In the Wheatstone's network given,`P = 10 Omega, Q = 20 Omega, R = 15 Omega, S = 30 Omega`.The current passing through the battery (of negligible internal resistance) is

Options

A. `0.36` A
B. `0` A
C. `0.18` A
D. `0.72` A

Answer

A

Explanation

Wheatstone's network is balance as per resistances of P,Q, R and S given.No current flow through galvanometer.P and R are in series in upper branch.
Resistance = (P+ R) = 10 + 15 = 25 `Omega`
Q and S are in series in lower branch.
`therefore` Resistance = Q + S = 20 + 30 = 50 `Omega`
The upper and the lower branches are in parallel.
`therefore` Resistance = `(25 xx 50)/(25 + 50) = (25 xx 50)/(75 = (50)/(3) Omega`
`therefore` `Current I = V/R = (6)/(50)/(3) = 0.36 A`