Basic Electricity
23

The current I will flow through 7 ohms resistor also.The power absorbed by it will be

(62) (7) = 252 W (Ans).

Referring to Fig. 23-B and applying the currentdivision technique of equation (1.29).

12

1
2

I(R )
6
4

I
1.5 A

R
R
4
12

×

=
=
=

+
+

Current in 7 ohm resistor = current in 5 resistor= I2, as they are in series Power absorbed by 7 resistor = (1.52) 7

= 15.75 W (Ans).

Power absorbed by 5 resistor = (1.5)2 × 5

= 11.25 W (Ans).

Illustration 19

Three resistors of 2, 5 and 10 are joined in paralleland a total current of 24 A is passed through them.Find the current through each resistor.

Solution

The problem is dipicted in Fig. 1.24.

Fig. 1.23.

60 V

7 ?

3 ?

Fig. 1.23B. Circuit for illustration 18 with currentdirections marked, for evaluation of I2

The current I hence is, 60/10 = 6 A

The power delivered by the source is the product ofits V and I. That is 60 × 6 = 360 W (Ans).

Fig. 1.24. Circuit for illustration 19

2 ?

5 ?

10 ?

24 A

1 ?

60 V

60 V

7 ?

7 ?

5 ?

5 ?

7 ?

7 ?

6 ?

R1

12 ?

4 ?

I

I

I1

I1

I2

I2
III

R2

Fig. 1.23A.