## Network Theory MCQ Set 1

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?

a) i_{c} = ce^{-t/RC}

b) i_{c} = ce^{t/RC}

c) i_{c} = ce^{-t/RC}

d) i_{c} = ce^{t/RC}

### View Answer

_{c}= ce

^{-t/RC}.

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?

a) i_{p} = V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ+tan^{-1}(1/ωRC))

b) i_{p}= -V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ-tan^{-1}(1/ωRC))

c) i_{p} = V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ-tan^{-1}(1/ωRC))

d) i_{p} = -V/√(R^{2}+(1/ωC)^{2} ) cos(ωt+θ+tan^{-1}(1/ωRC))

### View Answer

_{p}= V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ+tan

^{-1}(1/ωRC)).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = V/R cosθ+V/√(R^{2}+(1/(ωC))^{2} ) cos(θ+tan^{-1}(1/ωRC))

b) c = V/R cosθ+V/√(R^{2}+(1/(ωC))^{2} ) cos(θ-tan^{-1}(1/ωRC))

c) c = V/R cosθ-V/√(R^{2}+(1/(ωC))^{2} ) cos(θ-tan^{-1}(1/ωRC))

d) c = V/R cosθ-V/√(R^{2}+(1/(ωC))^{2} ) cos(θ+tan^{-1}(1/ωRC))

### View Answer

^{2}+(1/(ωC))

^{2}) cos(θ+tan

^{-1}(1/ωRC)).

4. The complete solution of the current in the sinusoidal response of R-C circuit is?

a) i = e^{-t/RC}[V/R cosθ+V/√(R^{2}+(1/(ωC))^{2}) cos(θ+tan^{-1}(1/ωRC))+V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

b) i = e^{-t/RC}[V/R cosθ-V/√(R^{2}+(1/ωC)^{2}) cos(θ+tan^{-1}(1/ωRC))-V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

c) i = e^{-t/RC}[V/R cosθ+V/√(R^{2}+(1/ωC)^{2}) cos(θ+tan^{-1}(1/ωRC))-V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

d) i = e^{-t/RC}[V/R cosθ-V/√(R^{2}+(1/(ωC))^{2}) cos(θ+tan^{-1}(1/ωRC))+V/√(R^{2}+(1/ωC)^{2}) cos(ωt+θ+tan^{-1}(1/ωRC)].

### View Answer

^{-t/RC}[V/R cosθ-V/√(R

^{2}+(1/(ωC))

^{2}) cos(θ+tan

^{-1}(1/ωRC))+V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ+tan

^{-1}(1/ωRC)).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complementary function of the solution of ‘i’ is?

a) i_{c} = c exp (-t/10^{-10})

b) i_{c} = c exp(-t/10^{10})

c) i_{c} = c exp (-t/10^{-5})

d) i_{c} = c exp (-t/10^{5})

### View Answer

^{-5}) )i=-500sin(1000t+π/4). The complementary function is i

_{c}= c exp (-t/10

^{-5}).

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.

a) i_{p} = (4.99×10^{-3}) cos(100t+π/4-89.94^{o})

b) i_{p} = (4.99×10^{-3}) cos(100t-π/4-89.94^{o})

c) i_{p} = (4.99×10^{-3}) cos(100t-π/4+89.94^{o})

d) i_{p} = (4.99×10^{-3}) cos(100t+π/4+89.94^{o})

### View Answer

_{p}= A cos (ωt + θ) + B sin (ωt + θ) we get i

_{p}= V/√(R

^{2}+(1/ωC)

^{2}) cos(ωt+θ-tan

^{-1}(1/ωRC)) where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get i

_{p}= (4.99×10

^{-3}) cos(100t+π/4+89.94

^{o}).

