Resistance P, Q, R, S in the four sides of Whearstone bridge have respective values `10 Omega, 30 Omega, 20 Omega and 60 Omega`. A cell connected across one diagonal has emf 5 V and internal resistance 2 `Omega`. If resistance of galvanometer is `60 Omega` then current passing the cell is .....

Options

A. 0.2 A
B. 0.15 A
C. 0.17 A
D. 2A

Answer

C

Explanation

0.17 A
Resistance of Wheatstone bridge in the balanced condition,
`R = ((10 + 30)(20 + 60))/((10 + 30) + (20 + 60)) = (40 xx 80)/(120) = (80)/(3) Omega`
`therefore ` Current through the battery.
`I = (5)/(R + 2) = (5)/((80)/(3)+2) `
`therefore I = (5 xx 3)/(86) = 0.17446 A `
`therefore I = 0.17 A `