The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are `10Omega, 30Omega, 30Omega` and `90Omega`, respectively. The emfs and internal resistances of the cell are 7V and `5Omega` respectively. If the galvanometer resistance is `50Omega`, the current drawn from the cell will be

Options

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

Answer

B

Explanation

Effective resistance, `R_("eff")=(40xx120)/(120+40)=(4800)/(160)=30 Omega`
`therefore` Current `l=(7)/((30+5))=(7)/(35)=0.2 A " " [because l=(E )/(R+r)]`