Question

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

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

Answer 1

Correct Answer - B
`E=pil=(Vl)/L=(iR)/LxxIimpliesE=E/(R+R_(h)+r) xxR/L xxl`
`implies E=10/(5+4+1) xx5/5xx3=3V`