9/12/17 Mesh Analysis II

Mesh Analysis ||

Date : 9/12/17

Overview 

           Use mesh analysis to analysis the circuit. Find V1  and I1. 

Pre-lab 

1) Predict I1 

loop I0 
        -3 + 20k (I0) + 4.7k (I0 - I1) = 0
        24.7k (I0) - 4.7k (I1) = 3

loop I1
        4.7k (I1 - I0) + 10k (I1) + 6.8k (I1 - I2) = 0
        -4.7k (I0)  +21.5k (I1) - 6.8k (I2) = 0

loop I2 
         6.8k (I2 - I1) + 5 = 0
         -6.8k ( I1) + 6.8k (I2) = -5

Matrix of Loop I0, I1, I2

I0 =  60.4 μA
I1 = -321 μA
I2 = -1056 μA

2) Predict V1

V1 = -321μA (6800) - (-1056μA) (6800)
V1 = 4.998  volts
V1 = 5 volts

Also, the voltage source of 5 volts into parallel resistors is the same for each resistor, meaning 5 volts goes through each resistor, resulting in 5 volts for V1  which also matches our calculated value. 

Lab 

V1 = 5.19 volts 

Percentage error between calculated and actual for V1
            ((5.19 - 5.0) / (5.0)) * 100  = % 3.8 error

Comments 

Noted that the Voltage source was at 5.2 volts instead of 5 volts which resulted in percentage error. Also, our recorded 5.19 volts matches the voltage source of 5.2 volts, where 5.2 volts is rounded up to 5.2 on the Power Source.
 
Below is the circuit created in the Every Circuit App. Our calculated value of -321 matches the simulated value shown below, with the negative indicating counter clockwise from our original clockwise direction.

EveryCircuitApp

TinkerCadApp