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For the circuit shown in the above figure, show that the charge on the capacitor in the steady state is [RC / (R+R 0 ) ] (E - E 0 )```
8 years ago

Ramesh V
70 Points
```										using kirchoffs laws, let I be current in circuit
here at steady state, surrent through capacitor is zero, so discard that loop with capacitor
and let current be I in the loop,
then E - I(R+Ro) - Eo = 0
so I = (E-Eo)/(R+Ro)   .... (1)
now for charging the capicitor, the potenntial difference across capacitor is same as pot. diff. across resistance R as in both loops we have a battery with e.m.f of E,
so, potential drop = I.R = R*(E-Eo)/(R+Ro)
as charge in capacitor is q=C.V = R.C.(E-Eo)/(R+Ro)
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8 years ago
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