JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: $\frac{dI}{dt} = -1 \, \text{A/s}$, $I = 2 \, \text{A}$, $R = 12 \, \Omega$, $L = 6 \, \text{H}$, and source emf $= 12 \, \text{V}$.

Find: The value of $V_{AB} = V_A - V_B$.

Applying Kirchhoff's Voltage Law:

$$V_A - IR - L\frac{dI}{dt} - 12 = V_B$$

Substituting the given values:

$$V_A - 2 \times 12 - 6(-1) - 12 = V_B$$

Rearranging,

$$V_A - V_B = 24 - 6 + 12 = 30 \, \text{V}$$

Therefore, the value of $V_{AB}$ at this instant is $30 \, \text{V}$.

Common mistakes

• Using the sign of the inductor term incorrectly. Since $\frac{dI}{dt}$ is negative, the term $L\frac{dI}{dt}$ is also negative, and its sign in the KVL equation must be handled carefully. Write the loop equation first, then substitute values.

• Ignoring the polarity while evaluating $V_{AB}$. The question asks for $V_A - V_B$, not the magnitude alone. Always define $V_{AB}$ before solving.

• Forgetting to include the battery emf in the loop equation. The resistor drop, inductor contribution, and source emf all affect the terminal voltage. Do not use only $IR$ or only $L\frac{dI}{dt}$.