QJEEM23Ph61be: JEE Main 2023 Physics - Class 12 JEEM23Ph61be Solution: Two identical cells each of emf are connected in series across…

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 $5\,\Omega$ .