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 would be the reading of the ammeter?

• A. 0.25 A
• B. 0.50 A
• C. 0.75 A
• D. 0.70 A

Answer 1

The correct option is A 0.25 A

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 ammeter shows a reading of 0.25 amperes.