The resistance of the four arms `P, Q, R` and `S` in a Wheatstone's bridge are `10 ohm 30 ohm` and `90 ohm` rerspectively. The e.m.f. and internal resistance of the cell are `7 volt` and `5 ohm` respectively. If the galvanometer resistance is `50 ohm`, the current drawn for the cell will be

Options

A. 2.0 A
B. 1.0 A
C. 0.2 A
D. 0.1 A

Answer

C

Explanation

Balanced bridge

`R_(eq) = (40 xx 120)/(40+ 120)`
`= (40 xx 120)/160 = 30 Omega`
Current `I = 7/(30 + 5) = 1/5 = 0.2 A`