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Physics 104 26.3. IDENTIFY: The emf of the battery remains constant, but changing the resistance across it changes its power output. V2 . SET UP: The power output across a resistor is P R EXECUTE: With just R1, P1 R2 added, Rtot 400. P 26.4. Assignment 26 V2 and V P1R1 (360W)(250) 300 V is the battery voltage. With R1 V 2 (300 V)2 225 W. Rtot 400 EVALUATE: The two resistors in series dissipate electrical energy at a smaller rate than R1 alone. IDENTIFY: For resistors in parallel the voltages are the same and equal to the voltage across the equivalent resistance. 1 1 1 SET UP: V IR. . Req R1 R2 1 1 EXECUTE: (a) Req 32 20 (b) I 1 123 . V 240 V 195 A. Req 123 V 240 V V 240 V 75 A; I 20 12 A. R 32 R 20 EVALUATE: More current flows through the resistor that has the smaller R. IDENTIFY: First do as much series-parallel reduction as possible. SET UP: The 45.0- and 15.0- resistors are in parallel, so first reduce them to a single equivalent resistance. Then find the equivalent series resistance of the circuit. EXECUTE: 1/Rp 1/(45.0 ) 1/(15.0 ) and Rp 11.25 . The total equivalent resistance is (c) I 32 26.7. 26.8. 18.0 11.25 3.26 32.5 . Ohm’s law gives I (25.0 V)/(32.5 ) 0.769 A. EVALUATE: The circuit appears complicated until we realize that the 45.0- and 15.0- resistors are in parallel. IDENTIFY: Eq. (26.2) gives the equivalent resistance of the three resistors in parallel. For resistors in parallel, the voltages are the same and the currents add. (a) SET UP: The circuit is sketched in Figure 26.8a. EXECUTE: parallel 1 1 1 1 Req R1 R2 R3 1 1 1 1 Req 160 240 480 Req 0800 Figure 26.8a (b) For resistors in parallel the voltage is the same across each and equal to the applied voltage; V1 V2 V3 280 V V IR so I1 I2 V1 280 V 175 A R1 160 V2 280 V V 280 V 117 A and I 3 3 58 A R2 240 R3 48 1 (c) The currents through the resistors add to give the current through the battery: I I1 I 2 I3 175 A 117 A 58 A 350 A EVALUATE: Alternatively, we can use the equivalent resistance Req as shown in Figure 26.8b. IReq 0 I Req 280 V 350 A, which checks 0800 Figure 26.8b (d) As shown in part (b), the voltage across each resistor is 28.0 V. (e) IDENTIFY and SET UP: We can use any of the three expressions for P:P VI I 2R V 2 /R. They will all give the same results, if we keep enough significant figures in intermediate calculations. (280 V) 2 (280 V) 2 490 W, P2 V22 /R2 327 W, and EXECUTE: Using P V 2 /R, P1 V12 /R1 160 240 P3 V32 /R3 (28.0 V)2 163W. 4.80 EVALUATE: The total power dissipated is Pout P1 P2 P3 980 W. This is the same as the power Pin I (280 V)(350 A) 980 W delivered by the battery. 26.10. (f) P V 2 /R. The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance. (a) IDENTIFY: The current, and hence the power, depends on the potential difference across the resistor. SET UP: P V 2 /R EXECUTE: (a) V PR (50 W)(15,000) 274 V (b) P V 2 /R (120 V)2 /(9,000) 16 W SET UP: (c) If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe. Therefore the maximum power in the larger resistor must be 2.00 W. Use P I 2 R to find the maximum current through the series combination and use Ohm’s law to find the potential difference across the combination. EXECUTE: P I 2 R gives I P/R (200 W)/(150 ) 0115 A. The same current flows through both resistors, and their equivalent resistance is 250 . Ohm’s law gives V IR (0.115 A)(250 ) 28.8 V. Therefore P150 2.00 W and P100 I 2R (0.115 A)2 (100 ) 1.32 W. EVALUATE: If the resistors in a series combination all have the same power rating, it is the largest resistance that limits the amount of current. 1 1 1 26.13.IDENTIFY: For resistors in parallel, the voltages are the same and the currents add. so Req R1 R2 R1R2 , For resistors in series, the currents are the same and the voltages add. Req R1 R2 . R1 R2 SET UP: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in Figure 26.13. 600 V EXECUTE: Req 500. In Figure 26.13c, I 120 A. This is the current through each of the 500 resistors in Figure 26.13b. V12 IR12 (120 A)(200) 240 V. V34 IR34 (120 A)(300) 360 V. Req Note that V12 V34 600 V. V12 is the voltage across R1 and across R2 , so I1 2 V12 240 V 800 A and R1 300 I2 V12 240 V V 360 V 400 A. V34 is the voltage across R3 and across R4 , so I 3 34 300 A and R2 600 R3 120 I4 V34 360 V 900 A. R4 400 EVALUATE: Note that I1 I 2 I3 I 4 . Figure 26.13 26.17.IDENTIFY: Apply Ohm’s law to each resistor. SET UP: For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents are the same and the voltages add. EXECUTE: The current through the 2.00- resistor is 6.00 A. Current through the 1.00- resistor also is 6.00 A and the voltage is 6.00 V. Voltage across the 6.00- resistor is 12.0 V 6.0 V 18.0 V. Current through the 6.00- resistor is (18.0 V)/(6.00 ) 3.00 A. The battery emf is 18.0 V. EVALUATE: The current through the battery is 6.00 A 3.00 A 9.00 A. The equivalent resistor of the resistor network is 2.00 , and this equals (18.0 V)/(9.00 A). 