Equivalent networks and superposition pre lab 3

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 10³*I₁ + 2x10³ (I₁ – I₂) =0 à (1)

• Loop (2):

-1x10³*I₂ – 2x10³ (I₂ – I₁) – 5.1x10³ (I₂ – I₃) = 0 à (2)

• Loop (3):

5.1x10³ (I₃ – I₂) + 2x10³*I₃ – 1.5 = 0 à (3)

• Solve equations (1), (2) and (3) to find the values of I₁, I₂ and I₃ by using the calculator:

I₁ = 0.002342 A

I₂ = 0.000813 A

I₃ = 0.000373 A

V_L = I₂ 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).

Table 1: The Values of V_L

• The % difference formula:

Cite This work.

To export a reference to this article please select a referencing stye below.