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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.
1748675_f48e9d5427df4e17b57ade6489c1c9ae.png


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)
$$ln(\dfrac{I_1}{\varepsilon /R_1})=-(\dfrac{R_1+R_2}{R_1R_2})\times \dfrac{t}{C}$$
$$I_{through\, capacitor}=\dfrac{\varepsilon}{R_1}e^{-(\dfrac{R_1+R_2}{R_1R_2})\dfrac{t}{C}}$$

1623131_1748675_ans_25483dc96afd4dc69503f3d091447f91.png

Physics
NCERT
Standard XII

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