DIRECT-CURRENT CIRCUITS 26.7.IDENTIFY: First do as much series-parallel reduction as possible. SET UP: The 45.0-Ω and 15.0-Ω 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 Ω) + 1/(15.0 Ω) and Rp = 11.25 Ω. The total equivalent resistance is 18.0 Ω + 11.25 Ω + 3.26 Ω = 32.5 Ω. Ohm’s law gives I = (25.0 V)/(32.5 Ω) = 0.769 A. EVALUATE: The circuit appears complicated until we realize that the 45.0-Ω and 15.0-Ω resistors are in parallel. 26.8. 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 eq 1 2 3 1 1 1 1R R R R eq 1 1 1 11.60 2.40 4.80 R eq 0.800 R Figure 26.8a (b) For resistors in parallel the voltage is the same across each and equal to the applied voltage; 1 2 3 28.0 VV V V E 11 1 28. 0 V so 17. 5 A1.60 VV I R I R 232328.0 V 28.0 V11.7 A a nd 5.8 A2.40 4.8 VVIIRR (c) The currents through the resistors add to give the current through the battery: 1 2 3 1 7 . 5 A 1 1 . 7 A 5 . 8 A 3 5 . 0 AI I I I EVALUATE: Alternatively, we can use the equivalent resistance eqR as shown in Figure 26.8b. eq 0IRE eq 28 .0 V 35 .0 A ,0.8 00 I RE which checks 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 22: / .P P V I I R V R They will all give the same results, if we keep enough significant figures in intermediate calculations. 26 EXECUTE: Using 2/,P V R 2222 1 1 1 2 2 22 8 . 0 V 2 8 . 0 V/ 4 9 0 W , / 3 2 7 W , a n d1 . 6 0 2 . 4 0 P V R P V R 22 3 3 3 2 8 .0 V/ 1 6 3 W4 .8 0 P V R EVALUATE: The total power dissipated is o u t 1 2 3 9 8 0 W .P P P P This is the same as the power in 2 .8 0 V 3 5 .0 A 9 8 0 WPIE delivered by the battery. (f) 2/.P V R The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance. 26.10. (a) IDENTIFY: The current, and hence the power, depends on the potential difference across the resistor. SET UP: /P V R EXECUTE: (a) ( 5 . 0 W ) (1 5 , 0 0 0 ) 2 7 4 VV P R (b) / (1 2 0 V ) / ( 9 , 0 0 0 ) 1 .6 WP V R 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 2P I 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 R gives I = P/R = (2.00 W)/(150 Ω) = 0.0133 A. The same current flows through both resistors, and their equivalent resistance is 250 Ω. Ohm’s law gives V = IR = (0.0133 A)(250 Ω) = 3.33 V. Therefore P150 = 2.00 W and 100P I R = (0.0133 A)2(100 Ω) = 0.0177 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. 26.13. IDENTIFY: In both circuits, with and without R4, replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use 2P IR to calculate the power dissipated in each bulb. (a) SET UP: The circuit is sketched in Figure 26.13a. EXECUTE: 2 3 4, , and R R R are in parallel, so their equivalent resistance eqR is given by eq 2 3 4 1 1 1 1R R R R Figure 26.13a eqeq13 a nd 1.50 .4.50 RR The equivalent circuit is drawn in Figure 26.13b. 1 eq 0I R RE 1 eqI RR E Figure 26.13b 19 . 0 0 V 1 . 5 0 A a n d 1 . 5 0 A4 . 5 0 1 . 5 0 II Then 1 1 1 1 .5 0 A 4 .5 0 6 .7 5 VV I R e q e q e q e q1 . 5 0 A , 1 . 5 0 A 1 . 5 0 2 . 