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## Network Theory MCQ Set 1

1. The equation of the average power (Pavg) is?
a) (VmIm/2)cosθ
b) (VmIm/2)sinθ
c) VmImcosθ
d) VmImsinθ

Answer: a [Reason:] To find the average value of any power function we have to take a particular time interval from t1 to t2, by integrating the function we get the average power. The equation of the average power (Pavg) is Pavg = (VmIm/2)cosθ.

2. Average power (Pavg) =?
a) VeffImcosθ
b) VeffIeffcosθ
c) VmImcosθ
d) VmIeffcosθ

Answer: b [Reason:] To get average power we have to take the product of the effective values of both voltage and current multiplied by cosine of the phase angle between the voltage and current. The expression of average power is Average power (Pavg) = VeffIeffcosθ

3. In case of purely resistive circuit, the average power is?
a) VmIm
b) VmIm/2
c) VmIm/4
d) VmIm/8

Answer: b [Reason:] In case of purely resistive circuit, the phase angle between the voltage and current is zero that is θ=0⁰. Hence the average power = VmIm/2.

4. In case of purely capacitive circuit, average power = ____ and θ=___
a) 0, 0⁰
b) 1, 0⁰
c) 1, 90⁰
d) 0, 90⁰

Answer: d [Reason:] In case of purely capacitive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0.

5. In case of purely inductive circuit, average power = ____ and θ=___
a) 0, 90⁰
b) 1, 90⁰
c) 1, 0⁰
d) 0, 0⁰

Answer: a [Reason:] In case of purely inductive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0.

6. If a circuit has complex impedance, the average power is ______
a) power stored in inductor only
b) power stored in capacitor only
c) power dissipated in resistor only
d) power stored in inductor and power dissipated in resistor

Answer: c [Reason:] If a circuit has complex impedance, the average power is power dissipated in resistor only and is not stored in capacitor or inductor.

7. A voltage v (t) = 100sinωt is applied to a circuit. The current flowing through the circuit is i(t) = 15sin( ωt-30⁰). Find the effective value of voltage.
a) 70
b) 71
c) 72
d) 73

Answer: b [Reason:] The expression of effective value of voltage is Veff = Vm/√2. Given Vm = 100. On substituting the value in the equation we get effective value of voltage = 100/√2 = 71V.

8. Find the effective value of current in the question 7.
a) 9
b) 10
c) 11
d) 12

Answer: c [Reason:] The expression of effective value of current is Ieff = Im/√2. Given Im = 15. On substituting the value in the equation we get effective value of current = 15/√2=11V.

9. Determine the average power delivered to the circuit.
a) 620
b) 630
c) 640
d) 650

Answer: d [Reason:] The expression of average power delivered to the circuit is Pavg = VeffIeffcosθ, θ = 30⁰. We have Veff = 71, Ieff = 11. So the average power delivered to the circuit Pavg = 71 x 11 x cos 30⁰ = 650W.

10. Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30⁰.
a) 61.5
b) 62.5
c) 63.5
d) 64.5

Answer: b [Reason:] The expression of the average power delivered to the circuit is Pavg = Im2 R/2. Given Im = 5, R = 5.So the average power delivered to the circuit = 52×5/2 = 62.5W.

## Network Theory MCQ Set 2

1. The expression of power (P1) at lower half power frequency is?
a) (I2maxR)/8
b) (I2maxR)/4
c) (I2maxR)/2
d) I2maxR

Answer: c [Reason:] The upper and lower cut-off frequencies are sometimes called the half-power frequencies,. At these frequencies the power from the source is half of the power delivered at the resonant frequency. The expression of power (P1) at lower half power frequency is P1 = (I2maxR)/2.

2. At upper half power frequency, the expression for power (P2) is?
a) I2maxR
b) (I2maxR)/2
c) (I2maxR)/4
d) (I2maxR)/8

Answer: b [Reason:] At upper half power frequency, the expression for power (P2) is P2 = (I2maxR)/2. The response curve is also called the selectivity curve of the circuit.

