Download Problem 18 (a) Convert the circuit to the left of terminals AB in Figure

Thevenin and Norton Equivalent Circuits
Problem 18 (a) Convert the circuit to the left of terminals AB in Figure 1 (a) to an equivalent
Thevenin circuit by initially converting to a Norton equivalent circuit (b) Determine the
current flowing in the 1.8  resistor.
(a)
E1 =
12V
E2 =
24V
1.8
r1 = 3
r2 = 2
Figure 1
For the branch containing the 12 V
source, converting to a Norton
equivalent circuit gives ISC = 12/3
= 4 A and r1 = 3 . For the branch
containing the 24 V source,
converting to a Norton equivalent
circuit gives ISC2 = 24/2 =12 A and
r2=2 
.
Thus Figure 1b shows a network equivalent to Figure 1a . From Figure 1b the total shortcircuit current is 4 + 12 = 16A .
ISC =
4A
ISC =
12A
r1=
3
A
And the total resistance 3 x 2 = 1.2
3+2
R2=
Thus Figure 11.45(b) simplifies to Figure 1(c). The opencircuit voltage across AB of Figure 1(c),
2
B
FigureA1b
16A
r1=
1.2
B
E = (16)(l.2) = 19.2 V, and the resistance 'looking-in' at AB
is1.2. Hence the Thevenin equivalent circuit is as shown
in Figure 1(d).
(b)
When the 1.8  resistance is connected between
terminals A and B of Figure 1(d) the current I flowing is
given by:
Figure 1c
A
r1=
1.2
I
=
19.2 =
6.4A
1.2
+1.8
19.2 V
B
Figure 1d
29
Problem 19. Determine by successive conversions between Thevenin and Norton equivalent
networks a Thevenin equivalent circuit for terminals AB of Figure 2(a). Hence determine the
current flowing in the 200  resistance.
For the branch containing the 10 V source,
converting to a Norton equivalent network
gives
1 mA
E1 =
10V
600 
E2 =
6V
200 
2k
3 k
Figure 2
ISC = 6 = 2mA and r2 = 3k
3000
1 mA
E1 =
10V
600 
E2 =
6V
2k
ISC = 10 = 5mA and r1 = 2k
2000
For the branch containing the 6 V source,
converting to a Norton equivalent network
gives
A
3 k
Thus the network of Figure 2(a) converts
to Figure 2(b). Combining the 5 mA and
2 mA current sources gives the
equivalent network of Figure 2(c) where
the short-circuit current for the original
two branches considered is 7 mA and
the resistance is
B
Figure 2b
1 mA
600 
2 x 3 = 1.2 k.
2+3
Both of the Norton equivalent networks shown in Figure
2(c) may be converted to Thevenin equivalent circuits. The
open-circuit voltage across CD is (7 x l0-3)(1.2 x 10-3) = 8.4
V and the resistance 'looking-in' at CD is 1.2 k.
A
7mA
The open-circuit voltage across EF is (1 x 10-3)(600) = 0.6 V
and the resistance 'looking-in' at EF is 0.6 k. Thus Figure
2(c) converts to Figure 2(d). Combining the two Thevenin
circuits gives
1.2k
B
Figure 2c
E = 8.4 - 0.6 = 7.8 V and the resistance r = (1.2 + 0.6) k =
1.8 k.
0.6 V
600 
8.4 V
A
Thus the Thevenin equivalent circuit for terminals AB
of Figure 2(a) is as shown in Figure 2(e).
Hence the current I flowing in a 200 resistance
connected between A and B is given by:
1.2k
B
I = 7.8
= 7.8 = 3.9 mA
1800+200
2000
Figure 2d
30
A
7.8 V
1.8k
B
Figure 2e
31