Example Of Report On Lab 5

Part 1

1. Measured values. Construct the circuit shown in Figure 1. Measure all node voltages and record in your data table. If they do not agree with your calculated values, find and correct the errors before proceeding
Figure 1. Voltage sources only.
V2=V3+7

For a node, according to the first Kirchhoff law:

5-V1150+V2-V1200=V1100V2240+V3100+V3200=V2-V3300V2=V3+7Thus, we perform the transformations:
10-2V1+V2-V1=3V15V2+12V3+6V3-4V2=4V1V2=V3+7
6V1-V2=104V1+V2+18V3=0V2=V3+7
6V1-V2=104V1+19V3=-7V2=V3+7
6V1-V3=174V1+19V3=-7V2=V3+7
-12V1+2V3=-3412V1+57V3=-21V2=V3+7
V3=-17+6V1,12V1+57(-17+6V1)=-21,V2=V3+7.

The calculated voltages are obtained from the equations:

V3=-5559≈0.93V, V2=6.07V, V1=16.076=2.689V.

Part 2

2. Simulated values. Build the circuit in Figure 1 in MultiSim. Measure the voltage across each component and record in your data table.

The discrepancies in results appear due to simulation errors. This becomes evident with current analysis in V1:

∆I=25.2-8.44-16.6≈0.06.

Despite of the discrepancies the simulated and calculated results are close.

Part 2
1. Build the circuit in Figure 2 in MultiSim only. Determine the node voltages and the voltages across the 150, 300 and 200 ohm resistors. Fill in your table.
Figure 2. Current sources only.
I1∙900Om=I2∙200Om⇒I1=2I2
I1+I2=200Om

I1≈133mA

I2≈66.7mA
V3=100 Om ∙133 mA=13.3B
V2: V1-V2300-V2240=0.2
V1: V3100+V1-V2300=0.1

For a node:

4V1-4V2-5V2=2403V2+V1-V2=30
4V1-9V2=2404V1-V2=30 -8V2=2204V1=30+V2 ⇒ V2=-26.3V1=0.9

The simulated and calculated current values are close. Therefore, the calculations results prove the Multisim data.