Question

A flashlight cell of emf. $1.5volt$ gives $15$ ampere current when connected to an ammeter of resistance $0.04\Omega$. The internal resistance of the cell will be

• A. $0.04\Omega$
• B. $0.06\Omega$
• C. $0.10\Omega$
• D. $10\Omega$

Answer 1

The correct option is B $0.06\Omega$

The emf of cell is, $E=1.5V$
The current in circuit is, $I=15A$
The resistance of ammeter in circuit is, $R_A=0.04\Omega$
The equation for current in circuit is
$I=\frac{E}{r+R_A}$
where, $r$ is the internal resistance of cell.
$15=\frac{1.5}{r+0.04}$
$r+0.04=\frac{1.5}{15}=0.1$
$r=0.1-0.04=0.06\Omega$