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

Answer

B

Explanation

`because (P)/(Q)=(P)/(S)rArr`Balanced bridge
Equivalent resistance of bridge
`=(40xx120)/(40+120)=30Omega`
`therefore I_("battery")=(7)/(30+5)=0.2 A`