Question

# Switch S is closed in the circuit at time t=0. Find the current through the capacitor and the inductor at any time t.

Solution

## Circuit can be analyzed separately(i) Inductor $$\varepsilon=I_LR_3+L\dfrac{dI_L}{dt}$$on ontegrating , we get$$I_L=\dfrac{E}{R_3}(1-e^{\dfrac{R_2t}{L}})$$(ii) In loop ABCDEFA , $$\varepsilon IR_1+(I-I_1)R_2$$,$$\dfrac{\varepsilon+I_1R_2}{R_1+R_2}=I$$In loop ABGEFA , $$\varepsilon =IR_1+\dfrac{q}{C}$$$$\varepsilon=(\dfrac{(\varepsilon+I_2R_2)}{R_1+R_2})R_1+\dfrac{q}{c}$$Differentiating w.r.t time , we get $$0=0+\dfrac{dI_1}{dt}\times (\dfrac{R_1R_2}{R_1+R_2})+\dfrac{dq}{dt}\times \dfrac{1}{C}$$$$\dfrac{dI_1}{dt}=-(\dfrac{R_1+R_2}{R_1R_2})\times \dfrac{1}{C}\times I_1$$$$\dfrac{dI_1}{I_1}=(\dfrac{-R_1+R_2}{R_1R_2})\times\dfrac{1}{C}\times dt$$$$[ln\,I_1]^{I_1}_{\varepsilon / R_1}=(\dfrac{-R_1+R_2}{R_1R_2})\times \dfrac{[t]^t_0}{C}$$($$\therefore$$ t=0,C acts as a conducting wire)$$ln(\dfrac{I_1}{\varepsilon /R_1})=-(\dfrac{R_1+R_2}{R_1R_2})\times \dfrac{t}{C}$$$$I_{through\, capacitor}=\dfrac{\varepsilon}{R_1}e^{-(\dfrac{R_1+R_2}{R_1R_2})\dfrac{t}{C}}$$PhysicsNCERTStandard XII

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