Equivalent networks and superposition pre lab 3

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 _2kW =

= = 0.001285 A

I _2kW = 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 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:

For example:

% difference= x 100%= 0%