Basic Electrical Engineering Questions and Answers
Q1.
Find the value of V1 and V2.
a) 87.23V, 29.23V
b) 23.32V, 46.45V
c) 64.28V, 16.42V
d) 56.32V, 78, 87V
Ans: c
The nodal equations are:
0.3V1-0.2V2=16
-V1+3V2=-15
Solving these equations simultaneously,
we get V1=64.28V and V2=16.42V.
Q2.
Find the value of Vx due to the 16V source.
a) 4.2V
b) 3.2V
c) 2.3V
d) 6.3V
Answer: b
When we consider the 16V source, we short the 10V source and open the 15A and 3A source. From the resulting series circuit we can use voltage divider to find Vx.
Vx = 16*20/(20+80)=3.2A.
Q3.
Find Vx due to the 3A source.
a) 56V
b) 78V
c) 38V
d) 48V
Ans: d
Due to the 3A source, we short the 16V and 10V source and open the 15A source. From the resulting circuit, we can use current divider to find the current in the 20 ohm branch and then multiply it with the resistance to find the voltage.
I20 = 3*80/(20+80)=2.4A
Vx=20*2.4=48V.
Q4.
Find the value of Vx due to the 10V source.
a) 1V
b) 2V
c) 3V
d) 4V
Ans: b
Due to the effect of the 10V source, we short the 16V source and open the 3A and 15A source. From the resulting series circuit, we can use voltage divider to find the value of Vx.
Vx=10*20/(80+20)=2V.
Q5.
Find the voltage due to the 15A source.
a) 0V
b) 2V
c) 4V
d) 6V
Ans: a
Due to 15 A current source, 10V and 16V sources get shorted and the 3A source acts as an open circuit. Since the 10V source is shorted, it acts as a low resistance path and current flows only within that loop and do not flow to the 20 ohm resistor. Hence the voltage is 0V.
Q6.
Calculate the Thevenin resistance across the terminal AB for the following circuit.
a) 4.34 ohm
b) 3.67 ohm
c) 3.43 ohm
d) 2.32 ohm
Ans: b
Thevenin resistance is found by opening the circuit between the specified terminal and shorting all voltage sources.
When the 10V source is shorted, we get:
Rth=(1||2)+3=3.67 ohm.
Q7.
Calculate Vth for the given circuit.
a) 5.54V
b) 3.33V
c) 6.67V
d) 3.67V
Ans: c
4 ohm is removed and then v across 2 ohm is calculated by voltage divider 2*10/(2+1) = 6.67V. Voltage between A and B i.e. Vth is equal to voltage across 4 ohm resistance since no current flow through 3 ohm resistance. So, Vth = 6.67V.
Q8.
Calculate the current across the 4 ohm resistor.
a) 0.86A
b) 1.23A
c) 2.22A
d) 0.67A
Ans: a
Thevenin resistance is found by opening the circuit between the specified terminal and shorting all voltage sources.
When the 10V source is shorted, we get:
Rth=(1||2)+3=3.67 ohm.
Vth is calculated by opening the specified terminal.
Using voltage divider, Vth= 2*10/(2+1)=6.67V.
On drawing the Thevenin equivalent circuit, we get Rth, 4 ohm and Vth in series.
Applying Ohm’s law, I=Vth/(4+Rth) = 0.86A.
Q9.
Calculate the Norton resistance for the following circuit if 5 ohm is the load resistance.
a) 10 ohm
b) 11 ohm
c) 12 ohm
d) 13 ohm
Ans: c
Shorting all voltage sources and opening all current sources we have:
RN=(3||6)+10 = 12 ohm.
Q10.
Calculate the short circuit current is the 5 ohm resistor is the load resistance.
a) 0.72A
b) 0.32A
c) 0.83A
d) 0.67A
Ans: a
Since the 5 ohm is the load resistance, we short it and find the resistance through the short.
If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh.
The mesh equations are:
9I1-6I2=4
-6I1+16I2=6
On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.
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