`A_(3)` volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current (1) in circuit will be

Options

A. 1/3 A
B. 1A
C. 1.5A
D. 2A

Answer

C

Explanation

Given : Voltage of battery (V) = 2 V
and resistance of each resistors
` (R_(1)) = R_(2) = R_(3) = 3 Omega`
We know that in the given circuit , `R_(2) and R_(3)` are connected in series
`:.` Their equivalent resistance
`(R') = R_(2) + R_(3) = 3 + 3 = 6 Omega`
Now R is in parallel combination with `R_(1)`
`:.` Their equivalent resistance
`(R) = (R_(1)xxR')/(R_(1)R') = (3 xx 6)/( 3 + 6 ) = (18)/(9) `
and current in the circuit
`(I) = (V)/(R) = (3)/(2) = 1.5 ` A