Question

The given figure shows a part of a bigger circuit. The rate of change 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 $\frac{dV}{dt}=2V{s}^{-1}$, $\frac{d^{2}V}{d{t}^2}=\frac{1}{2}V{s}^{-2}$ and the current in the $4\Omega$ resistor is decreasing at the rate of $1mA{s}^{-1}$. What is the potential difference (in $mV$) across the inductor at this moment?

• A. $4$
• B. $3$
• C. $8$
• D. $6$

Answer 1

Answer: (A)

$4$

Charge transferred will be distributed between capacitor $Q_c=CV$

$i_c=\frac{d{Q}_c}{dt}=C\frac{dV}{dt}+V\frac{dC}{dt}$

$\frac{d{i}_c}{dt}=C\frac{d^{2}V}{dt}+2\frac{dV}{dt}\frac{dC}{dt}+V\frac{d^{2}C}{dt}=6\times \frac{1}{2}-2\times \frac{1}{2}-2\times \frac{1}{2}=1mA/s$

$\frac{d{i}_r}{dt}=1mA/s$

$\frac{d{i}_l}{dt}=\frac{d{i}_c}{dt}+\frac{d{i}_r}{dt}=2mA/s$

$V_L=L\frac{d{i}_l}{dt}=4mV$