Question

In the electric circuit shown in the figure, power dissipated in 4 $\Omega$ resistor is :

• A. 4 W.
• B. 2 W
• C. 1 W
• D. 6 W

Answer 1

The correct option is A 4 W.

The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.
$Rp=\frac{(8\times 8)}{(8+8)}=4ohms$
This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then
Rt =R1+Rp=4+4 = 8 ohms.
Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,
$I=\frac{V}{R}=\frac{8}{8}=1A$.
Hence, the current flowing through the resistor R1 (4 ohms) is 1A.
Now the power dissipated in the resistor R1 (4 ohms) is given as follows.
$P=I^{2}R=1^2\times 4=1\times 4=4W$.
Hence, the power dissipated in the resistor R1 (4 ohms) = 4W.