Question

The figure shows part of a bigger circuit. The capacity of the capacitor is 6µF and decreasing at the constant rate 0.5 µF s⁻¹. The potential difference across the capacitor is changing as under

$\frac{dV}{dt} = 2Vs^{-1}, \frac{d^2V}{dt^2} = -\frac{1}{2}Vs^{-2}$.

The current in the 4Ω resistor is decreasing at the rate of 1 µA s⁻¹. What is the potential difference (in micro volts) across the inductor at this moment ?

Answer 1

-2 µV

Answer 2

In the series branch consisting of an inductor (L) and a resistor (R), by Kirchhoff’s Voltage Law the voltage is

$$V = iR + L\frac{di}{dt}.$$

Here the current through the resistor (and hence the branch) is given to be decreasing at a rate

$$\frac{di}{dt} = -1\times10^{-6}\, \text{A/s}.$$

Thus the voltage drop across the inductor is

$$V_L = L\frac{di}{dt} = 2\,\text{H} \times \left(-1\times10^{-6}\,\text{A/s}\right) = -2\times10^{-6}\, \text{V}.$$

Converting to microvolts,

$$-2\times10^{-6}\,\text{V} = -2\,\mu\text{V}.$$