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?

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 $$\Rightarrow L\dfrac{di}{dt}=\varepsilon-iR$$$$\Rightarrow \int^i_{i_0}\dfrac{di}{\varepsilon-iR}=\int^t_0\dfrac{dt}{L}$$$$\Rightarrow ln(\dfrac{\varepsilon-iR}{\varepsilon-i_0R})=-\dfrac{Rt}{L}$$$$\Rightarrow \varepsilon-iR=(\varepsilon-i_0R)e^{-Rt/L}$$$$\Rightarrow i=\dfrac{\varepsilon-(\varepsilon-i_0R)e^{-Rt/L}}{R}$$PhysicsNCERTStandard XII

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