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

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Solution

Circuit can be analyzed separately
(i) Inductor
$ε = I_{L} R_{3} + L \frac{d I_{L}}{d t}$
on ontegrating , we get
$I_{L} = \frac{E}{R_{3}} (1 - e^{\frac{R_{2} t}{L}})$
(ii) In loop ABCDEFA , $ε I R_{1} + (I - I_{1}) R_{2}$ ,
$\frac{ε + I_{1} R_{2}}{R_{1} + R_{2}} = I$
In loop ABGEFA , $ε = I R_{1} + \frac{q}{C}$
$ε = (\frac{(ε + I_{2} R_{2})}{R_{1} + R_{2}}) R_{1} + \frac{q}{c}$
Differentiating w.r.t time , we get
$0 = 0 + \frac{d I_{1}}{d t} \times (\frac{R_{1} R_{2}}{R_{1} + R_{2}}) + \frac{d q}{d t} \times \frac{1}{C}$
$\frac{d I_{1}}{d t} = - (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) \times \frac{1}{C} \times I_{1}$
$\frac{d I_{1}}{I_{1}} = (\frac{- R_{1} + R_{2}}{R_{1} R_{2}}) \times \frac{1}{C} \times d t$
$[l n I_{1}]_{ε / R_{1}}^{I_{1}} = (\frac{- R_{1} + R_{2}}{R_{1} R_{2}}) \times \frac{[t]_{0}^{t}}{C}$

( $∴$ t=0,C acts as a conducting wire)
$l n (\frac{I_{1}}{ε / R_{1}}) = - (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) \times \frac{t}{C}$
$I_{t h r o u g h c a p a c i t o r} = \frac{ε}{R_{1}} e^{- (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) \frac{t}{C}}$

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