Unformatted text preview: Problem Solutions
18.2 (a) R eq = R1 + R 2 + R 3 = +4. Ω + 8. Ω + 12 Ω = 24 Ω 0 0 (b) The same current exists in all resistors in a series combination. I= ΔV 24 V 0 = = 1. A R eq 24 Ω (c) If the three resistors were connected in parallel,
−1 ⎛1 1 1⎞ 1 1⎞ ⎛1 + + =⎜ + + R eq = ⎜ 18 ⎟ ⎟ = 2. Ω 0 0 ⎝ 4. Ω 8. Ω 12 Ω ⎠ ⎝ R1 R 2 R 3 ⎠ −1 Resistors in parallel have the same potential difference across them, so I= 4 ΔV 24 V 24 V 24 V 0 = = 6. A , I = 0 0 = 3. A , and I 2 = = 2. A 8 1 R 4 4. Ω 0 8. Ω 0 12 Ω
7.00 W 9.00 W 18.5 (a) The equivalent resistance of the two parallel resistors is 4.00 W 1⎞ ⎛1 Rp = ⎜ 12 + ⎟ = 4. Ω 00 0 ⎝ 7. Ω 10. Ω ⎠
Thus, −1 10.0 W a b R ab = R 4 + R p + R 9 = ( 4. + 4. + 9. ) Ω = 17. Ω 00 12 00 1
(b) I = ab ( ΔV )ab 34. V 0
R ab = 17. Ω 1 99 99 = 1. A , so I = I = 1. A 4 9 Also, 99 12 18 ( ΔV ) p = IabR p = ( 1. A ) ( 4. Ω ) = 8. V
I= 7 Then, ( ΔV ) p
R7 = 8. V 18 = 1. A 17 7. Ω 00
8. V 18 818 = 0. A 10. Ω 0 and I= 10 ( ΔV ) p
R10 = 18.8 (a) The rules for combining resistors in series and parallel are used to reduce the circuit to an equivalent resistor in the stages shown below. The result is R eq = 5. Ω . 13
10.0 W 3.00 W 4.00 W 3.00 W 10 3 W 4.00 W 3.00 W Figure 2 5.00 W 3.00 W Figure 1
22 3 W 3.00 W 3.00 W 3.00 W Figure 3
2 66 31 W Req = 5.13 W Figure 4 Figure 5 (b) From P = ( ΔV ) R eq , the emf of the power source is ΔV = P⋅R eq = 00 13 53 ( 4. W )( 5. Ω ) = 4. V 18.13 The resistors in the circuit can be combined in the stages shown below to yield an equivalent resistance of R ad = ( 63 11) Ω .
6.0 W 3.0 W c 6.0 W 2.0 W 12 W e I d a 3.0 W I1 I2 3.0 W c b e 3.0 W 2.0 W d 3.0 W a 3.0 W b I1 I2 I12 4.0 W I Figure 1 6.0 W a I 3.0 W I1 I2 b 5.0 W d a I 3W b
30 11 18 V Figure 2 W 18 V d a I 63 11 W d 18 V 18 V Figure 3 Figure 4 18 V Figure 5 From Figure 5, I= ( ΔV )ad
R ad = 18 V 14 = 3. A ( 63 11) Ω Then, from Figure 4, 14 57 ( ΔV )bd = IR bd = ( 3. A ) ( 30 11 Ω ) = 8. V Now, look at Figure 2 and observe that I= 2
so ( ΔV )bd
3. Ω + 2. Ω 0 0 = 8. V 57 71 = 1. A 5. Ω 0 71 0 14 ( ΔV )be = I2R be = ( 1. A ) ( 3. Ω ) = 5. V Finally, from Figure 1, I= 12 ( ΔV )be 5. V 14
R12 = 12 Ω 43 = 0. A 18.14 The resistance of the parallel combination of the 3. Ω and 1. Ω resistors is 00 00 2.00 W 1⎞ ⎛1 Rp = ⎜ 750 + ⎟ = 0. Ω 00 00 ⎝ 3. Ω 1. Ω ⎠
The equivalent resistance of the circuit connected to the battery is −1 18.0 V 4.00 W 3.00 W 1.00 W 2.00 W R eq = 2. Ω + R p + 4. Ω = 6. Ω 00 00 75
and the current supplied by the battery is
18.0 V 0.750 W I= ΔV 18. V 0 = = 2. A 67 R eq 6. Ω 75 4.00 W The power dissipated in the 2. Ω resistor is 00 P2 = I2 R 2 = ( 2. A ) ( 2. Ω ) = 14. W 67 00 2