26.25.IDENTIFY: Apply Kirchhoff’s point rule at point a to find the current through R. Apply Kirchhoff’s loop rule to loops (1) and (2) shown in Figure 26.25a to calculate R and . Travel around each loop in the direction shown. (a) SET UP: Figure 26.25a EXECUTE: Apply Kirchhoff’s point rule to point a: I 0 so I 400 A 600 A 0 I 2.00 A (in the direction shown in the diagram). (b) Apply Kirchhoff’s loop rule to loop (1): (600 A)(300) (200 A) R 280 V 0 180 V (200) R 280 V 0 280 V 180 V 500 200 A (c) Apply Kirchhoff’s loop rule to loop (2): (600 A)(300) (400 A)(600) 0 180 V 240 V 420 V EVALUATE: Can check that the loop rule is satisfied for loop (3), as a check of our work: 280 V (400 A)(600) (200 A)R 0 R 3 280 V 420 V 240 V (200 A)(500) 0 520 V 420 V 100 V 520 V 520 V, so the loop rule is satisfied for this loop. (d) IDENTIFY: If the circuit is broken at point x there can be no current in the 600- 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 26.25b. Figure 26.25b EXECUTE: 280 V (300)I (500)I 0 280 V 350 A 800 EVALUATE: Breaking the circuit at x removes the 42.0-V emf from the circuit and the current through the 300- resistor is reduced. 26.27.IDENTIFY: Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop rule to three loops to calculate 1, 2 and R. (a) SET UP: The circuit is sketched in Figure 26.27. I Figure 26.27 EXECUTE: Apply the junction rule to point a: 300 A 500 A I3 0 I3 800 A Apply the junction rule to point b: 200 A I 4 300 A 0 I 4 100 A Apply the junction rule to point c: I3 I 4 I5 0 I5 I3 I 4 800 A 100 A 700 A EVALUATE: As a check, apply the junction rule to point d: I5 200 A 500 A 0 I5 700 A (b) EXECUTE: Apply the loop rule to loop (1): 1 (300 A)(400) I3 (300) 0 1 120 V (800 A)(300) 360 V Apply the loop rule to loop (2): 2 (500 A)(600) I3 (300) 0 4 2 300 V (800 A)(300) 540 V (c) Apply the loop rule to loop (3): (200 A) R 1 2 0 2 1 540 V 360 V 900 200 A 200 A EVALUATE: Apply the loop rule to loop (4) as a check of our calculations: (200 A)R (300 A)(400) (500 A)(600) 0 (200 A)(900) 120 V 300 V 0 180 V 180 V 0 IDENTIFY: Use Kirchhoff’s rules to find the currents. SET UP: Since the 10.0-V battery has the larger voltage, assume I1 is to the left through the 10-V battery, I 2 R 26.28. is to the right through the 5 V battery, and I 3 is to the right through the 10- resistor. Go around each loop in the counterclockwise direction. EXECUTE: Upper loop: 100 V (200 300 ) I1 (100 400 ) I 2 500 V 0. This gives 50 V (500 ) I1 (500 ) I 2 0, and I1 I 2 100 A. Lower loop: 500 V (100 400 ) I 2 (100 ) I3 0. This gives 500 V (500 ) I 2 (100 ) I3 0, and I 2 2I3 100 A. Along with I1 I 2 I3 , we can solve for the three currents and find: I1 0800 A, I 2 0200 A, I3 0600 A. (b) Vab (0200 A)(400 ) (0800 A)(300 ) 320 V. EVALUATE: Traveling from b to a through the 400- and 300- resistors you pass through the resistors in the direction of the current and the potential decreases; point b is at higher potential than point a. 26.40.IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each time. SET UP: The voltage across a capacitor is VC q /C. EXECUTE: (a) At the instant the circuit is completed, there is no voltage across the capacitor, since it has no charge stored. (b) Since VC 0, the full battery voltage appears across the resistor VR 245 V. (c) There is no charge on the capacitor. 245 V (d) The current through the resistor is i 00327 A. Rtotal 7500 (e) After a long time has passed the full battery voltage is across the capacitor and i 0. The voltage across the capacitor balances the emf: VC 245 V. The voltage across the resistor is zero. The capacitor’s charge is 26.41. q CVC (460 106 F) (245 V) 113 103 C. The current in the circuit is zero. EVALUATE: The current in the circuit starts at 0.0327 A and decays to zero. The charge on the capacitor starts at zero and rises to q 113 103 C. IDENTIFY: The capacitor discharges exponentially through the voltmeter. Since the potential difference across the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases exponentially with the same time constant as the charge. SET UP: The reading of the voltmeter obeys the equation V V0e –t /RC , where RC is the time constant. EXECUTE: (a) Solving for C and evaluating the result when 4.00s gives t 400 s C 8.49 107 F R ln(V /V0 ) 120 V 6 (340 10 )ln 300 V (b) RC (3.40 106 )(8.49 107 F) 2.89 s EVALUATE: In most laboratory circuits, time constants are much shorter than this one. 26.46.IDENTIFY: The charge is increasing while the current is decreasing. Both obey exponential equations, but they are not the same equation. 5 SET UP: The charge obeys the equation Q Qmax (1 et /RC ), but the equation for the current is I I maxe–t /RC . EXECUTE: When the charge has reached 1 4 of its maximum value, we have Qmax /4 Qmax (1– e–t /RC ), which says that the exponential term has the value e – t /RC 34 . The current at this time is I I maxe–t /RC Imax (3/4) (3/4)[(10.0 V)/(12.0 )] 0.625 A. EVALUATE: Notice that the current will be 3 , 4 not 1 , 4 of its maximum value when the charge is 1 4 of its maximum. Although current and charge both obey exponential equations, the equations have different forms for a charging capacitor. 6