This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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 . F + 20 . F = 35 F R eff = 50 + 30 = 80 = 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 : V ( t ) = V (0) exp( t/ ) and I ( t ) = I (0) exp( t/ ) . We can find t since we know V (0) and V ( t ). exp( t/ ) = V ( t ) V (0) t = 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 F capacitor is charging through a 14 . 0 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??...
View Full Document
This note was uploaded on 02/09/2009 for the course PHYS 021 taught by Professor Hickman during the Fall '08 term at Lehigh University .
- Fall '08