2 5 VI V I R For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so 2 3 4 2.25 V.V V V 2 3 42 3 4 2 3 4 2.2 5 V 0.5 00 A , 0.5 00 A , 0.5 00 A4.50 V V VI I IR R R EVALUATE: Note that 2 3 4 1.50 A,I I I which is eq.I For resistors in parallel the currents add and their sum is the current through the equivalent resistor. (b) SET UP: 2P I R EXECUTE: 2 1 1.50 A 4.50 10.1 WP 22 3 4 0.500 A 4.50 1.125 W ,P P P which rounds to 1.12 W. 1R glows brightest. EVALUATE: Note that 2 3 4 3.37 W.P P P This equals 22e q e q e q (1 . 5 0 A ) (1 . 5 0 ) 3 . 3 7 W ,P I R the power dissipated in the equivalent resistor. (c) SET UP: With 4R removed the circuit becomes the circuit in Figure 26.13c. EXECUTE: 23 and RR are in parallel and their equivalent resistance eqR is given by eq 2 3 1 1 1 24.50 R R R and eq 2.25 R Figure 26.13c The equivalent circuit is shown in Figure 26.13d. 1 eq 0I R RE 1 eqI RR E 9 .0 0 V 1 .3 3 3 A4 .5 0 2 .2 5 I Figure 26.13d 1 1 1 11 . 3 3 A , 1 . 3 3 3 A 4 . 5 0 6 . 0 0 VI V I R e q e q e q e q 2 31 . 3 3 A , 1 . 3 3 3 A 2 . 2 5 3 . 0 0 V a n d 3 . 0 0 V .I V I R V V 23233. 00 V 0. 667 A , 0. 667 A4.50 VVIIRR (d) SET UP: 2P I R EXECUTE: 1 (1 .3 3 3 A ) ( 4 .5 0 WP 23 ( 0 .6 6 7 A ) (4 .5 0 ) 2 .0 0 W .PP (e) EVALUATE: When R4 is removed, P1 decreases and P2 and P3 increase. Bulb R1 glows less brightly and bulbs R2 and R3 glow more brightly. When R4 is removed the equivalent resistance of the circuit increases and the current through R1 decreases. But in the parallel combination this current divides into two equal currents rather than three, so the currents through R2 and R3 increase. Can also see this by noting that with R4 removed and less current through R1 the voltage drop across R1 is less so the voltage drop across R2 and across R3 must become larger. 6.14.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: From Ohm’s law, the voltage drop across the 6.00 resistor is V = IR = (4.00 A)(6.00 ) = 24.0 V. The voltage drop across the 8.00 resistor is the same, since these two resistors are wired in parallel. The current through the 8.00 resistor is then I = V/R = 24.0 V/8.00 = 3.00 A. The current through the 25.0 resistor is the sum of these two currents: 7.00 A. The voltage drop across the 25.0 resistor is V = IR = (7.00 A)(25.0 ) = 175 V, and total voltage drop across the top branch of the circuit is 175 V + 24.0 V = 199 V, which is also the voltage drop across the 20.0 resistor. The current through the 20.0 resistor is then / 1 9 9 V / 2 0 9 .9 5 A .I V R EVALUATE: The total current through the battery is 7 .0 0 A 9 .9 5 A 1 6 .9 5 A. Note that we did not need to calculate the emf of the battery. 26.15. 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 2.00- resistor is 6.00 A. Current through 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). 26.22. IDENTIFY: Apply the loop rule and junction rule. SET UP: The circuit diagram is given in Figure 26.22. 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: 12 0 . 0 V ( 1 . 0 0 A ) ( 1 . 0 0 ) ( 1 . 0 0 A ) ( 4 . 0 0 ) ( 1 . 0 0 A ) ( 1 . 0 0 ) ( 1 . 0 0 A ) ( 6 . 