# Question

Discussion in 'Homework Help' started by ecetstudent, Feb 23, 2009.

1. ### ecetstudent Thread Starter New Member

Feb 23, 2009
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When the switch is thrown to position 2 @ time t=0 in the attached circuit, is the current through R1 4.0A?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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No the current is 12/11 Amps. At the instant the switch closes consider the capacitor to be a short circuit and the inductor an open circuit.

3. ### mik3 Senior Member

Feb 4, 2008
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The current will be (12-2)/11=0.91 Amps

If you assume that with the switch at position 1 the circuit was in a steady state then the voltage across the capacitor will be 2V. At the instant the switch moves to position 2 the capacitor will act a short circuit and the inductor as an open circuit. However, the voltage across the capacitor is 2V and it will subtract from the 12V battery voltage due to its polarity.

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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I agree with mik3. I disagree with t n k.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yes t_n_k is wrong!