Three resistors `1Omega, 2 Omega` and `3 Omega` are connected to form a triangle. Across `3Omega` a 3 V battery is connected. The current through `3Omega` resistor is

Options

A. 0.75 A
B. 1A
C. 2A
D. 1.5 A

Answer

B

Explanation

The resistance `1Omega` and `2Omega` are in series of arm ABC, are in parallel to resistance `3Omega` in arm AC. Their effective resistance is
`((1+2)xx3)/((1+2)+3)=3/2 Omega`
Current in the circuit
`I=3/((3//2)Omega)=2A`
Current in arm `AC, I_(1)=(2(Axx3Omega))/(3Omega+3)=1A`