hw15 - homework 15 – LEE, BENJAMIN – Due: Oct 9 2007,...

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Unformatted text preview: homework 15 – LEE, BENJAMIN – Due: Oct 9 2007, 4:00 am 1 Question 1, chap 28, sect 7. part 1 of 3 10 points Consider the circuit shown below, where the capacitor is initially uncharged. C R 1 R 2 V S b a After S is switched to position “ a ”, the initial current through R 1 is 1. I t =0 = V o R 2 2. I t =0 = R 1 R 2 R 1 + R 2 V o 3. I t =0 = V o R 1 + R 2 4. I t =0 = R 1 V o 5. I t =0 = V o R 1 correct 6. I t =0 = ( R 1 + R 2 ) V o 7. I t =0 = R 2 V o 8. I t =0 = R 1 + R 2 R 1 R 2 V o Explanation: When the switch is in position “ a ” at t = 0, there is no potential drop across the capacitor. Note: There is no current flowing through R 2 , so the entire potential drop, V o , is across the resistor R 1 . From Ohm’s law, I = V o R 1 . Question 2, chap 28, sect 7. part 2 of 3 10 points Leave the switch at position “ a ” for a long time, then move the switch from “ a ” to “ b ”. When the switch is in position “ b ”, what is the time constant, τ , of the circuit? 1. τ = R 1 C 2. τ = R 2 C 3. τ = ( R 1 + R 2 ) C 4. τ = R 1 + R 2 R 1 R 2 C 5. τ = ( R 1 + R 2 ) C correct 6. τ = R 1 R 2 ( R 1 + R 2 ) C 7. τ = R 1 + R 2 R 1 R 2 C 8. τ = R 1 R 2 R 1 + R 2 C 9. τ = R 2 C 10. τ = R 1 C Explanation: When the switch is in position “ b ”, R 1 and R 2 are now in series so the equivalent resis- tance is R = R 1 + R 2 . By definition, the time constant is τ = R C = ( R 1 + R 2 ) C . Question 3, chap 28, sect 7....
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This note was uploaded on 04/07/2008 for the course PHY 303K taught by Professor Turner during the Fall '07 term at A.T. Still University.

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hw15 - homework 15 – LEE, BENJAMIN – Due: Oct 9 2007,...

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