7. The current flowing in the circuit at t = 0 in the question 5 is?

a) 1.53

b) 2.53

c) 3.53

d) 4.53

### View Answer

8. The complete solution of ‘i’ from the information provided in the question 5.

a) i = c exp (-t/10^{-5}) – (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

b) i = c exp (-t/10^{-5}) + (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

c) i = -c exp(-t/10^{-5}) + (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

d) i = -c exp(-t/10^{-5}) – (4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

### View Answer

^{-5}) + (4.99×10

^{-3}) cos(100t+π/2+89.94

^{o}).

9. The value of c in the complementary function of ‘i’ in the question 5 is?

a) c = (3.53-4.99×10^{-3}) cos(π/4+89.94^{o} )

b) c = (3.53+4.99×10^{-3}) cos(π/4+89.94^{o} )

c) c = (3.53+4.99×10^{-3}) cos(π/4-89.94^{o} )

d) c = (3.53-4.99×10^{-3}) cos(π/4-89.94^{o})

### View Answer

^{-3}) cos(π/4+89.94

^{o}).

10. The complete solution of ‘i’ in the question 5 is?

a) i = [(3.53-4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)+4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

b) i = [(3.53+4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)+4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

c) i = [(3.53+4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)-4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

d) i = [(3.53-4.99×10^{-3})cos(π/4+89.94^{o})] exp(-t/0.00001)-4.99×10^{-3}) cos(100t+π/2+89.94^{o} )

### View Answer

^{-3})cos(π/4+89.94

^{o})] exp(-t/0.00001)+4.99×10

^{-3}) cos(100t+π/2+89.94

^{o}).

## Network Theory MCQ Set 2

1. The Laplace transform of kf(t) is?

a) F(s)

b) kF(s)

c) F(s)/k

d) k^{2} F(s)

### View Answer

2. The Laplace transform of f_{1} (t) + f_{2} (t) is?

a) F_{1}(s) + F_{2}(s)

b) F_{1}(s) – F_{2}(s)

c) F_{1}(s) – 2F_{2}(s)

d) F_{1}(s) + 2F_{2}(s)

### View Answer

_{1}(t) + f

_{2}(t)) = F

_{1}(s) + F

_{2}(s).

3. Find the Laplace transform of the function f (t) = 4t^{3} + t^{2} -6t +7.

a) 24/s^{4} +2/s^{3} +6/s^{2} +7/s

b) 24/s^{4} -2/s^{3} -6/s^{2} +7/s

c) 24/s^{4} +2/s^{3} -6/s^{2} +7/s

d) 24/s^{4} -2/s^{3} +6/s^{2} +7/s

### View Answer

^{3}+ T

^{2}-6t +7) = 4L (t

^{3}) + L( t

^{2})-6L (t) +7L(1) = 4×3!/s

^{4}+2!/s

^{3}-6 ( 1!)/( s

^{2})+71/s = 24/s

^{4}+2/s

^{3}-6/s

^{2}+7/s.

4. Find the Laplace transform of the function f(t) = cos2t.

a) (2s^{2}+4)/2s(s^{2}-4)

b) (2s^{2}-4)/2s(s^{2}-4)

c) (2s^{2}-4)/2s(s^{2}+4)

d) (2s^{2}+4)/2s(s^{2}+4)

### View Answer

^{2}+4)/2s(s

^{2}+4).

5. Find the Laplace transform of the function f (t) = 3t^{4} -2t^{3} +4e^{-3t} -2sin5t +3cos2t.

a) 72/s^{5} -12/s^{4} +4/(s+3)+10/(s^{2}+25)+3s/(s^{2}+4)

b) 72/s^{5} -12/s^{4} +4/(s+3)-10/(s^{2}+25)+3s/(s^{2}+4)

c) 72/s^{5} -12/s^{4} -4/(s+3)+10/(s^{2}+25)+3s/(s^{2}+4)

d) 72/s^{5} -12/s^{4} -4/(s+3)-10/(s^{2}+25)+3s/(s^{2}+4)

### View Answer

^{4}-2t

^{3 }+4e

^{-3t}-2sin5t +3cos2t) =3 L (t

^{4})-2L (t

^{3})+4L (e

^{-3t})-2L (sin5t) +3L (cos2t) = 72/s

^{5}) -12/s

^{4}+4/(s+3)-10/(s

^{2}+25)+3s/(s

^{2}+4).

