21. The basic elements of an electric circuit are _________
A. R, L, and C
B. Voltage
C. Current
D. Voltage as well as current
Answer: A
The elements which show their behavior only when excited is called basic circuit elements. Here resistance, inductance, and capacitance show their behavior only when excited. Hence they are the basic elements of an electric circuit.
22. In the circuit given below, the value of the maximum power transferred through RL is ___________
A. 0.75 W
B. 1.5 W
C. 2.25 W
D. 1.125 W
Answer: A
I + 0.9 = 10 I
I = 0.1 A
VOC = 3 × 10 I = 30 I
VOC = 3 V
Now, ISC = 10 I = 1 A
Rth = 3/1 = 3 Ω
Vth = VOC = 3 V
RL = 3 Ω
Pmax = 32/4×3 = 0.75 W.
23. The energy stored in the magnetic field at a solenoid 100 cm long and 10 cm diameter wound with 1000 turns of wire carrying an electric current of 10 A, is ___________
A. 1.49 J
B. 0.49 J
C. 0.1 J
D. 1 J
Answer: B
L = \(\frac{N^2 μ_0 A}{l}\)
= \(\frac{10^6.4π.10^{ − 7}.\frac{π}{4}.(100 X 10^{ − 4})}{1}\)
= \(\frac{π^2 X 10^{ − 3}}{1}\)
Energy = 0.5 LI2
= 0.49 J.
24. The resistance of a strip of copper of rectangular cross-section is 2 Ω. A metal of resistivity twice that of copper is coated on its upper surface to a thickness equal to that of a copper strip. The resistance of composite strip will be _________
A. 3/4 Ω
B. 3/4 Ω
C. 3/4 Ω
D. 6 Ω
Answer: B
Given that copper and coated metal strip have a resistance of 2 ohms respectively. These two strips are connected in parallel.
Hence, the resistance of the composite strip
= (2 × 4)/(2 + 4)
= 8/6 = 4/3 Ω.
25. In the circuit shown below what is the value of RL for which maximum power is transferred to RL?
A. 2.4 Ω
B. 8/3 Ω
C. 4 Ω
D. 6 Ω
Answer: C
Maximum power is transferred to RL when the load resistance equals the Thevenin resistance of the circuit.
RL = RTH = VOC/ISC
Due to open − circuit, VOC = 100 V; ISC = I1 + I2
Applying KVL in lower loop, 100 – 8I1 = 0
Or, I1 = 100/8 = 25/2
And VX = − 4I1 = − 4 × 25/2 = − 50V
KVL in upper loop, 100 + VX – 4I2 = 0
I2 = (f100 − 50)/4 = 25/2
Hence, ISC = I1 + I2 = 25/2 + 25/2 = 25
RTH = VOC/ISC = 100/25 = 4 Ω
RL = RTH = 4 Ω.
26. A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear resistance characterized by the relation VNL = I2NL. The power dissipated in the resistance is ___________
A. 1 W
B. 1.5 W
C. 2.5 W
D. 3 W
Answer: A
3 = 2I + I2
∴ I = 1 A; VNL = 1V
∴ Power dissipated in RNL = 1 × 1 = 1 W.
27. The two windings of a transformer have an inductor of 2 H each. If the mutual inductor between them is also 2 H, then which of the following is correct?
A. turns ratio of the transformer is also 2
B. Transformer is an ideal transformer
C. It is a perfect transformer
D. It is a perfect as well as an ideal transformer
Answer: C
We know that
K = \(\frac{M}{\sqrt{L_1 L_2}}\)
= \(\frac{2}{\sqrt{2 X 2}}\)
Hence, it is a perfect transformer.
28. In the circuit given below, what is the amount of maximum power transfer to R?
30. The form factor of sinusoidal alternating electric current is ___________
A. 0
B. 1
C. 1.11
D. 1.15
Answer: C
We know that for alternating electric current form factor is defined as the ratio of RMS value and the average value of alternating current.
Now the RMS value of alternating electric current = 0.07 × the maximum value of alternating current.
The average value of alternating electric current = 0.637 × the maximum value of alternating current.
∴ Form factor = 0.707/0.637 = 1.11.
31. The average power delivered to the 6 Ω load in the circuit in the figure below is ___________
A. 8 W
B. 76.68 W
C. 625 kW
D. 2.50 kW
Answer: B
I2 = V2/6, I1 = I2/5 = V2/30
V1 = 5V2
50 = 400(I1 – 0.04V2) + V1
V2 = 21.45 V
∴ PL = V22/6
= 21.452/6
= 460.1025/6 = 76.68 W.
32. The RMS value of the sine wave is 100 A. Its peak value is ____________
A. 70.7 A
B. 141.4 A
C. 150 A
D. 282.8 A
Answer: B
We know that for sinusoidal alternating electric current the peak factor or amplitude factor can be expressed as the ratio of maximum or peak value and RMS value of alternating current.
So the peak value = rms value of alternating electric current × peak factor of alternating electric current
= 100 × 1.414 = 141.4 A.
33. Potential of the earth is – 50 V. If the potential difference between anode and cathode (eartheD. is measured as 150 V, actual voltage on anode is __________
A. 0 V
B. 100 V
C. 200 V
D. 250 V
Answer: C
Given that, the potential difference between anode and cathode (eartheD. is measured as 150 V and the potential of the earth is 50 V.
So, actual voltage on anode, V = 150 – ( − 50)
= 150 + 50
= 200 V.
34. An alternating voltage V = 150 sin(314)t is supplied to a device that offers a resistance of 20 Ω in the forwarding direction of electric current while preventing the flow of electric current in the reverse direction. The form factor is ___________
A. 0.57
B. 0.318
C. 1.414
D. 1.57
Answer: D
From the voltage equation, we can get
Vm = 150 V and Im = 150 / 20 = 7 A
RMS value of the current
Irms = Im / 2 = 7/2 = 3.5 A
Average value of the current
Iavg = Im / π = 2.228 A
Form factor = Irms / Iavg
= 3.5 / 2.228 = 1.57.
35. A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm2. The inductor of the coil corresponding to a magnetizing electric current of 3 A will be?
A. 37.68 μH
B. 47.68 μH
C. 113.04 μH
D. 120.58 μH
Answer: C
The inductance of the coil
L = \(\frac{μ_0 n^2 A}{l}\)
= \(\frac{4π X 10^{ − 7} X 300 X 300 X 300 X 10^{ − 6}}{300 X 10^{ − 3}}\)
