Figure 1: Mesh Current Method |
goal of this lab assignment. Three are three
independent DC voltage sources in the circuit.
There are five nodes and three branches.
KVL in the left clockwise loop is (i1):
5 - (10k)i1 - (20k)*(i1+i2) +5 = 0
KVL in the right counterclockwise loop is (i2):
-3 - (6.8k)*(i2) - (20k)*(i2*i1) +5 = 0
These two KVLs make a system equation with two unknown variable i1 and i2.
(30k)*i1 + (20k)*i2 = 10
(20k)*i1 + (26.8k)*i2 = 2
i1 = 0.564 mA , i2 = -0.347 mA
According to Krichhoff's Current Law (KCL) at node a: I1 = I2 + I3 so I2 = i1 + i2 , I2 = 0.217 mA I1 = i2, I3 = i1
5 - (10k)*(0.564mA) = Va , Va = - 0.64 volts
Va - (20k)*(I2) + 5 = 0 , V2 = (20k)*I2 = Va + 5 = 4.34 v
Va - (6.8k)*I1 + 3 = 0 , V1 = (6.8k)*(I1) = Va + 3 = 2.36 v
Every resistor has a specific tolerance, or a interval accuracy. Resistance of the resistors in the circuit are measured by a DMM. R1 = 9.81k , R2 = 6.7k , R3 = 22k
According to these resistors (Fig 2) V1 = 2.41 v , V2 = 4.4 v
Percent Error = [(Theory - Measurement) / Theory] * 100%
Percent Error (V1) = [(2.36 - 2.41) / 2.36]*100% = 2.12%
Percent Error (V2) = [(4.34 - 4.4) / 4.34]*100% = 1.38%
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