1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground.
1.15 Through repeated application of Thévenin's theorem, find the Thévenin equivalent of the circuit in Fig. Pl.15 between node 4 and ground, and hence find the current that flows through a load resistance of 3 k2 connected between node 4 and ground.
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
Transcribed Image Text:1.15 Through repeated application of Thévenin's theorem,
find the Thévenin equivalent of the circuit in Fig. Pl.15
between node 4 and ground, and hence find the current that
flows through a load resistance of 3 k2 connected between
node 4 and ground.
1 20 k2
2 20 k2
3 20 k2
4
10 V
20 k2
20 ΚΩ
20 kΩ
Figure P1.15

Transcribed Image Text:20 kN
20 kN
20 kN
4
10 V
20 k2
20 k2
20 k2
12.31 kN
10 kN
2
20 k2
20 kN
3
4
30 kΩ
0.77 V
3 kN
20 kN
20 k2
20
10 X
20 + 20
= 5 V
Now, when a resistance of 3 k2 is connected
(20//30) = 12 klN 3
-어
20 k2
4
between node 4 and ground,
0.77
I =
12.31 +3
= 0.05 mA
32 k2
20
20 kΩ
5X
20 + 30
= 2V
Thévenin equivalent: (20//32)
= 12.31 kN
20
2 X
20 +32
= 0.77 V
(cont d on the next page)
PIC•COLLAGE
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