Basic Electrical Engineering Questions and Answers
Q1.
Calculate the energy in the 10 ohm resistance in 10 seconds.
a) 400J
b) 40kJ
c) 4000J
d) 4kJ
Ans: b
Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V2/R. P=2002/10= 4000W. E=Pt = 4000*10=40000Ws = 40000J = 40kJ.
Q2.
Calculate the energy in the 5 ohm resistor in 20 seconds.
a) 21.5kJ
b) 2.15kJ
c) 2.15J
d) 21.5J
Ans: a
The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/(5+10)=14.67A. P=I2R = 14.672*5=1075.8W. E=Pt = 1075.8*20 = 21516J=21.5kJ.
Q3.
A wire has the same resistance as the one given in the figure. Calculate its resistivity if the length of the wire is 10m and its area of cross section is 2m.
a) 16 ohm-metre
b) 8 ohm-metre
c) 16 kiloohm-metre
d) 8 kiloohm-metre
Ans: b
From the given circuit, R=V/I = 200/5 = 40ohm.
Resistivity= Resistance*Area of cross section/ Length of the wire.
Resistivity= 40*2/10= 8 ohm-metre.
Q4.
Find the value of I1, I2 and I3.
a) -0.566A, 1.29A, -1.91A
b) -1.29A, -0.566A, 1.91A
c) 1.29A, -0.566A, -1.91A
d) 1.91A, 0.566A, 1.29A
Ans: c
Using the matrix method:
Matrix(3,-2,0) (I1)=(5)
(-2,9,-4) (I2)=(0)
(0,-4,9) (I3)=(-15)
Solving this matrix equation,
we get I1 = 1.29A, I2 = -0.566A and I3 = -1.91A.
Q5.
Find the value of V, if the value of I3= 0A.
a) 1.739 V
b) 6.5 V
c) 4.5V
d)2.739V
Ans: a
5-3I1+2I2=0, 9I2-2I1=0, -4I2+V=0
On solving, V=1.739V.
Q6.
Find the value of R if the power in the circuit is 1000W.
a) 10 ohm
b) 9 ohm
c) 8 ohm
d) 7 ohm
Ans: c
To find the value of I:
VI=P =>100I=1000 => I=10A.
Voltage across the 2 ohm resistor = 20V.
Voltage across the R resistor = 100-20= 80V.
R=V/I => R=80/10 = 8A.
Q7.
Find the current in the 4 ohm resistor.
a) 5A
b) 0A
c) 2.2A
d) 20A
Ans: b
The 4 ohm resistor gets shorted since current always prefers the low resistance path. All the current flows to the branch which is connected in parallel to the 4 ohm branch, hence no current flows in the 4 ohm resistance.
Q8.
What is the current in the circuit?
a) 0A
b) 15A
c) 5A
d) 10A
Ans: a
If we move in the clockwise direction, we get the total voltage to be equal to: -10-20+30 = 0V. Since I=V/R = 0/4=0, I=0A.
Q9.
Does the 15A source have any effect on the circuit?
a) Yes
b) No
c) Cannot be determined
d) Yes, only when the 10V source is removed
Ans: b
The 15A current source has a lower resistance path associated with it and hence it keeps moving in that particular loop. It does not leave that loop and enter the circuit, hence the circuit is not affected by it.
Q10.
Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively.
a) 1.54A, -0.189A, -1.195A
b) 2.34A, -3.53A, -2.23A
c) 4.33A, 0.55A, 6.02A
d) -1.18A, -1.17A, -1.16A
Ans: a
The three mesh equations are:
-3I1+2I2-5=0
2I1-9I2+4I3=0
4I2-9I3-10=0
Solving the equations,
we get I1= 1.54A, I2=-0.189 and I3= -1.195A.
0 Comments