0 0 ) 0E 1 2 0 . 0 V 1 . 0 0 V 4 . 0 0 V 1 . 0 0 V 6 . 0 0 V 1 8 . 0 VE . The loop rule applied to loop (2) gives: 22 0 . 0 V ( 1 . 0 0 A ) ( 1 . 0 0 ) ( 2 . 0 0 A ) ( 1 . 0 0 ) ( 2 . 0 0 A ) ( 2 . 0 0 ) ( 1 . 0 0 A ) ( 6 . 0 0 ) 0E 2 2 0 . 0 V 1 . 0 0 V 2 . 0 0 V 4 . 0 0 V 6 . 0 0 V 7 . 0 VE . Going from b to a along the lower branch, ( 2 . 0 0 A ) ( 2 . 0 0 ) 7 . 0 V ( 2 . 0 0 A ) ( 1 . 0 0 )baVV. 13.0 VbaVV ; point b is at 13.0 V lower potential than point a. EVALUATE: We can also calculate baVVby going from b to a along the upper branch of the circuit. ( 1 . 0 0 A ) ( 6 . 0 0 ) 2 0 . 0 V ( 1 . 0 0 A ) ( 1 . 0 0 ) and 13.0 VbaVV . This agrees with baVV calculated along a different path between b and a. Figure 26.22 26.23. 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 12, and .REE (a) SET UP: The circuit is sketched in Figure 26.23. Figure 26.23 EXECUTE: Apply the junction rule to point a: 33.00 A 5.00 A 0I 3 8.00 AI Apply the junction rule to point b: 42.00 A 3.00 A 0I 4 1.00 AI Apply the junction rule to point c: 3 4 5 0I I I 5 3 4 8 .0 0 A 1 .0 0 A 7 .0 0 AI I I EVALUATE: As a check, apply the junction rule to point d: 5 2.00 A 5.00 A 0I 5 7.00 AI (b) EXECUTE: Apply the loop rule to loop (1): 133 .0 0 A 4 .0 0 3 .0 0 0IE 1 1 2 .0 V 8 .0 0 A 3 .0 0 3 6 .0 VE Apply the loop rule to loop (2): 235 .0 0 A 6 .0 0 3 .0 0 0IE 2 3 0 .0 V 8 .0 0 A 3 .0 0 5 4 .0 VE (c) Apply the loop rule to loop (3): 122.00 A 0R EE 21 5 4 .0 V 3 6 .0 V 9 .0 0 2 .0 0 A 2 .0 0 AR EE EVALUATE: Apply the loop rule to loop (4) as a check of our calculations: 2 . 0 0 A 3 . 0 0 A 4 . 0 0 5 . 0 0 A 6 . 0 0 0R 2 .0 0 A 9 .0 0 1 2 .0 V 3 0 .0 V 0 18.0 V 18.0 V 0 26.38. IDENTIFY: An uncharged capacitor is placed into a circuit. Apply the loop rule at each time. SET UP: The voltage across a capacitor is /CV qC . EXECUTE: (a) At the instant the circuit is completed, there is no voltage over the capacitor, since it has no charge stored. (b) Since 0CV , the full battery voltage appears across the resistor 125 V.RV E (c) There is no charge on the capacitor. (d) The current through the resistor is to tal 1 2 5 V 0 .0 1 6 7 A .7 5 0 0 i RE (e) After a long time has passed the full battery voltage is across the capacitor and 0i . The voltage across the capacitor balances the emf: 125V.CV The voltage across the resister is zero. The capacitor’s charge is 64( 4 . 6 0 1 0 F ) ( 1 2 5 V ) 5 . 7 5 1 0 C .Cq C V The current in the circuit is zero. EVALUATE: The current in the circuit starts at 0.0167 A and decays to zero. The charge on the capacitor starts at zero and rises to 45.75 10 Cq . 