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Exercise 10.8 · Q3

Q.The equation of electromotive force for an electric circuit containing resistance and self-inductance is E=Ri+LdidtE=Ri+L\dfrac{di}{dt}, where EE is the electromotive force given to the circuit, RR the resistance and LL, the coefficient of induction. Find the current ii at time tt when E=0E=0.

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✓ Free question

Setting E=0E=0 turns the circuit equation into a simple separable decay equation for the current.

Step 1. Set E=0E=0 in E=Ri+LdidtE=Ri+L\dfrac{di}{dt}. 0=Ri+Ldidt ⟹ Ldidt=−Ri0=Ri+L\dfrac{di}{dt}\ \Longrightarrow\ L\dfrac{di}{dt}=-Ri.

Step 2. Separate variables. dii=−RL dt\dfrac{di}{i}=-\dfrac{R}{L}\,dt.

Step 3. Integrate. ln⁡i=−RLt+C1 ⟹ i=Ce−Rt/L\ln i=-\dfrac{R}{L}t+C_1\ \Longrightarrow\ i=Ce^{-Rt/L}.

Step 4. Write CC as the current at t=0t=0. Denoting the initial current i0i_0, i(t)=i0e−Rt/Li(t)=i_0e^{-Rt/L}.

✓Final answer

i(t)=i0e−Rt/Li(t)=i_0e^{-Rt/L} — the current decays exponentially once the EMF is switched off.

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