Direct Current and Transient Analysis
121
R.2.83 Let us gain some experience by using the different network theorems and techniques just presented by solving the following example.
For the circuit shown in Figure 2.15, fi nd the voltage across R L = 5 Ω (Vaa’) 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.
R TH
I N = V TH /R TH
R TH
V TH
a
a ′
a
a
FIGURE 2.15
Network of R.2.83.
R L = 5 Ω
R = 15 Ω
I s = 12 A
V s = 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 = 20I 1 − 5I 2
and since
I 2 = −12 A
CRC_47760_CH002.indd 121
CRC_47760_CH002.indd 121
7/23/2008 1:38:39 PM
7/23/2008 1:38:39 PM
121
R.2.83 Let us gain some experience by using the different network theorems and techniques just presented by solving the following example.
For the circuit shown in Figure 2.15, fi nd the voltage across R L = 5 Ω (Vaa’) 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.
R TH
I N = V TH /R TH
R TH
V TH
a
a ′
a
a
FIGURE 2.15
Network of R.2.83.
R L = 5 Ω
R = 15 Ω
I s = 12 A
V s = 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 = 20I 1 − 5I 2
and since
I 2 = −12 A
CRC_47760_CH002.indd 121
CRC_47760_CH002.indd 121
7/23/2008 1:38:39 PM
7/23/2008 1:38:39 PM