6. Find the Laplace transform of e^{at}sinbt.

a) b/((s-a)^{2}+b^{2} )

b) b/((s+a)^{2}+b^{2} )

c) b/((s+a)^{2}-b^{2} )

d) b/((s-a)^{2}-b^{2} )

### View Answer

^{2}+b

^{2}). So the Laplace transform of e

^{at}sinbt is L(exp(at) sinbt)=b/((s-a)

^{2}+b

^{2}).

7. Find the Laplace transform of (t + 2)^{2} e^{t}.

a) 2/(s-1)^{3} -2/(s-1)^{2} +4/(s-1)

b) 2/(s-1)^{3} -2/(s-1)^{2} -4/(s-1)

c) 2/(s-1)^{3} +2/(s-1)^{2} +4/(s-1)

d) 2/(s-1)^{3} +2/(s-1)^{2} -4/(s-1)

### View Answer

^{2}+2t+4 is L(t

^{2}+2t+4)=2/(s)

^{3}+2/(s)

^{2}+4/s. So the Laplace transform of (t + 2)

^{2}e

^{t}is L((t + 2)

^{2}e

^{t}) = 2/(s-1)

^{3}+2/(s-1)

^{2}+4/(s-1).

8. Find the Laplace transform of ramp function r (t) = t.

a) 1/s

b) 1/s^{2}

c) 1/s^{3}

d) 1/s^{4}

### View Answer

9.Find the Laplace transform of the function f (t) = tsin2t.

a) 4s/(s^{2}+4)^{2}

b) -4s/(s^{2}+4)^{2}

c) -4s/(s^{2}-4)^{2}

d) 4s/(s^{2}-4)^{2}

### View Answer

^{2}+4). So the Laplace transform of the function f (t) = tsin2t is L(tsin2t)=-d/ds [2/(s

^{2}+4)]= 4s/(s

^{2}+4)

^{2}.

10.If u (t) = 1 for t >= 0 and u (t) = 0 for t < 0, determine the Laplace transform of [u (t) – u (t – a)].

a) 1/s(1+e^{(-as)})

b) 1/s(1-e^{(-as)})

c) 1/s(1+e^{as})

d) 1/s(1-e^{as})

### View Answer

^{(-as)}1/s = 1/s (1-e

^{(-as)}).

## Network Theory MCQ Set 3

1.A signal generator supplies a sine wave of 20V, 5 kHz to the circuit as shown. Determine total current IT.

a) 0.21∠33⁰

b) 0.22∠33⁰

c) 0.23∠33⁰

d) 0.24∠33⁰

### View Answer

^{3}×0.2×10

^{-6})=159.2Ω.Current in the resistance branch I

_{R}=V

_{S}/R=20/100 = 0.2A.Current in capacitive branch I

_{C}=V

_{S}/X

_{C}= 20/159.2 = 0.126A.Total current I

_{T}= (I

_{R}+j I

_{C}) A = (0.2+j0.13) A. In polar form, I

_{T}= 0.24∠33⁰ A.

2. Find the phase angle in the circuit shown above.

a) 31⁰

b) 32⁰

c) 33⁰

d) 34⁰

### View Answer

_{T}= 0.24∠33⁰ A, the phase angle between applied voltage and current is 33⁰. The phase angle between the applied voltage and total current is 33⁰.

3. Determine the total impedance in the circuit.

a) 73.3∠33⁰

b) 83.3∠-33⁰

c) 83.3∠33⁰

d) 73.3∠-33⁰

### View Answer

_{S}/I

_{T}= 20∠0⁰ / 0.24∠33⁰ = 83.3∠-33⁰Ω. The phase angle indicates that the total line current is 0.24 A and leads the voltage by 33⁰.

