Part (A):
- For Node (1):
(V1-10)/1000+ (V1-V2)/1000+ V1/(5.1 x 1000)=0 (1)
- For Node (2):
(V2-V1)/1000+ V2/(3.3 x 1000)=0 (2)
- Solve equations (1) and (2) to find the values of V1 and V2 by using the calculator:
V1= 6.99968 V
V2= 5.37185 V
Vth = V2 = 5.37185 V
- The circuit is open so there is no current in a & b :
R_{eq1} = 1 kW // 5.1 kW
= 836.066 W
R_{eq2} = R_{eq1} + 1 kW
= 836.066 + 1000 = 1836.07 W
R_{eq3} = R_{eq2} // 3.3 kW
= = 1179.7 W
R_{eq4} = R_{eq3} + 1 kW
= 1179.7 + 1000 = 2179.7 W
Rth = R_{eq4} = 2179.7 W
- The current through the 2 kW resistor:
I_{ 2k}_{W} =
= = 0.001285 A
I_{ 2k}_{W} = 1.285 mA
Part (B):
Figure 1 below shows circuit (B) done by the PSPICE.
Figure 1: Circuit (B)
Part (C):
- My Netlist:
V-V1 N001 0 10V
R_R1 N002 N001 1k W
R_R2 N003 N002 1k W
R_R3 N004 N003 1k W
R_R4 0 N002 5.1k W
R_R5 0 N003 3.3k W
R_RL 0 N004 2k W
- The PSPICE Netlist:
- The difference between the PSPICE Netlist and my Netlist is the choice of nodes.
Part (D):
- By using the Mesh current method:
- Loop (1):
15 + 5.1 x 10^{3}*I_{1} + 2x10^{3} (I_{1} – I_{2}) =0 à (1)
- Loop (2):
-1x10^{3}*I_{2} – 2x10^{3} (I_{2} – I_{1}) – 5.1x10^{3} (I_{2} – I_{3}) = 0 à (2)
- Loop (3):
5.1x10^{3} (I_{3} – I_{2}) + 2x10^{3}*I_{3} – 1.5 = 0 à (3)
- Solve equations (1), (2) and (3) to find the values of I_{1}, I_{2} and I_{3} by using the calculator:
I_{1} = 0.002342 A
I_{2} = 0.000813 A
I_{3} = 0.000373 A
V_{L} = I_{2} x 1 kW
V_{L} = -0.000813 x 1000 = 0.813 V
Part (E):
Figure 2 below shows circuit (E)
Figure 2: Circuit (E)
Part (F):
Figure 3 below shows circuit (F)
Figure 3: Circuit (F)
Table 1 below shows the calculation for V_{L} for part (D, E, F).
V_{L} (voltage over the 1 kW resistor) |
Calculation Part D (No diode) |
PSPICE Part E (No diode) |
PSPICE Part F (with diode) |
1.V1 & V2 Present |
0.813 |
0.813 |
3.275 |
2.V1 only |
1.091 |
1.091 |
2.881 |
3.V2 only |
-0.278 |
-0.2782 |
7.151x10^{-6} |
4.Add line 2&3 |
0.813 |
0.8128 |
2.88099 |
5.% difference between line1 and line4 |
0% |
0.0246 % |
26.5347% |
Table 1: The Values of V_{L}
- The % difference formula:
This problem has been solved.
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