In the circuit shown in figure E1=2E2=20V,R1=R2=10kΩ and C=1μF. The sum of current through R1,R2 and C when S has been kept connected to A for a long time.
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Solution
When S is closed for long time, the capacitor C is fully charged and no current will pass through it. Thus, IC=0
The current drawn from battery E1 is I=E1R1+R2=20(10+10)103=1mA
As the resistances are in series so the current through them will be equal to I