The current in the given circuit is increasin | vedclass

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Question

Given $\frac{d I}{d t}=4 A / S$ when convesrt $I=2 A$

we $KVL$ acron loop$:-$

$E-I R-L \frac{d I}{d t}-\frac{9}{c}=0$ (sum of are voltangers acr

Vedclass · Accepted Answer

$6$

Vedclass · Answer

Given $\frac{d I}{d t}=4 A / S$ when convesrt $I=2 A$

we $KVL$ acron loop$:-$

$E-I R-L \frac{d I}{d t}-\frac{9}{c}=0$ (sum of are voltangers across a loop in zero)

put value,

$\Rightarrow 4-2 	imes 1-1 	imes 4-\frac{9}{3}=0 \Rightarrow 9=6 \mu C$

The current in the given circuit is increasing with $a$ rate $a = 4\, amp/s$. The charge on the capacitor at an instant when the current in the circuit is $2 \,amp$ will be :......$\mu\, C$

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