in series with the pair, the total circuit resistance is (4 + R) = /I = 9 giving R = 5 b. ecause the voltages of the two resistors in parallel are equal we have 6I 1 = 1I and I 1 + I = giving i. 4/3 ii. /3 c. Summing the potential differences form point X gives V X + IR = 0 + ( )(5 ) = V = 10 V. ontinuing along gives V 0 V = V = 10 V. nd V + (/3 )(1 ) = V = V d. P = I = 40 W a. P = V R gives R = 40 b. ulbs Y and Z in parallel have an equivalent resistance of 10. dding bulb X in series with the pair gives R = 360 c. P T = /R T = 40 W d. I = /R = 1/3 e. V X = IR X = 80 V f. The current splits equally through Y and Z. V Z = I Z R Z = (1/6 )(40 ) = 40 V a. The equivalent resistance of R1 and R is (1 4)/(1 + 4) = 3. dding R 3 in series with the pair gives R = b. 1 = IRT = 4.8 V c. The voltage across resistor 1 (equal to the voltage across R) is the emf of the battery minus the drop across R 3 which is 4.8 V (0.4 )(9 ) = 1. V d. P = V /R = 0.36 W e. Q = It = (0.4 /s)(60 s) = 4 158

8 19883 a. On the right we have two resistors in series: 10 + = 1. This is in parallel with the 4 resistor which is an equivalent resistance of 3 and adding the remaining main branch resistor in series gives a total circuit resistance of 9. The current is then I = /RT = 8 b. The voltage remaining for the parallel branches on the right is the emf of the battery minus the potential dropped across the 6 resistor which is 7 V (8 )(6 ) = 4 V. Thus the current in the 10 resistor is the current through the whole 1 branch which is I = V/R = (4 V)/(1 ) = c. V 10 = I 10 R 10 = 0 V d. When charged, the capacitor is in parallel with the 10 resistor so V = V 10 = 0 V and Q = V = 60 e. U = ½ V = J a. i. P = I R = ( ) (10 ) = 40 W ii. P = Fv = mgv = 0 W (using g = 10 m/s ) iii. P = P R + P M = 40 W + 0 W = 60 W b. i. V = IR = 0 V ii. V = P/I = (0 W)/( ) = 10 V iii. = V R + V M = 30 V c. Since the speed is increased by 3/, the voltage drop increases by the same value and is now (3/)(10 V) = 15 V d. The new voltage across the resistor is found from V R = V M = 15 V and I = V R /I = (15 V)/( ) = a. The 4 and 8 are in series so their equivalent resistance is 1. nother 1 resistor in parallel makes the equivalent resistance (1 1)/(1 + 1) = 6 b. dding the remaining resistors in series throughout the circuit gives a total circuit resistance of 1 and the total current (which is also the current in the 5 resistor) = /R = c. V = Ir = V d. The current divides equally between the two branches on the right so P1 = I R = (1 ) (1 ) = 1 W e. From to you only have to pass through the 1 resistor which gives V = (1 )(1 ) = 1 V f. P = V /R external = ( V) /11 = 44 W a/b. VXY = Ir and using data from the graph we can find two equations to solve simultaneously 4 V = (1 )r and 3 V = (3 )r will yield the solutions = 4.5 V and r = 0.5 c. VXY = IR which gives 3 V = (3 )R and R = 1 d. I max occurs for R = 0 and V XY = 0 which gives = I max r and I max = 9 (this is the x intercept of the graph) e. 159

9 1995 a. P = V /R gives R = 4 b. = Pt where t = (30 days)(4 h/day)(3600 sec/h) gives = c. 7 J The bulb, needing only 1 V must have a resistor in series with it and the toaster, requiring 10 V must be connected directly to the power supply. d. The current through the bulb is I = P/V = 0.5, which is also the current in the resistor, which must have 108 V across it to provide the light bulb only 1 V. R = V/I = (108 V)/(0.5 ) = 16 e. i. If the resistance of the resistor is increased, the current through the branch will decrease, decreasing the brightness of the bulb. ii. Since the toaster operates in its own parallel branch, nothing will change for the toaster a. For the smallest current, place the resistors in series b. For the largest current, place the resistors in parallel c. i. The 0 and 30 resistors combine in series as a 50 resistor, which is in parallel with the 100 resistor making their effective resistance dding the 10 resistor in the main branch in series gives a total circuit resistance of 43. The current in the 10 resistor is the total current delivered by the battery /R = 0.8 ii. P = /R = 3.35 W d. = Pt, or t = /P = (10 10 J)/(3.35 W) = seconds 160

