Direct Current and Transient Analysis 121
R.2.83 Let us gain some experience by using the different network theorems and tech-
niques just presented by solving the following example.
For the circuit shown in Figure 2.15, fi nd the voltage across RL = 5 Ω (Va a’) by
hand calculation by using
a. Loop equations
b. Node equations
c. Source transformation—voltage to current source
d. Source transformation—current to voltage source
e. Superposition
f. Thevenin’s theorem
g. Norton’s theorem
FIGURE 2.14
Thevenin’s source transformation into a Norton’s equivalent circuit.
RTH
IN = VTH/RTH
RTH
VTH
a
a′ a
a
FIGURE 2.15
Network of R.2.83.
RL = 5 Ω
R = 15 Ω
Is = 12 A
Vs = 60 V
a
a′
ANALYTICAL Solutions
a. The loop equation solution is illustrated in the circuit diagram of Figure 2.16. The
loop currents I 1 and I 2 are indicated in Figure 2.16. Then the loop equation for loop
# 1 is given by
60 V = 20 I 1 − 5 I 2
and since
I 2 = −12 A