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Question

Figure shows a part of a bigger circuit.The capacity of the capacitor is 6 mF and is decreasing at the constant rate $$ 0.5 mF s^{-1}$$. The potential difference across the capacitor at the shown moment is changing as follows
$$\dfrac{dV}{dt}=2\, Vs^{-1},\dfrac{d^2V}{dt^2}=\dfrac{1}{2}Vs^{-2}$$
The current in the $$4 \Omega$$ resistor is decreasing at the rate of $$1 m A s^{-1}$$. What is the potential difference (in mV) across the inductor at this moment?

Solution
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$$q=CV$$

$$I=\dfrac{dq}{dt}$$

$$=>C\dfrac{dv}{dt}$$

$$=>V\dfrac{dC}{dt} $$

$$I_L=I_0-1$$
$$\dfrac{dI_L}{dt}=\dfrac{dI_0}{dt}-[c\dfrac{d^2V}{dt^2}+\dfrac{dV}{dt}\dfrac{dC}{dt}+\dfrac{dV}{dt}\dfrac{dC}{dt}+V\dfrac{d^2C}{dt^2}]$$
$$=-1\times 10^{-3}-[6\times 10^{-3}\times \dfrac{1}{2}+2(-0.5 \times 10^{-3})+2(-0.5\times 10^{-3})+0]$$
$$V_L=L\dfrac{dI_L}{dt}=2\times 2\times 10^{-3}$$

$$=4\times 10^{-3}V=4 V$$

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