JEE Main 2023 — Current Electricity Question with Solution

Let the internal resistance of each cell be $$r$$.

Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $$2r$$. The total EMF of the series combination of cells is $$1.5\text{V}+1.5\text{V}=3\text{V}$$.

Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:

$${\text{EMF}}_{total}-I(R+2r)=0$$

We are given that the voltage across the $$10\Omega$$ resistor, as measured by the ideal voltmeter, is $$1.5\text{V}$$. According to Ohm's law, the current in the circuit can be determined as:

$$I=\frac{V}{R}=\frac{1.5\text{V}}{10\Omega }=0.15\text{A}$$

Now, substitute the given values into the KVL equation:

$$3\text{V}-0.15\text{A}(10\Omega +2r)=0$$

Solve for $$2r$$:

$$3\text{V}-1.5\text{V}=0.15\text{A}\cdot 2r$$

$$1.5\text{V}=0.3\text{A}\cdot r$$

Now, solve for the internal resistance $$r$$:

$$r=\frac{1.5\text{V}}{0.3\text{A}}=5\Omega$$

So, the internal resistance of each cell is $$5\Omega$$.