The total power dissipated in watts in the circuit shown here is

Options

A. 4W
B. 16 W
C. 40 W
D. 54 W

Answer

D

Explanation

The resistance of `6 Omega` and `3 Omega` are in parallel in the given circuit, their equivalent resistance is `1/(R_(1))=1/6+1/3=(1+2)/6=1/2`
or `R_(1)=2Omega`
Again `R_(1)` is in series with `4 Omega` resistance hence
`R=R_(1)+4=2+4 =6Omega`
Thus, the total dissipated in the circuit
`p=(V^(2))/R`
Here `V=18 V, R=6 Omega`
Thus, `p=((18)^(2))/6=54W`