Introduction to Electric Circuits

2.3 Circuit elements 21

Figure 2.13

20A

O V

1

I2

R1 = lOf~ R2 = 40t2

Example 2.10

The circuit of Fig. 2.14 is a series-parallel circuit. Calculate (1) the current

drawn from the supply (I); (2) the potential difference across the resistor R 4

(V4); (3) the current through the resistor R 6 (/6)"

(-

I R1 = 10t2

IOOV

R2 = 5f~

I 1

R3 = 20f~

l

R4 = 25t2

I!

R5 = 10~

Figure 2.14

Solution

The equivalent resistance of the parallel combination of resistors R5 and R6 is
given by

R56 = RsR6/(R 5 + R6) = 10 • 30/(10 + 30) = 300/40 = 7.5 12

For the parallel combination of the resistors R2, R3 and R4 the equivalent
resistance is given by

1/R234 = 1/R2 + 1/R3 + 1/R4 = 0.2 + 0.05 + 0.04 = 0.29 S

Therefore R234 = 1/0.29 - 3.45 ~.

The equivalent resistance of the whole series-parallel circuit is given by
Req -- R 1 + R234 -F R56 so Req - 10 + 3.45 + 7.5 = 20.95 II

I = V/Req = 100/20.95 = 4.77 A

V 4 : IR234 : 4.77 • 3.45 = 16.46 V

I 6 : R5I/(R 5 Jr" R6): 10 X 4.77/40 = 1.19 A