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hw9 - Physics 21 Fall 2008 Solution to HW-9 26-41 In the...

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Physics 21 Fall, 2008 Solution to HW-9 26-41 In the circuit shown in the figure both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 15.0 V? (b) What will be the current at that time? First find the effective capacitance and resistance of the circuit, then use those to determine the time constant. C eff = 15 . 0 µ F + 20 . 0 µ F = 35 µ F R eff = 50 Ω + 30 Ω = 80 Ω τ = R eff C eff = 2 . 8 ms (a) The voltage across the capacitors and the current through the resistors both decay exponentially with time constant τ : V ( t ) = V (0) exp( t/τ ) and I ( t ) = I (0) exp( t/τ ) . We can find t since we know V (0) and V ( t ). exp( t/τ ) = V ( t ) V (0) t = τ ln V (0) V ( t ) = 3 . 1 ms (b) The current goes down by the same factor as the voltage, and the initial value of the current is V (0) /R eff : I ( t ) = V ( t ) V (0) V (0) R eff = V ( t ) R eff = 0 . 19 A . 26-44 A 1.70 µ F capacitor is charging through a 14 . 0 Ω resistor using a 12.0 V battery. (a) What will be the current when the capacitor has acquired 1/4 of its maximum charge Q max ? (b) Will it be 1/4 of the maximum current?
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