4. A 50Ω resistor is connected in parallel with an inductive reactance of 30Ω. A 20V signal is applied to the circuit. Find the line current in the circuit.

a) 0.77∠-58.8⁰

b) 0.77∠58.8⁰

c) 0.88∠-58.8⁰

d) 0.88∠58.8⁰

### View Answer

_{R}= V

_{S}/R = 20∠0⁰/50∠0⁰=0.4A. Current in the inductive branch is I

_{L}= V

_{S}/X

_{L}= 20∠0⁰/30∠90⁰= 0.66∠-90⁰. Total current is I

_{T}= 0.4-j0.66 = 0.77∠-58.8⁰.

5. Determine total impedance in the circuit shown above.

a) 25∠-58.8⁰

b) 25∠-58.8⁰.

c) 26∠-58.8⁰.

d) 26∠58.8⁰.

### View Answer

_{S}/I

_{T}= 20∠0⁰ / 0.77∠-58.8⁰ = 25.97∠-58.8⁰Ω. So the total impedance in the circuit shown is 25.97∠-58.8⁰Ω.

6. Determine Z in the figure shown below.

a) 26∠-20.5⁰

b) 26∠20.5⁰

c) 25∠-20.5⁰

d) 25∠20.5⁰

### View Answer

_{L}=6.28 x 50 x 0.1 = 31.42Ω. In figure the 10Ω resistance is in series with the parallel combination of 20Ω and j31.42Ω. Z

_{T}= 10+ (20)(j31.42)/(20+j31.42)=24.23+j9.06=25.87∠20.5⁰

7. Find I_{T} in the figure shown above.

a) 0.66∠-20.5⁰

b) 0.66∠20.5⁰

c) 0.77∠20.5⁰

d) 0.77∠-20.5⁰

### View Answer

_{T}= V

_{S}/Z

_{T}= 20/25.87∠20.5

^{o}=0.77∠-20.5

^{o}.

8. Find the phase angle θ in the circuit shown above.

a) 20.5⁰

b) 20⁰

c) 19.5⁰

d) 19⁰

### View Answer

_{T}= 0.77∠-20.5

^{o}, the phase angle θ between the voltage and the current in the circuit is 20.5⁰.

9. Find the impedance in the circuit shown below.

a) 25

b) 26

c) 27

d) 28

### View Answer

_{C}= 1/2πfC = 1/(6.28×50×100×10

^{-6})=31.83Ω. Capacitive susceptance B

_{C}= 1/X

_{C}= 1/31.83 = 0.031S. Conductance G=1/R = 1/50 = 0.02S. Total admittance Y=√(G

^{2}+B

_{c}

^{2})=√(0.02

^{2}+0.031

^{2})=0.037S. Total impedance Z = 1/Y = 1/0.037 = 27.02Ω.

10. Determine the phase angle in the circuit shown above.

a) 56⁰

b) 56.5⁰

c) 57.5⁰

d) 57⁰

### View Answer

^{-1}(R/X

_{c}). Resistance R = 50Ω and capacitive reactance X

_{c}= 31.83Ω. So the phase angle in the circuit = tan

^{-1}(50/31.83)=57.52⁰.

## Network Theory MCQ Set 4

1. For the circuit shown below, determine its resonant frequency.

a) 6.12

b) 7.12

c) 8.12

d) 9.12

### View Answer

_{r}= 1/(2π√LC). Given L = 5H and C = 100uf. On substituting the given values in the equation we get resonant frequency = 1/(2π√(5×100×10

^{-6})) = 7.12 Hz.

2. Find the quality factor of the circuit shown in the question 1.

a) 2.24

b) 3.34

c) 4.44

d) 5.54

### View Answer

_{L}/R = 2πf

_{r}L/R. Given f = 7.12 Hz and L = 5H and R = 100. On substituting the given values in the equation we get the quality factor = (6.28×7.12×5)/100 = 2.24.

