Question

5) Determine the power dissipated in 5 Ohm resistor for the circuit shown below.​

Answer 1

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=

(8+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=

R

V

=

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

×4=1×4=4W.

Hence, the power dissipated in the resistor R1 (4 ohms) = 4W.