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

Options

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

Answer

C

Explanation

(c ) 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) = 2 Omega`
Again, `R_(1)` is in series with `4 Omega` resistance, hence ltbgt `R = R_(1) + 4 = 2 + 4 = 6 Omega`
Thus, the total power dissipatesin the circuit
`P = (V^(2))/(R )`
Here, `V = 18 V, R = 6 Omega`
Thus, `P = ((18)^(2))/(6) = 54 W`