2 and that dissipated in the 4. Ω resistor is 00 P4 = I2 R 4 = ( 2. A ) ( 4. Ω ) = 28. W 67 00 4
2 00 00 The potential difference across the parallel combination of the 3. Ω and 1. Ω resistors is
67 750 00 ( ΔV ) p = IR p = ( 2. A ) ( 0. Ω ) = 2. V Thus, the power dissipation in these resistors is given by P3 = ( ΔV ) p ( 2. V )2 00
2 R3
2 = 3. Ω 00 = 1. W 33 and P1 = ( ΔV ) p ( 2. V )2 00
R1 = 1. Ω 00 = 4. W 00 18.30 The time constant is τ = R C . Considering units, we find ⎛ V ols ⎞ ⎛ C oul bs⎞ ⎛ C oul bs⎞ t om om R C → ( O hm s)( Far ) = ⎜ ads ⎟⎜ ⎟ ⎟=⎜ A m per ⎠ ⎝ V ols ⎠ ⎝ A m per ⎠ es t es ⎝ ⎛ ⎞ Coul bs om =⎜ ⎟ = Second om ⎝ Coul bs Second ⎠
or τ = R C has units of time. 18.31 (extra) (a) τ = R C = ( 2. × 106 Ω ) ( 6. × 10−6 F ) = 12 s 0 0 0 2 (b) Q m ax = C ε = ( 6. × 10−6 F) ( 20 V ) = 1. × 10− 4 C 18.32 (a) τ = R C = ( 100 Ω ) ( 20. × 10−6 F ) = 2. × 10−3 s = 2. m s 0 00 00 0 00 80 (b) Q m ax = C ε = ( 20. × 10−6 F ) ( 9. V ) = 1. × 10−4 C = 180 μ C
(c) 1⎞ ⎛ Q = Q m ax ( 1− e−tτ ) = Q m ax ( 1 − e−τ τ ) = Q m ax ⎜ 1− ⎟ = 114 μ C e⎠ ⎝ 18.33 Q m ax = C ε = ( 5. × 10−6 F ) ( 30 V ) = 1. × 10−4 C , and 0 5 τ = R C = ( 1. × 106 Ω ) ( 5. × 10−6 F ) = 5. s 0 0 0
Thus, at t= 10 s = 2τ Q = Q m ax ( 1− e−tτ ) = ( 1. × 10−4 C ) ( 1 − e−2 ) = 1. × 10−4 C 5 3 18.38 (a) The equivalent resistance of the parallel combination is
−1 ⎛1 1 1⎞ 1 1⎞ ⎛1 R eq = ⎜ + + + + ⎟ =⎜ ⎟ = 15 Ω ⎝ 150 Ω 25 Ω 50 Ω ⎠ ⎝ R1 R 2 R 3 ⎠ −1 so the total current supplied to the circuit is Iotal = t ΔV 120 V 0 = = 8. A R eq 15 Ω (b) Since the appliances are connected in parallel, the voltage across each one is ΔV = 120 V . (c) Iam p = l ΔV 120 V 80 = = 0. A R lam p 150 Ω (d) Pheater = ( ΔV ) 2 R heater = ( 120 V )
25 Ω 2 8 = 5. × 102 W 18.39 From P = ( ΔV ) R , the resistance of the element is
2 R= ( ΔV )
P 2 = ( 240 V )
3000 W 2 = 19. Ω 2 When the element is connected to a 120V source, we find that (a) I= ΔV 120 V 25 = = 6. A , and R 19. Ω 2 (b) P = ( ΔV ) I= ( 120 V )( 6. A ) = 750 W 25 18.40 The maximum power available from this line is Pm ax = ( ΔV ) I ax = ( 120 V ) ( 15 A ) = 1800 W m
Thus, the combined power requirements (2 400 W) exceeds the available power, and you cannotoper e t t o applances t at he w i oget er . h 18.45 The resistive network between a an b reduces, in the stages shown below, to an 5 equivalent resistance of R eq = 7. Ω .
2.4 W a 1.8 W b 3.6 W 3.5 W b 5.1 W a 1.8 W 3.6 W 8.6 W b 2.4 W a 1.5 W 3.6 W 2.4 W a 7.5 W b ...
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 Fall '08
 Abdulrazzaq
 Physics, Current, Resistor, Harshad number, Ω, Series and parallel circuits, τ

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