Question

The current in a circuit with an external resistance of $3.75\Omega$ is $0.5A$. When a resistance of $1\Omega$ is introduced into the circuit, the current becomes $0.4A$. The emf of the power source is

• A. $1V$
• B. $2V$
• C. $3V$
• D. $4V$

Answer 1

Answer: (B)

$2V$

$\epsilon =I(R+r)$

$r=\frac{\epsilon }{I}-R$

At first event, $I=0.5A,R=3.45\Omega$

$r=\frac{\epsilon }{0.5}-3.45......\left(1\right)$

At second event, $I=0.4A,R=1+3.45=4.45\Omega$

$r=\frac{\epsilon }{0.4}-4.45........\left(2\right)$

From (1) and (2)

$\frac{\epsilon }{0.5}-3.45=\frac{\epsilon }{0.4}-4.45$

$\epsilon =2V$

Hence, emf is $2V$