3. Find the bandwidth of the circuit shown in the question 1.

a) 1

b) 2

c) 3

d) 4

### View Answer

_{r}/Q. we obtained f

_{r}= 7.12 Hz and Q = 2.24. On substituting the given values in the equation we get the bandwidth = 7.12/2.24 = 3.178Hz.

4. The magnification in resonance considering the voltage across inductor is?

a) V/V_{L}

b) V_{L}/V

c) V x V_{L}

d) V_{L}

### View Answer

_{L}/V.

5. Considering the voltage across the capacitor, the magnification in resonance is?

a) V_{C}

b) V x V_{C}

c) V_{C}/V

d) V/V_{C}

### View Answer

_{C}/V.

6. The value of ω_{r} in parallel resonant circuit is?

a) 1/(2√LC)

b) 1/√LC

c) 1/(π√LC)

d) 1/(2π√LC)

### View Answer

_{L}= X

_{L}. The frequency at which the resonance occurs is called the resonant frequency. The value of ω

_{r}in parallel resonant circuit is ω

_{r}= 1/√LC.

7. The expression of resonant frequency for parallel resonant circuit is?

a) 1/(2π√LC)

b) 1/(π√LC)

c) 1/(2√LC)

d) 1/√LC

### View Answer

_{L}= X

_{L}. The expression of resonant frequency for parallel resonant circuit is f

_{r}= 1/(2π√LC).

8. Find the resonant frequency in the ideal parallel LC circuit shown in the figure.

a) 7.118

b) 71.18

c) 711.8

d) 7118

### View Answer

_{r}= 1/(2π√LC). Given L = 50mH and C = 0.01uF. On substituting the given values in the equation we get the resonant frequency = 1/(2π√(50×10

^{-3})×0.01×10

^{-6}) ))=7117.6 Hz.

9. If the value of Q of the circuit is high, then its effect on bandwidth is?

a) large bandwidth

b) small bandwidth

c) no effect on bandwidth

d) first increases and then decreases

### View Answer

10. If in a circuit, if Q value is decreased then it will cause?

a) small bandwidth

b) no effect on bandwidth

c) first increases and then decreases

d) large bandwidth

### View Answer

## Network Theory MCQ Set 5

1. The phase difference between voltage and current in case of resistor is?

a) in phase

b) out of phase

c) 45⁰ out of phase

d) 90⁰ out of phase

### View Answer

2. In case of inductor, the voltage?

a) is in phase with the current

b) lags behind the current by 90⁰

c) leads the current by 90⁰

d) is out of phase with the current

### View Answer

3. For inductor, the current?

a) is out of phase with the current

b) leads the current by 90⁰

c) is in phase with the current

d) lags behind the current by 90⁰

### View Answer

4. The value of inductance reactance is?

a) R

b) ωL

c) 1/ ωL

d) ωC

### View Answer

5. The current in pure capacitor?

a) lags behind the voltage by 90⁰

b) is in phase with the voltage

c) lags behind the voltage by 45⁰

d) leads the voltage by 90⁰

### View Answer

6. In case of pure capacitor, the voltage?

a) leads the voltage by 90⁰

b) lags behind the voltage by 45⁰

c) lags behind the voltage by 90⁰

d) is in phase with the voltage

### View Answer

7. The impedance value of a pure capacitor is?

a) ωC

b) 1/ ωC

c) ωL

d) R

### View Answer

8. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. How much is first sine wave shifted in phase from the 0⁰ reference?

a) leads reference angle by 10⁰

b) lags reference angle by 10⁰

c) leads reference angle by 15⁰

d) lags reference angle by 15⁰

### View Answer

9. How much is second sine wave shifted in phase from the 0⁰ reference?

a) lags reference angle by 15⁰

b) leads reference angle by 15⁰

c) lags reference angle by 10⁰

d) leads reference angle by 10⁰

### View Answer

10. What is the phase angle between two sine waves mentioned above?

a) 10⁰

b) 15⁰

c) 20⁰

d) 25⁰