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Question

Figure shows a circuit consisting of an ideal cell,an inductor ' L' and a resistor 'R', connected in series. Let switch 'S' be closed at t=0.Suppose at t=0 the current in the inductor is $$i_0$$, then find out the equation of current as a function of time?
1748204_db07cb063a67420ba9411ab1ccd31776.png


Solution

Let at an instant t, the current in the circuit is i which is increasing at the rate of time , i.e. di/dt.
According to  KVL rule  along the circuit , we have 
$$\Rightarrow L\dfrac{di}{dt}=\varepsilon-iR$$

$$ \Rightarrow \int^i_{i_0}\dfrac{di}{\varepsilon-iR}=\int^t_0\dfrac{dt}{L}$$

$$\Rightarrow ln(\dfrac{\varepsilon-iR}{\varepsilon-i_0R})=-\dfrac{Rt}{L}$$

$$\Rightarrow \varepsilon-iR=(\varepsilon-i_0R)e^{-Rt/L}$$

$$\Rightarrow i=\dfrac{\varepsilon-(\varepsilon-i_0R)e^{-Rt/L}}{R}$$

1619380_1748204_ans_02cb233737e842f285635f707677f16a.png

Physics
NCERT
Standard XII

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