3. Determine the resonant frequency for the specifications: R = 10Ω, L = 0.1H, C = 10µF.
a) 157
b) 158
c) 159
d) 160

Answer: c [Reason:] The frequency at which the resonance occurs is called resonant frequency. The expression of the resonant frequency is given by fr = 1/(2π√LC). On substituting the given values we get resonant frequency = 1/(2π√(0.1×10×10-6))=159.2 Hz.

4. The expression for lower half power frequency is?
a) (-R+√(x2+4LC))/4πL
b) (–R-√(x2+4LC))/4πL
c) (R-√(x2+4LC))/4πL
d) (R+√(x2+4LC))/4πL

Answer: a [Reason:] Selectivity indicates how well a resonant circuit responds to a certain frequency and eliminates all other frequencies. At lower power frequency, XC > XL(1/2πf1C)-2πf1L=R. f1 = (-R+√(x2+4LC))/4πL.

5. The expression for upper half power frequency is?
a) (R+√(x2+4LC))/4πL
b) (R-√(x2+4LC))/4πL
c) (–R-√(x2+4LC))/4πL
d) (-R+√(x2+4LC))/4πL

Answer: a [Reason:] The narrower the bandwidth the greater the selectivity. At upper half power frequency, XC < XL => -(1/2πf2C)+2πf2L=R. f2 = (R+√(x2+4LC))/4πL.

6. The expression for bandwidth is?
a) R/πL
b) R/2πL
c) R/4πL
d) R/8πL

Answer: b [Reason:] The bandwidth of any system is the range of frequencies for which the current or output voltage is equal to 70.7% of its value at the resonant frequency. The expression of bandwidth is BW = f2 – f1 = R/2πL.

7. In series circuits, the expression for quality factor is?
a) fr
b) BW
c) fr/BW
d) BW/ fr

Answer: c [Reason:] The quality factor is the ratio of the reactive power in the inductor or capacitor to the true power in the resistance in series with the coil or capacitor. The expression of quality factor is Q = fr/BW.

8. In a series circuit having resistance and inductance, the quality factor is?
a) ωL/R
b) R/ωL
c) ωL
d) R

Answer: a [Reason:] Quality factor Q = ωL/R. A higher value of circuit Q results in smaller bandwidth and a lower value of Q causes a larger bandwidth.

9. If a series circuit contains resistor and capacitor, the expression for quality factor is?
a) C
b) ωRC
c) ωC
d) 1/ωRC

Answer: d [Reason:] The expression of quality factor is Q = 1/ωRC. The ratio of voltage across either L or C to the voltage applied the resonance can be defined as magnification.

10. The quality factor of the coil for a series circuit having R = 10Ω, L = 0.1H, C = 10µF.
a) 1
b) 5
c) 10
d) 15

Answer: c [Reason:] The resonant frequency is given by fr = 1/(2π√LC)=1/(2π√(0.1×10×10-6))=159.2 Hz. The relation between quality factor, resonant frequency and bandwidth is Q = fr/BW = 2πfrL/R = (6.28×159.2×0.1)/10=10.

## Network Theory MCQ Set 3

1. The resistance element __________ while going from the time domain to frequency domain.
a) does not change
b) increases
c) decreases
d) increases exponentially

Answer: a [Reason:] The s-domain equivalent circuit of a resistor is simply resistance of R ohms that carries a current I ampere seconds and has a terminal voltage V volts-seconds. The resistance element does not change while going from the time domain to the frequency domain.

2. The relation between current and voltage in case of inductor is?
a) v=Ldt/di
b) v=Ldi/dt
c) v=dt/di
d) v=di/dt

Answer: b [Reason:] Consider an inductor with an initial current Io. The time domain relation between current and voltage is v=Ldi/dt.

3. The s-domain equivalent of the inductor reduces to an inductor with impedance?
a) L
b) sL
c) s2L
d) s3L

Answer: b [Reason:] If the initial energy stored in the inductor is zero, the equivalent circuit of the inductor reduces to an inductor with impedance sL ohms.

4. The voltage and current in a capacitor are related as?
a) i=Cdt/dv
b) v=Cdv/dt
c) i=Cdv/dt
d) v=Cdt/dv

Answer: c [Reason:] Consider an initially charged capacitor and the initial voltage on the capacitor is Vo. The voltage current relation in the time domain is i=Cdv/dt.

