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 __________ $\Omega$.
Answer (integer)
5
Solution
Let the internal resistance of each cell be $r$.
<br/><br/>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 $1.5\,\text{V} + 1.5\,\text{V} = 3\,\text{V}$.
<br/><br/>Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:
<br/><br/>$\text{EMF}_{total} - I(R + 2r) = 0$
<br/><br/>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:
<br/><br/>$I = \frac{V}{R} = \frac{1.5\,\text{V}}{10\,\Omega} = 0.15\,\text{A}$
<br/><br/>Now, substitute the given values into the KVL equation:
<br/><br/>$3\,\text{V} - 0.15\,\text{A}(10\,\Omega + 2r) = 0$
<br/><br/>Solve for $2r$:
<br/><br/>$3\,\text{V} - 1.5\,\text{V} = 0.15\,\text{A} \cdot 2r$
<br/><br/>$1.5\,\text{V} = 0.3\,\text{A} \cdot r$
<br/><br/>Now, solve for the internal resistance $r$:
<br/><br/>$r = \frac{1.5\,\text{V}}{0.3\,\text{A}} = 5\,\Omega$
<br/><br/>So, the internal resistance of each cell is $5\,\Omega$.
About this question
Subject: Physics · Chapter: Current Electricity · Topic: EMF and Internal Resistance
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