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

(b) `E = xl = (V)/(l) = (iR)/(L) xx l`
`implies E = (e)/((R + R_(h) + r)) xx (R )/(L) xx l`
`E = (10)/((5 + 4 + 1)) xx (5)/(5) xx 3 = 3 V`