5. The s-domain equivalent of the capacitor reduces to an capacitor with impedance?
a) sC
b) C
c) 1/C
d) 1/sC

Answer: d [Reason:] The s-domain equivalent of the capacitor can be derived for the charged capacitor and it reduces to an capacitor with impedance 1/sC.

6. From the circuit shown below, find the value of current in the loop. a) (V/R)/(s+1/RC)
b) (V/C)/(s+1/R)
c) (V/C)/(s+1/RC)
d) (V/R)/(s+1/R)

Answer: a [Reason:] Applying Kirchhoff’s law around the loop, we have V/s=1/sC I+RI. Solving above equation yields I=CV/(RCS+1)=(V/R)/(s+1/RC).

7. After taking the inverse transform of current in the circuit shown in question 6, the value of current is?
a) i=(V/C)e-t/R
b) i=(V/C)e-t/RC
c) i=(V/R)e-t/RC
d) i=(V/R)e-t/R

Answer: c [Reason:] We had assumed the capacitor is initially charged to Vo volts. By taking the inverse transform of the current, we get i=(V/R) e-t/RC.

8. The voltage across the resistor in the circuit shown in question 6 is?
a) Vet/R
b) Ve-t/RC
c) Ve-t/R
d) Vet/RC

Answer: b [Reason:] We can determine the voltage v by simply applying the ohm’s law from the circuit. And applying the Ohm’s law from the circuit v = Ri = Ve-t/RC.

9. The voltage across the resistor in the parallel circuit shown is? a) V/(s-1/R)
b) V/(s-1/RC)
c) V/(s+1/RC)
d) V/(s+1/C)

Answer: c [Reason:] The given circuit is converted to parallel equivalent circuit. By taking the node equation, we get v/R+sCv=CV. Solving the above equation, v=V/(s+1/RC).

10. Taking the inverse transform of the voltage across the resistor in the circuit shown in question 9.
a) Ve-t/τ
b) Vet/τ
c) Ve
d) Ve-tτ

Answer: a [Reason:] By taking the inverse transform, we get v=Ve-t/RC=Ve-t/τ, where τis the time constant and τ = RC. And v is the voltage across the resistor.

## Network Theory MCQ Set 4

1. Potential difference in electrical terminology is known as?
a) Voltage
b) Current
c) Resistance
d) Conductance

Answer: a [Reason:] Potential difference in electrical terminology is known as Voltage and is denoted either by V or v. It is expressed in terms of energy per unit charge.

2. The circuit in which current has a complete path to flow is called ______ circuit.
a) short
b) open
c) closed
d) open loop

Answer: c [Reason:] The circuit in which current has a complete path to flow is called closed circuit. When the current path is broken so that current cannot flow, the circuit is called an open circuit.

3. If the voltage-current characteristics is a straight line through the origin, then the element is said to be?
a) Linear element
b) Non-linear element
c) Unilateral element
d) Bilateral element

Answer: a [Reason:] If the voltage-current characteristic is a straight line through the origin, then the element is said to be Linear element. The difference in potential energy of charges is called Potential difference.

4. The voltage across R1 resistor in the circuit shown below is? a) 10
b) 5
c) 2.5
d) 1.25

Answer: b [Reason:] According to voltage divider rule, 10v is divide equally across resistors R1 and R2. So the voltage across R1 will be 5v.

5. The energy stored in the inductor is?
a) Li²/4
b) Li²/2
c) Li²
d) Li²/8

Answer: b [Reason:] The energy stored in the inductor the area under the power of the inductor and is given by W= ʃpdt = ʃLidi = Li²/2.

6. How many types of dependent or controlled sources are there?
a) 1
b) 2
c) 3
d) 4

Answer: d [Reason:] There are 4 dependent or controlled sources. They are VCVS(Voltage Controlled Voltage Source), VCCS(Voltage Controlled Current Source, CCVS(Current Controlled Voltage Source, CCCS(Current Controlled Current Source).

