A part of a complete circuit is shown in the figure. At some instant, the value of current $I$ is $1$ A and it is decreasing at a rate of $10^2\ \text{A s}^{-1}$. The value of the potential difference $V_P - V_Q$ (in volts) at that instant is ______.
Numerical value type. Enter your answer.
Answer: 33
The current flows from Q towards P (as marked) and is decreasing.
**Resistor:** going from Q along the current, the potential drops by $IR = 1\times2 = 2$ V.
**Cell:** its positive (longer) plate faces the inductor, so crossing it from the resistor side the potential rises by $30$ V.
**Inductor:** the induced emf opposes the decrease, so it tries to keep the current flowing towards P. Crossing it along the current the potential rises by $L\left|\dfrac{dI}{dt}\right| = 50\times10^{-3}\times100 = 5$ V.
$$V_P = V_Q - 2 + 30 + 5 \Rightarrow V_P - V_Q = 33\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics