Question

A battery of e.m.f. 3.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. The internal resistance of the battery is

• A. $4\Omega$.
• B. $1.2\Omega$.
• C. $0.2\Omega$.
• D. $2\Omega$.

Answer 1

The correct option is C $0.2\Omega$.

The total resistance of a circuit connected to the given cell of emf 1.8 V is the sum of all the resistances from various sources like, resistors, ammeter, voltmeter, etc., connected in series along with the internal resistance of the cell.
It is given that the current is 1.5 A, the voltmeter across the high resistance reads 2.7 V.
Therefore, the resistance of that high resistance is calculated from the formula R=V/I.
That is, $R=\frac{V}{I}=\frac{2.7V}{1.5A}=1.8\Omega$.
Given that the emf is 3.0 V and the current is 1.5 A as it remain the same throughout the serially connected circuit. The overall resistance of the circuit is given as $R_{total}=\frac{V}{I}=\frac{3.0V}{1.5A}=2\Omega$.
The value of the internal resistance can be obtained by subtracting the resistance of the external resistance from the overall resistance.
That is,internal resistance r= total resistance R[tot]-resistance of the resistor R.
$r=2\Omega -1.8\Omega =0.2\Omega$.
Hence, the value of the internal resistance is 0.2 ohms.