during charging of capacitor find time when potential drop across capacitor is equal to potential drop across resistance
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In the circuit shown, calculate the potential drop across the capacitor
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When capacitor is fully charged, no currents draws from the source. So we can remove the capacitor and the circuit becomes open as shown in figure.
Applying KVL , 18−9=4I+2I or I=9/6=3/2A
here, V
A
−V
B
=9+2I=9+2(3/2)=12V
The potential across capacitance 12−9=3V
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