In the given circuit, `R_(1)` = `10Omega` `R_(2) = 6Omega` and E = 10V, then correct statement is

Options

A. Effective resistance of the circuit is `10Omega`
B. Reading of `A_(1)` is 1/2 amp.
C. Reading of `A_(2)` is 1/2 amp
D. Reading of `A_(3)` is 1/8 amp.

Answer

B

Explanation

Potential difference across `R_(2)` is zero, therefore curent in three branches is zero, therefore current in two branch containing `R_(1)` will be same, and the circuit will be simplified as shown.
Effective resistance of the circuit `R|_(ef)` =`(40xx40)/(40+40)`=20Omega`
Current through the circuit I = (10)/(20)`= `(1//2)` amp
Hence eading of `A_(1) = 1//2 amp`. Hence reading of `A_(2) = 1//4 amp`.