A resistance of `4 Omega` and a wire of length 5 meters and resistance `5 Omega` are joined in series and connected to a cell of e.m.f. `10 V` and internal resistance `1 Omega`. A parallel combination of two identical cells is balanced across `300 cm` of wire. The e.m.f. `E` of each cell is

Options

A. 1.5 V
B. 3.0 V
C. 0.67 V
D. 1.33 V

Answer

B

Explanation

`E=pil=(Vl)/(L)=(iR)/(L)xxl`
`impliesE=(E)/(R+R_(h)+r)xx(R)/(L)xxl`
`impliesR=(10)/(5+4+1)xx(5)/(5)xx3=3V`