Question

Determine the potential (VC) and current (iC) in the capacitor in the figure ten seconds after starting to discharge. The initial potential of the capacitor is V0 = 50 V.

72 C-IF VO 50 2139

Answer 1

The 13 ohm resistor is in series with 7 ohm resistor, their equivalent resistance is R1 = 20 ohm

This R1 is in parallel to 5 ohm resistor, their equivalent resistance is .

1 R=G+ )"+(6+1) - 4ohm

Therefore, equivalent resistance across the capacitor is R = 4 ohm.

Now, when a charged capacitor is connected across resistor,

due to the decay of charge from the capacitor, charge on the capacitor at any time instant is given by $q=Qe^{-\frac{t}{RC}}$ -----(i) [where Q = charge given to capacitor].

We have C = 1F, R = 4 ohm, Q = .

putting the values in (i), we get

$q=Qe^{-\frac{t}{RC}}$

$\Rightarrow q=50*e^{-\frac{t}{4*1}}$

$\Rightarrow q=50*e^{-\frac{t}{4}}$-------(ii)

Now, after time t = 10 s,

potential difference across capacitor is $V_{C}= \frac{q}{C}=\frac{50*e^{-\frac{10}{4}}}{1}=4.10V$ [answer]

The current in the branch consisting capacitor is $i_{C}=|\frac{dq}{dt}|=50*\frac{1}{4}*e^{-\frac{t}{4}}$

therefore, after t = 10 s, the value of $i_{C}=50*\frac{1}{4}*e^{-\frac{10}{4}}=1.03 A$ [answer]