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Grade: 12


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 )

11 years ago

Answers : (1)

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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