7. Find the voltage Vx in the given circuit. a) 10
b) 20
c) 30
d) 40

Answer: a [Reason:] From the circuit applying Kirchhoff’s voltage law, we can write 50= 15+ 10+ 15+Vx => Vx= 10V.

8. If the resistances 1Ω, 2Ω, 3Ω, 4Ω are parallel, then the equivalent resistance is?
a) 0.46Ω
b) 0.48Ω
c) 0.5Ω
d) 0.52Ω

Answer: b [Reason:] The equivalent resistance 1/Rt= (1/R1)+(1/R2)+(1/R3)+(1/R4). And R1, R2, R3, R4 are 1Ω, 2Ω, 3Ω, 4Ω respectively. => Rt= 0.48Ω.

9. Find total current(mA) in the circuit. a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] R2 is parallel to R3. So equivalent resistance of R2 and R3 is 1K. The total resistance in the circuit is (1+1+1)K= 3K.Current in the circuit is 3V/3KΩ= 1mA.

10. If the resistances 3Ω, 5Ω, 7Ω, 9Ω are in series, then their equivalent resistance(Ω) is?
a) 9
b) 20
c) 24
d) 32

Answer: c [Reason:] If the resistances are in series, then equivalent resistance is the sum of all the resistances that are in series. Equivalent resistance is (3+5+7+9)Ω= 24Ω.

## Network Theory MCQ Set 5

1. The solution of differential equations for networks is of the form?
a) i(t)=Kn e⁡(sn t)
b) i(t)=Kn e⁡(-sn t)
c) i(t)=-Kn e⁡(-sn t)
d) i(t)=-Kn e⁡(sn t)

Answer: a [Reason:] The solution of differential equations for networks is of the form i(t)=Kn e⁡(sn t) where Sn is a complex number which is a root of the characteristic equation.

2. The real part of the complex frequency is called?
b) neper frequency
c) sampling frequency
d) angular frequency

Answer: b [Reason:] The complex number consists of two parts,the real part and the imaginary part. The real part of the complex frequency is called neper frequency.

3. The imaginary part of the complex frequency is called?
a) angular frequency
b) sampling frequency
c) neper frequency

Answer: d [Reason:] The complex number consists of two parts, the real part of the complex frequency is called radian frequency. The radian frequency is expressed in radian/sec and is related to the frequency or the periodic time.

4. The ratio of transform voltage to the transform current is defined as _________ of the resistor.
a) transform voltage
b) transform current
c) transform impedance

Answer: c [Reason:] Transform impedance of the resistor is defined as the ratio of transform voltage to the transform current and is expressed as ZR(s) = VR(s)/IR(s) =R.

5. The ratio of transform current to the transform voltage is defined as ________ of the resistor.
b) transform impedance
c) transform current
d) transform voltage

Answer: a [Reason:] Transform admittance of the resistor is defined as the ratio of transform current to the transform voltage and it is also defined as the reciprocal of transform impedance. YR(s) = IR(s)/VR(s) =G.

6. The transform impedance of the inductor is?
a) L
b) 1/L
c) sL
d) 1/sL

Answer: c [Reason:] Considering the sum of the transform voltage and the initial current voltage as V1(s) we have the transform impedance of the inductor. The transform impedance of the inductor is ZL(s) = V1(s)/IL(s) = sL.

7. The transform admittance of the inductor is?
a) 1/sL
b) sL
c) 1/L
d) L

Answer: a [Reason:] The transform admittance of the inductor is YL(s) = I1(s)/VL(s) = 1/sL where I1(s) is the total transform current through the inductor L.

8. The equivalent transform circuit contains an admittance of value ____ and equivalent transform current source.
a) 1/L
b) 1/sL
c) L
d) sL

Answer: b [Reason:] The time domain representation of inductor L has initial current iL(0+). The equivalent transform circuit contains an admittance of value 1/sL and equivalent transform current source.

9. The transform impedance of the capacitor is?
a) C
b) 1/C
c) sC
d) 1/sC

Answer: d [Reason:] The transform impedance of the capacitor is the ratio of the transform voltage V1(s) to the transform current IC(s) and is ZC(s) = 1/Cs.

10. The transform admittance of the capacitor is?
a) 1/sC
b) sC
c) 1/C
d) C