Question

A cell of e.m.f. 1.8 V and internal resistance $2\Omega$ is connected in series with an ammeter of resistance $0.7\Omega$ and a resistor of $4.5\Omega$ as shown in Fig. What is the potential difference across the terminals of the cell?

• A. 1.3 V
• B. 1.8 V
• C. 1 V
• D. 2 V

Answer 1

The correct option is A 1.3 V

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.
In this case, the total resistance is given as $2\Omega +0.7\Omega +4.5\Omega =7.2\Omega$.
From the ohm's law the total current can be calculated from the formula $I=V/R$.
That is, $I=\frac{V}{R}=\frac{1.8V}{7.2A}=0.25A$.
Hence, the current flowing through the circuit is $0.25$ amperes.
The voltage output of a device is measured across its terminals and is called its terminal voltage $V$.

Terminal voltage is given by the equation $V=emf-Ir$ where, $r$ is the internal resistance and $I$ is the current flowing at the time of the measurement.
Therefore, $V=1.8-(0.25\times 2)=1.3V$.
Hence, the potential difference across the terminals of the cell is $1.3V$.