# hw9 - Physics 21 Fall 2004 Solution to HW-9 26 P 80...

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Physics 21 Fall, 2004 Solution to HW-9 26 P44 Two 6 . 0 μ F capacitors, two 2 . 2 kΩ resistors, and a 12.0 V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1 . 50 mA? 2200 2200 6.0 µ F 6.0 µ F 12 V Replace the C ’s by equivalent value 1 C ef = 1 6 . 0 μ F + 1 6 . 0 μ F C ef =3 . 0 μ F , and the R ’s by equivalent value R ef =2 . 2kΩ+2 . 2 kΩ = 4400 Ω . The time constant is R ef C ef =(3 . 0 × 10 - 6 F)(4400 Ω) = 13 . 2ms At the instant the current starts, the capacitors are un- charged, so the voltage drop is entirely across the resistors. Hence the initial value of the current is I 0 = 12 V 4400 Ω . 727 mA . We must solve exp( t/ 13 . 2ms)= 1 . 50 mA 2 . 727 mA =0 . 55 . The result is t =13 . × ln µ 1 . 55 =7 . 9ms 26 P80 Determine the current in each resistor of the cir- cuit shown in Fig. 26-57. 1 2 3 I 4 I 5 I 6 I 12 node Each current is labelled by the resistor it ﬂows through. node: I 5 + I 4 = I 6 + I 12 loop 1: 5 + 10 5 I 5
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