The current in a circuit with an external resistance of `3.75 Omega` is `2.5A`. When a resistance of `Omega` is introduced into the circuit, the current becomes `0.4 A`. The emf of the power source is

Options

A. `1V`
B. `2V`
C. `3V`
D. `4V`

Answer

B

Explanation

`0.5=E/(r+3.75)` ……….i
`0.4=E/(r+4.75)`………….ii
Solving these two equations, we get
`E=2V`