10 19974 a. i. In series R T = 90 and P = V /R = 6.4 W ii. In parallel RT = 10 and P = 57.6 W b. The fastest heating occurs with a parallel connection 003 a. The resistance of the 6 and 3 resistors in parallel is (6 3)/(6 + 3) =. dding the 3 resistor in the main branch gives a total circuit resistance of 5. The current in bulb in the main branch is the total current delivered by the battery I = /R = (9 V)/(5 ) = 1.8 b. ulb is the brightest. In the main branch, it receives the most current. You can also calculate the power of each resistor where P = 9.7 W, P =. W and P = 4.3 W c. i. Removing ulb from the circuit changes the circuit to a series circuit, increasing the total resistance and decreasing the total current. With the total current decreased, bulb is dimmer. ii. Since bulb receives less current, the potential drop is less than the original value and being in a loop with bulb causes the voltage of bulb to increase, making bulb brighter. The current through bulb is greater since it is no longer sharing current with bulb. iii. The current through bulb is zero, bulb goes out a. b. > > = ulb has the largest current through it, making it brightest. The voltage across bulb is the same as that across bulbs and combined, so it is next brightest, leaving and as least bright ulbs and are in series, and thus have the same current through them, so they must be equally bright. c. i. The brightness of bulb decreases. The total resistance of the circuit increases so the current in bulb decreases. ii. The brightness of bulb increases. The current (and the voltage) across increases. ven though the total current decreases, it is no longer splitting to do through the branch with bulb. nother way to look at it is since has less current, the potential difference across is decreased, this allows a larger share of the battery voltage to be across and. 161

11 0003 a. The equivalent resistance of the two resistors in parallel is R/, which if ½ the resistance of the resistor in the main branch, so the parallel combination will receive half the potential difference of the main branch resistor. The 30 V of the battery will then divide into 0 V for the main branch resistor (and across the voltmeter) and 10 V each for the resistors in parallel. b. fter the switch has been closed for a long time, the voltage across the capacitor will be 30 V. Q = V = c. i. The 30 V battery is still connected across the capacitor so the potential difference remains 30 V. ii. = 0 inside a conductor in electrostatic equilibrium iii. iv. = V/d and you can use the entire gap or just one of the two gaps; = 30 V/(0.5 mm) or 15 V/(0.5 mm) = 60 V/mm or 60,000 V/m 003 a. i. P = V /R gives R = 480 and V = IR gives I = 0.5 ii. P = V /R gives R = 360 and V = IR gives I = 0.33 b. i./ii. The resistances are unchanged = 480 and 360. The total resistance in series is = 840 making the total current I = V/R = 0.14 which is the same value for both resistors in series c. The bulbs are brightest in parallel, where they provide their labeled values of 40 W and 30 W. In series, it is the larger resistor (the 30 W bulb) that glows brighter with a larger potential difference across it in series. This gives the order from top to bottom as d. i. In parallel, they each operate at their rated voltage so they each provide their rated power and P T = 30 W + 40 W = 70 W ii. In series PT = V T /R T = 17 W 003 a. For two capacitors in series the equivalent capacitance is (6 1)/(6 + 1) = 4 F b. The capacitors are fully charged so current flows through the resistors but not the capacitors. R T = 30 and I = V/R = 0. c. The potential difference between and is the voltage across the 0 resistor. V = IR = 4 V d. The capacitors in series store the same charge as a single 4 F capacitor. Q = V = (4 F)(4 V) = 16 e. Remains the same. No current is flowing from to P to therefore braking the circuit at point P does not affect the current in the outer loop, and therefore will not affect the potential difference between and. 16

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