JEE Physics 2023 Question with Solution

Given: A resistor network is connected to a $9 \, \text{V}$ battery. The resistances are $2 \, \Omega$, $4 \, \Omega$, $4 \, \Omega$, and $2 \, \Omega$. Find: The magnitude of potential difference between points A and B.

The two paths between the battery terminals act as two series branches in parallel.

Left branch resistance:

$$R_L = 2 + 4 = 6 \, \Omega$$

Right branch resistance:

$$R_R = 4 + 2 = 6 \, \Omega$$

Since these two equal branches are in parallel, the equivalent resistance is

$$R_{eq} = \frac{6 \times 6}{6 + 6} = 3 \, \Omega$$

Hence the total current supplied by the battery is

$$I = \frac{9}{3} = 3 \, \text{A}$$

Because both parallel branches have equal resistance, the current divides equally:

$$I_L = I_R = \frac{3}{2} = 1.5 \, \text{A}$$

Potential drop from the battery terminal to point A through the $2 \, \Omega$ resistor is

$$V_A = 1.5 \times 2 = 3 \, \text{V}$$

Potential drop from the battery terminal to point B through the $4 \, \Omega$ resistor is

$$V_B = 1.5 \times 4 = 6 \, \text{V}$$

Therefore,

$$V_{AB} = |V_B - V_A| = |6 - 3| = 3 \, \text{V}$$

So, the required magnitude of voltage difference is $3 \, \text{V}$.

Given: The two opposite series branches are $2 + 4$ and $4 + 2$. Find: The voltage difference between A and B.

Both branches have the same total resistance:

$$2 + 4 = 4 + 2 = 6 \, \Omega$$

So the branch currents are equal. With equal current in each branch, the potential drops are proportional to the resistances encountered from the same battery side.

At A, the drop is across $2 \, \Omega$:

$$V_A = 1.5 \times 2 = 3 \, \text{V}$$

At B, the drop is across $4 \, \Omega$:

$$V_B = 1.5 \times 4 = 6 \, \text{V}$$

Thus,

$$|V_B - V_A| = 3 \, \text{V}$$

Therefore, the answer is $3$.