26.41. IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors. SET UP: Since V is proportional to Q, V must obey the same exponential equation as Q, V = V0 e–t/RC. The current is I = (V0 /R) e–t/RC. EXECUTE: (a) Solve for time when the potential across each capacitor is 10.0 V: t = RC ln(V/V0) = –(80.0 Ω)(35.0 µF) ln(10/45) = 4210 µs = 4.21 ms (b) I = (V0 /R) e–t/RC. Using the above values, with V0 = 45.0 V, gives I = 0.125 A. EVALUATE: Since the current and the potential both obey the same exponential equation, they are both reduced by the same factor (0.222) in 4.21 ms. 26.46. IDENTIFY: Both the charge and energy decay exponentially, but not with the same time constant since the energy is proportional to the square of the charge. SET UP: The charge obeys the equation Q = Q0 e–t/RC but the energy obeys the equation U = Q2/2C = (Q0 e–t/RC)/2C = U0 e–2t/RC. EXECUTE: (a) The charge is reduced by half: Q0/2 = Q0 e–t/RC. This gives t = RC ln 2 = (175 Ω)(12.0 µF)(ln 2) = 1.456 ms = 1.46 ms. (b) The energy is reduced by half: U0/2 = U0 e–2t/RC. This gives t = (RC ln 2)/2 = (1.456 ms)/2 = 0.728 ms. EVALUATE: The energy decreases faster than the charge because it is proportional to the square of the charge. 26.49. IDENTIFY: For each circuit apply the loop rule to relate the voltages across the circuit elements. (a) SET UP: With the switch in position 2 the circuit is the charging circuit shown in Figure 26.49a. At t = 0, q = 0. Figure 26.49a EXECUTE: The charge q on the capacitor is given as a function of time by Eq.(26.12): /1 t RCq C eE 54f 1 . 5 0 1 0 F 1 8 . 0 V 2 . 7 0 1 0 C .QC E 59 8 0 1 .5 0 1 0 F 0 .0 1 4 7 sRC Thus, at 0 . 0 1 0 0 s / 0 . 0 1 4 7 s40 . 0 1 0 0 s , 2 . 7 0 1 0 C 1 1 3 3 C .t q e (b) 51 3 3 C 8 .8 7 V1 .5 0 1 0 FC qv C The loop rule says 0CRvvE 1 8 .0 V 8 .8 7 V 9 .1 3 VRCvvE (c) SET UP: Throwing the switch back to position 1 produces the discharging circuit shown in Figure 26.49b. The initial charge 0Q is the charge calculated in part (b), 0 133 C.Q Figure 26.49b EXECUTE: 51 3 3 C 8 .8 7 V ,1 .5 0 1 0 FC qv C the same as just before the switch is thrown. But now 0 , s o 8 .8 7 V .C R R Cv v v v (d) SET UP: In the discharging circuit the charge on the capacitor as a function of time is given by Eq.(26.16): /0 .t RCq Qe EXECUTE: 0.0147 s,RC the same as in part (a). Thus at ( 0 . 0 1 0 0 s ) / ( 0 . 0 1 4 7 s )0 . 0 1 0 0 s , (1 3 3 C ) 6 7 . 4 C .t q e EVALUATE: 10.0 mst is less than one time constant, so at the instant described in part (a) the capacitor is not fully charged; its voltage (8.87 V) is less than the emf. There is a charging current and a voltage drop across the resistor. In the discharging circuit the voltage across the capacitor starts at 8.87 V and decreases. After 10.0 mst it has decreased to / 4.49 V.Cv q C 26.79. IDENTIFY and SET UP: Zero current through the galvanometer means the current 1I through N is also the current through M and the current 2I through P is the same as the current through X. And it means that points b and c are at the same potential, so 12IN I P . EXECUTE: (a) The voltage between points a and d is E , so 1I NME and 2I PXE . Using these expressions in 12IN I P gives NP N M P XEE . ( ) ( )N P X P N M. NX PM and /X MP N . (b) ( 8 5 0 . 0 ) ( 3 3 . 4 8 ) 1 8 9 7 1 5 . 0 0 MPX N EVALUATE: The measurement of X does not require that we know the value of the emf. Neil Goldman