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. 0.1 A
B. 2.0 A
C. 1. 0 A
D. 0.2 A

Answer

D

Explanation

The situation is described through a circuit diagram as shown below.

For a balanced Wheatstone's bridge,
`(P)/(Q) = (R )/(S)`
`therefore (10 Omega)/(30 Omega) = (30 Omega)/(90 Omega) or (1)/(3) = (1)/(3)`
The equivalent resistance of the circuit is
`R_(eq) = 5Omeaga `
`+((10Omega+30Omega)(30Omega+90Omega))/((10Omega+30Omega)+(30Omega+90Omega))`
` = 5 Omega + ((40 Omega)(120 Omega))/(40 Omega + 120 Omega))`
` = 50 Omega + 30 Omega = 35 Omega`
Current drawn from the cell is
`l = (7V)/(35 Omega) = (1)/(5) A = 0.2 A `