`AB` and `CD` are two uniform resistance wires of lengths `100 cm` and `80 cm` respectively. The connections are shown in the figure. The cell of emf `5 V` is ideal while the other cell of `E` has internal resistance `2 Omega`. A length of `20 cm` of wire `CD` is balanced by `40 cm` of wire `AB`. Find the emf `E` in volt, if the reading of the ideal ammeter is `2 A`. The other connecting wires have negligible resistance

Options

A. 10
B. 12
C. 8
D. zero

Answer

B

Explanation

(b) Potential difference across wire `AB = 5 V`
p.d. across `40 cm` of this wire `= (5)/(100) xx 40 = 2` volts
So potential difference across `20 cm` of wire `CD = 2 V`.
`:.` p.d. across wire `CD = (2)/(20) xx 80 = 8` volt
p.d. across `2 Omega` resistor ` = 2 xx 2 = 4` volt
`:. "Emf of the cell" = 8 + 4 = 12` volt.