The response of a series R-C circuit is given by$I\left( s \right) = \frac{{\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}}}{{R\left( {s + \frac{1}{{RC}}} \right)}}$Where q0 is the initial charge on the capacitor. What is the final value of the current?

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{1}{R}\left( {\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}} \right)$

TuteeHUB · Answer

$\frac{{{e^{\frac{t}{{RC}}}}}}{R}\left( {\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}} \right)$

TuteeHUB · Answer

Infinity

TuteeHUB · Answer

Zero

1.	The response of a series R-C circuit is given by\(I\left( s \right) = \frac{{\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}}}{{R\left( {s + \frac{1}{{RC}}} \right)}}\)Where q0 is the initial charge on the capacitor. What is the final value of the current?
A.	\(\frac{1}{R}\left( {\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}} \right)\)
B.	\(\frac{{{e^{\frac{t}{{RC}}}}}}{R}\left( {\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}} \right)\)
C.	Infinity
D.	Zero
Answer» E.

The response of a series R-C circuit is given by\(I\left( s \right) = \frac{{\frac{{2V}}{\pi } - \frac{{{q_0}}}{C}}}{{R\left( {s + \frac{1}{{RC}}} \right)}}\)Where q0 is the initial charge on the capacitor. What is the final value of the current?

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