Question

The figure shows part of a bigger circuit. At the moment, the capacity of the capacitor is $6\mu F$ and decreasing at the constant rate $0.5\mu F{s}^{-1}$. The potential difference across the capacitor at this moment is changing as under [direction of current is shown in figure]
$\frac{dV}{dt}=2V{s}^{-1},\frac{d^{2}V}{d{t}^2}=\frac{1}{2}V{s}^{-2}$ ___

The current in the $4\Omega$ resistor is decreasing at the rate of $1\mu A{s}^{-1}$. What is the potential difference (in micro volts) across the inductor at this moment?

Answer 1

$V_L=\left|(\frac{d{i}_R}{dt}-\frac{d^{2}q}{d{t}^2})\right|=L|\frac{d}{dt}{i}_R-(C\frac{d^{2}V}{d{t}^2}+\frac{2dC}{dt}\frac{dV}{dt})|$
= 8 volts