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Suppose that the switch has been closed for a length of time sufficiently long enough for the capacitor to become fully charged. (R = 11.0 k, R2 = 16.0 k, R3 = 5.00 k, V = 9.40 V) http://www.webassign.net/sercp/p18-50alt.gif
Find the steady-state current for each resistor.

2007-02-18 07:40:18 · 2 answers · asked by Shelia 1 in Science & Mathematics Physics

2 answers

After the capacitor is fully charged, no more current will flow through it. So the steady state current for R3 is zero.

This means that all of the current (in the steady state) will be flowing through R and R2. The current through R will be the same as the current in R2.

Current in R = Current in R2 = (9.4 V) / (11.0 k + 16.0 k) = 0.348 A

2007-02-18 16:10:37 · answer #1 · answered by Bill C 4 · 0 0

We anticipate that the inductor is right, i.e. has no resistance. So the secure state present day is 600 milliamperes. The time consistent is LR, that's one millisecond; the 0.5 value is this circumstances the organic log of two.

2016-11-23 17:05:48 · answer #2 · answered by ? 4 · 0 0

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