Introduction to Electric Circuits

3. 11 Self-assessment test 61

R23 = R2 + R3 + RzR3/R~ - 5 + 10 + 50/10 = 20 1~

R31 = R 3 + R 1 + R3R1/R 2 = 10 + l0 + 100/5 - 40 D,

This leads to the circuit of Fig. 3.38 and then to Fig. 3.39 by replacing the two
parallel connected 40 ~ resistors by their 20 11 equivalent resistor.

40~

! I

1

L

)+ov

40f~ I 20f~

l I o A ~ I

20~ ~--~20~ l'~20f~

0 0

O

80v

+l

Figure 3.38 Figure 3.39

11

l J 20o

The equivalent resistance between the terminals A and B in Fig. 3.39 is given

by

RAB = (20 X 40)/60 = (80//6) 11. It follows that

I = V/RAB = 80 • 6/80 = 6A.

By current division

I 1 = (20/60)I "--(20/60) • 6 = 2 A

The current through each 40 f~ resistor in Fig. 3.38 is therefore 1A and since the

top one of these is the resistor R in Fig. 3.36, the required answer is 1A.

3.11 SELF-ASSESSMENT TEST

1 State Kirchhoff's first (current) law (KCL).

2 Define a node.

3 Define a mesh and state the difference between a mesh and a loop.

4 State Kirchhoff's second (voltage) law (KVL).

5 Explain the usefulness of the principle of superposition in electrical circuit

analysis.

6 State Thevenin's theorem.

7 State Norton's theorem.

8 What is the condition for the maximum power transfer from a source to a

load?