A current of `3 A` flows through the `2 Omega` resistor as shown in the circuit. The power dissipated in the `5 Omega` resistor is

Options

A. `4 W`
B. `2 W`
C. `1 W`
D. `5 W`

Answer

D

Explanation

(d) Voltage across `2 Omega` is the same as voltage across arm containing `1 Omega` and `5 Omega` resistance. Voltage across `2 Omega` resistance,
`V_(1) = 6 V`
Current across `5 Omega, I = (6)/(1 + 5) = 1 A`
Thus, power across `5 Omega`,
`P = I^(2) R = (1)^(2) xx 5 = 5 W`