In the circuit shown in figure find `V_(ab)` at `1 s`
Text Solution
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Charge on capacitor is increasing. So, there is a current in the circuit from right to left. This current is given by `i=(dq)/(dt)=2A` At `1s, q=2C` So, at `1 s`, circuit is as shown in figure `V_a+2/2(2)(4)+10=V_b` `V_a-V_b` or `V_(ab)=-19vol t.`
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