Question

A voltmeter reads the potential difference across the terminals of an old battery as 1.40 V, while a potentiometer reads its voltage to be 1.55 V. The voltmeter resistance is $280\Omega$

• A. The emf of the battery is 1.4 V
• B. The emf of the battery is 1.55 V
• C. The internal resistance r of the battery is $30\Omega$.
• D. The internal resistance r of the battery is $5\Omega$

Answer 1

Answer: (B), (C)

The emf of the battery is 1.55 V
The internal resistance r of the battery is $30\Omega$.

The potentiometer measures the exact value of emf of a battery.

Therefore, $E=1.55V$,

Also, $1.4=I\left(280\right)$ or $I=0.005A$

Also, $V=E-Ir$ or $r=\frac{(E-V)}{I}=\frac{1.55-1.40}{0.005}=30\Omega$