300+ Network Theorem MCQ – Objective Question Answer for Network Theorem Quiz

201. The maximum power is delivered from a source to its load when the load resistance is ______ the source resistance.

A. greater than
B. less than
C. equal to
D. less than or equal to

Answer: C

The maximum power is delivered from a source to its load when the load resistance is equal to the source resistance. The maximum power transfer theorem can be applied to both dc and ac circuits.

 

202. If source impedance is complex, then maximum power transfer occurs when the load impedance is _______ the source impedance.

A. equal to
B. negative of
C. complex conjugate of
D. negative of the complex conjugate of

Answer: C

The maximum power transfer theorem can be applied to complex impedance circuits. If the source impedance is complex, the maximum power transfer occurs when the load impedance is the complex conjugate of the source impedance.

 

203. If the source impedance is complex, then the condition for maximum power transfer is?

A. ZL  = ZS
B. ZL  = ZS*
C. ZL  = − ZS
D. ZL  = − ZS

Answer: B

The maximum power is transferred when the load resistance is equal to the source resistance. The condition for maximum power transfer is ZL  = ZsS *.

 

204. If ZL  = ZS * , then?

A. RL  = 1
B. RL  = 0
C. RL  = − RS
D. RL  = RS

Answer: D

If ZL  = ZS * , then RL  = RS.

This means that the maximum power transfer occurs when the load impedance is equal to the complex conjugate of source impedance ZS.

 

205. For ZL  = ZS ×, the relation between XL and XS is?

A. XL  = XS
B. XL  = 0
C. XL  = 1
D. XL  = − XS

Answer: D

For ZL  = ZS* , the relation between XL and XS is XL  = − XS.

Maximum power transfer is not always desirable since the transfer occurs at a 50 percent efficiency.

 

206. In the circuit shown below, find the value of load impedance for which source delivers maximum power.

In the circuit shown below, find the value of load impedance for which source delivers maximum power.

A. 15 − j20
B. 15 + j20
C. 20 − j15
D. 20 + j15

Answer: A

The maximum power transfer occurs when the load impedance is equal to the complex conjugate of source impedance

ZS. ZL  = ZS *   = (15 − j20)Ω.

 

207. What is the load current in the following circuit by applying the maximum Power transfer theorem?

In the circuit shown below, find the value of load impedance for which source delivers maximum power.

A. 1.66∠90°
B. 1.66∠0°
C. 2.66∠0°
D. 2.66∠90°

Answer: B

The load current is the ratio of voltage to the impedance.

So the load current is

I = (50∠0°)/(15 + j20 + 15 − j20) = 1.66∠0° A.

 

208. The maximum power delivered by the source in the below circuit shown?

In the circuit shown below, find the value of load impedance for which source delivers maximum power.

A. 39.33
B. 40.33
C. 41.33
D. 42.33

Answer: C

The term power is defined as the product of the square of the current and the impedance.

So the maximum power delivered by the source in the circuit is

P = I2R×Z = 1.662×15 = 41.33W.

 

209. For the circuit shown, the resistance R is variable from 2Ω to 50Ω. What value of RS results in maximum power transfer across terminals ‘ab’.

For the circuit shown, the resistance R is variable from 2Ω to 50Ω. What value of RS results in maximum power transfer across terminals ‘ab’.

A. 1
B. 2
C. 3
D. 4

Answer: B

As RL is fixed, the maximum power transfer theorem does not apply. Maximum current flows in the circuit when RS is minimum. So RS  = 2Ω.

 

210. Find the maximum power delivered by the source in the following circuit.

For the circuit shown, the resistance R is variable from 2Ω to 50Ω. What value of RS results in maximum power transfer across terminals ‘ab’.

A. 96.6
B. 97.6
C. 98.6
D. 99.6

Answer: C

ZT  = RS – j5 + RL

= 2 − j5 + 20 = 22.56∠ − 12.8⁰Ω.

I = VS/ZT  = − 50∠0⁰/22.56∠ − 12.8⁰ = 2.22∠ − 12.8⁰A.

P = I2R = (2.22)2×20 = 98.6W.

 

211. 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 are 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.

 

212. In the circuit given below, the value of the maximum power transferred through RL is ___________

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.

 

213. 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.

 

214. 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.

 

215. In the circuit shown below what is the value of RL for which maximum power is transferred to RL?

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.

 

216. 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.

 

217. 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.

 

218. In the circuit given below, what is the amount of maximum power transfer to R?

In the circuit given below, what is the amount of maximum power transfer to R?

A. 56 W
B. 76 W
C. 60 W
D. 66 W

Answer: D

Drop across V1Ω  = 5 × 1 = 5V

Also,

\(\frac{V − V_{1Ω}}{10} + \frac{V − 20 − V_{1Ω}}{2} + \frac{V − V_{OC}}{5}\) = 2

0.1 V – 0.1V1Ω  + 0.5V – 10 – 0.5V1Ω  + 0.2 – 0.2VOC = 2

0.8V – 0.6V1Ω  = 12 + 0.2VOC

0.8 V – 0.2VOC  = 12 + 3 = 15 (Putting V1Ω = 5)

(frac{VOC − V)/5) + 2 = 5

Or, 0.2VOC – 0.2V = 3

Again, RTH  = {(10||2) + 1} + 5

=   5 + 20/(12 + 1) = 7.67 Ω

Following the theorem of maximum power transfer

R = RTH  = 7.67 Ω

PMAX = \(\frac{V_{OC}^2}{4R} = \frac{45^2}{4×7.67}\) = 66 W

 

219. In the circuit given below, the value of RL for which it absorbs maximum power is ___________

In the circuit given below, the value of RL for which it absorbs maximum power is

A. 400/3 Ω
B. 2/9 kΩ
C. 350.38 Ω
D. 4/9 kΩ

Answer: C

5 = 200I – 50 × 2I

I = 5/100 = 0.05 A

VOC  = 100 × 3I + 200 × I = 25 V

V1 = \(\frac{\frac{5}{50}}{\frac{1}{50} + \frac{1}{200} + \frac{1}{100}} \)

\(\frac{0.1}{0.02 + 0.005 + 0.01} \) = 2.85 V

I = 2.85/100 = 0.0142 A = 14.2 mA

ISC = 2.85/100 + 3 × 0.0142 = 0.07135 A

∴ RTH = VOC/ISC = 25/0.07135 = 350.38 Ω.

 

220. 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.

 

221. The average power delivered to the 6 Ω load in the circuit in the figure below is ___________

average power delivered to the 6 Ω load in the circuit of 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.

 

222. 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.

 

223. The 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.

 

224. 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.

 

225. 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 = \(\frac{μ_0 n^2 A}{l}\)

= \(\frac{4π X 10^{ − 7} X 300 X 300 X 300 X 10^{ − 6}}{300 X 10^{ − 3}}\)

= 113.04 μH.

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