The power dissipated in `3 Omega` resistance in the following circuit is

Options

A. 1 W
B. 0.5 W
C. 0.25 W
D. 0.75 W

Answer

D

Explanation

The given circuit can be redrawn as shown in figure:

Equivalent resistance of middle branch in the circuit,
`R_(1)=2+(3 xx 6)/(3+6)=4 Omega`
Now `R_(1)=4 Omega`, and `4 Omega` are in parallel, so equivalent resistance of the circuit is given by
`R=1+(4 xx 4)/(4+4)=3 Omega`
Current drawn from battery, `I=(V)/(R)=(4.5)/(3)=(3)/(2) A`
`I_(1)=I-I_(1)` (both branches have same resistance)
`:. I_(1)=(I)/(2)=(3)/(4)A`
Current through `3 Omega` resistance
`I_(2)=(6)/(3+6)I_(1)=(6)/(9) xx (3)/(4)=(1)/(2) A`
Power dissipated in `3 Omega` resistance, `P=I_(2)^(2)R`
`=((1)/(2))^(2) xx 4=(3)/(4) W=0.75 W`