My original calculated values are below. Item Calculated Value I 1 23.75mA I 2 3
ID: 1835635 • Letter: M
Question
My original calculated values are below.
Item
Calculated Value
I1
23.75mA
I2
3.95mA
IR1
23.75mA
IR2
19.8mA
IR3
3.95mA
IR4
3.95mA
VR1
4.275V
VR2
7.72V
VR3
0.869V
VR4
1.856V
PV1
0.285W
PV2
0.01975W
The values are based on the diagram below which is mesh analysis:
1- I1 was measured to be 27.54 mA, and the voltage across R3 is 2.058 V. Both supply voltages were tested to be with the correct value. What is the fault in the circuit? (Hint: Use MultiSim if you have access).
2- If VR1 was measured to be 12 V, and the voltage across I1 is 66.667 mA, then what is the fault in the circuit?
Item
Calculated Value
I1
23.75mA
I2
3.95mA
IR1
23.75mA
IR2
19.8mA
IR3
3.95mA
IR4
3.95mA
VR1
4.275V
VR2
7.72V
VR3
0.869V
VR4
1.856V
PV1
0.285W
PV2
0.01975W
Explanation / Answer
Please ask if you have any doubt.I will help you.
1) Given the I1 = 27.54 mA and voltage across the R3 is 2.058V . Lets measure voltage across R1
V(R1) = (180)(27.54mA) = 4.957V and given voltage across R3 = 2.058V .
Now take the KVL across the 12V ,R1,R3,R4 and 5V.
12 - V(R1) - V(R3) - V(R4) -5 = 0 ,Gives 12 - 4.957 - 2.058V - V(R4) - 5 = 0 implies V(R4) = 0 .
Therefore R4 is shorted.
This can be cross checked by calculating voltage across the R2 with the given current and voltage values.The cuurent through R3 is 2.058/220 = 9.355 mA.Thus the current through R2 is 27.54 - 9.355 = 18.185 mA .Voltage across the R2 is (390)(18.185 mA) = 7V approximately.
2) Given V(R1) = 12V ,which should give I(R1) = 12V/180 = 66.67mA .Thus we can clerly see that the 390 is shoted,which will give exact required circuit matching given values.
Therefore R2 is shorted.
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