Question 409: Current Electricity

Question

Two identical cells each of emf $$1.5 \mathrm{~V}$$ are connected in series across a $$10 ~\Omega$$ resistance. An ideal voltmeter connected across $$10 ~\Omega$$ resistance reads $$1.5 \mathrm{~V}$$. The internal resistance of each cell is __________ $$\Omega$$.

Accepted Answer

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 $$1.5\,	ext{V} + 1.5\,	ext{V} = 3\,	ext{V}$$.
Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:
$$	ext{EMF}_{total} - I(R + 2r) = 0$$
We are given that the voltage across the $$10\,\Omega$$ resistor, as measured by the ideal voltmeter, is $$1.5\,	ext{V}$$. According to Ohm's law, the current in the circuit can be determined as:
$$I = \frac{V}{R} = \frac{1.5\,	ext{V}}{10\,\Omega} = 0.15\,	ext{A}$$
Now, substitute the given values into the KVL equation:
$$3\,	ext{V} - 0.15\,	ext{A}(10\,\Omega + 2r) = 0$$
Solve for $$2r$$:
$$3\,	ext{V} - 1.5\,	ext{V} = 0.15\,	ext{A} \cdot 2r$$
$$1.5\,	ext{V} = 0.3\,	ext{A} \cdot r$$
Now, solve for the internal resistance $$r$$:
$$r = \frac{1.5\,	ext{V}}{0.3\,	ext{A}} = 5\,\Omega$$
So, the internal resistance of each cell is $$5\,\Omega$$.

Current Electricity

Two identical cells each of emf $1.5 \mathrm{~V} are connected in series across a 10 ~\Omega resistance. An ideal voltmeter connected across 10 ~\Omega resistance reads 1.5 \mathrm{~V}. The internal resistance of each cell is __________ \Omega$.

Choose the correct answer: