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## Network Theory MCQ Set 1

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?
a) ic = ce-t/RC
b) ic = cet/RC
c) ic = ce-t/RC
d) ic = cet/RC

Answer: a [Reason:] From the R-c circuit, we get the characteristic equation as (D+1/RC)i=-Vω/R sin⁡(ωt+θ). The complementary function of the solution i is ic = ce-t/RC.

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?
a) ip = V/√(R2+(1/ωC)2 ) cos⁡(ωt+θ+tan-1(1/ωRC))
b) ip= -V/√(R2+(1/ωC)2 ) cos⁡(ωt+θ-tan-1(1/ωRC))
c) ip = V/√(R2+(1/ωC)2 ) cos⁡(ωt+θ-tan-1(1/ωRC))
d) ip = -V/√(R2+(1/ωC)2 ) cos⁡(ωt+θ+tan-1(1/ωRC))

Answer: a [Reason:] The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is ip = V/√(R2+(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/√(R2+(1/(ωC))2 ) cos⁡(θ+tan-1(1/ωRC))
b) c = V/R cosθ+V/√(R2+(1/(ωC))2 ) cos⁡(θ-tan-1(1/ωRC))
c) c = V/R cosθ-V/√(R2+(1/(ωC))2 ) cos⁡(θ-tan-1(1/ωRC))
d) c = V/R cosθ-V/√(R2+(1/(ωC))2 ) cos⁡(θ+tan-1(1/ωRC))

Answer: d [Reason:] Since the capacitor does not allow sudden changes in voltages, at t = 0, i =V/R cosθ. So, c = V/R cosθ-V/√(R2+(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/√(R2+(1/(ωC))2) cos⁡(θ+tan-1(1/ωRC))+V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC)].
b) i = e-t/RC[V/R cosθ-V/√(R2+(1/ωC)2) cos⁡(θ+tan-1(1/ωRC))-V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC)].
c) i = e-t/RC[V/R cosθ+V/√(R2+(1/ωC)2) cos⁡(θ+tan-1(1/ωRC))-V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC)].
d) i = e-t/RC[V/R cosθ-V/√(R2+(1/(ωC))2) cos⁡(θ+tan-1(1/ωRC))+V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC)].

Answer: d [Reason:] The complete solution for the current becomes i = e-t/RC[V/R cosθ-V/√(R2+(1/(ωC))2) cos⁡(θ+tan-1(1/ωRC))+V/√(R2+(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) ic = c exp (-t/10-10)
b) ic = c exp(-t/1010)
c) ic = c exp (-t/10-5)
d) ic = c exp (-t/105)

Answer: c [Reason:] By applying Kirchhoff’s voltage law to the circuit, we have (D+1/10-5) )i=-500sin⁡(1000t+π/4). The complementary function is ic = c exp (-t/10-5).

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.
a) ip = (4.99×10-3) cos⁡(100t+π/4-89.94o)
b) ip = (4.99×10-3) cos⁡(100t-π/4-89.94o)
c) ip = (4.99×10-3) cos⁡(100t-π/4+89.94o)
d) ip = (4.99×10-3) cos⁡(100t+π/4+89.94o)

Answer: d [Reason:] Assuming particular integral as ip = A cos (ωt + θ) + B sin (ωt + θ) we get ip = V/√(R2+(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 ip = (4.99×10-3) cos⁡(100t+π/4+89.94o).

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

Answer: c [Reason:] At t = 0 that is initially current flowing through the circuit is i = V/R cosθ = (50/10)cos(π/4) = 3.53A.

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

Answer: b [Reason:] The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = c exp (-t/10-5) + (4.99×10-3) cos⁡(100t+π/2+89.94o ).

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+89.94o )
b) c = (3.53+4.99×10-3) cos⁡(π/4+89.94o )
c) c = (3.53+4.99×10-3) cos⁡(π/4-89.94o )
d) c = (3.53-4.99×10-3) cos⁡(π/4-89.94o)

Answer: a [Reason:] At t = 0, the current flowing through the circuit is 3.53A. So, c = (3.53-4.99×10-3) cos⁡(π/4+89.94o ).

10. The complete solution of ‘i’ in the question 5 is?
a) i = [(3.53-4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o )
b) i = [(3.53+4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o )
c) i = [(3.53+4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)-4.99×10-3) cos⁡(100t+π/2+89.94o )
d) i = [(3.53-4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)-4.99×10-3) cos⁡(100t+π/2+89.94o )

Answer: a [Reason:] The complete solution for the current is the sum of the complementary function and the particular integral. So, i = [(3.53-4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o).

## Network Theory MCQ Set 2

1. The Laplace transform of kf(t) is?
a) F(s)
b) kF(s)
c) F(s)/k
d) k2 F(s)

Answer: b [Reason:] Operational transforms indicate how mathematical operations performed in either f(t) or F(s) are converted into the opposite domain. Linearity property states that L (kf (t)) = kF (s).

2. The Laplace transform of f1 (t) + f2 (t) is?
a) F1(s) + F2(s)
b) F1(s) – F2(s)
c) F1(s) – 2F2(s)
d) F1(s) + 2F2(s)

Answer: a [Reason:] Addition or subtraction in time domain translates into addition or subtraction in frequency domain. L (f1 (t) + f2 (t)) = F1(s) + F2(s).

3. Find the Laplace transform of the function f (t) = 4t3 + t2 -6t +7.
a) 24/s4 +2/s3 +6/s2 +7/s
b) 24/s4 -2/s3 -6/s2 +7/s
c) 24/s4 +2/s3 -6/s2 +7/s
d) 24/s4 -2/s3 +6/s2 +7/s

Answer: c [Reason:] L (4t3 + T2 -6t +7) = 4L (t3 ) + L( t2 )-6L (t) +7L(1) = 4×3!/s4 +2!/s3 -6 ( 1!)/( s2 )+71/s = 24/s4 +2/s3 -6/s2 +7/s.

4. Find the Laplace transform of the function f(t) = cos2t.
a) (2s2+4)/2s(s2-4)
b) (2s2-4)/2s(s2-4)
c) (2s2-4)/2s(s2+4)
d) (2s2+4)/2s(s2+4)

Answer: d [Reason:] The Laplace transform of the function f(t) = cos2t is L (cos2t) = L((1+cos2t)/2) = L(1/2)+L(cos2t/2) = 1/2[L(1)+L(cos2t)] = (2s2+4)/2s(s2+4).

5. Find the Laplace transform of the function f (t) = 3t4 -2t3 +4e-3t -2sin5t +3cos2t.
a) 72/s5 -12/s4 +4/(s+3)+10/(s2+25)+3s/(s2+4)
b) 72/s5 -12/s4 +4/(s+3)-10/(s2+25)+3s/(s2+4)
c) 72/s5 -12/s4 -4/(s+3)+10/(s2+25)+3s/(s2+4)
d) 72/s5 -12/s4 -4/(s+3)-10/(s2+25)+3s/(s2+4)

Answer: b [Reason:] L (3t4 -2t3 +4e-3t -2sin5t +3cos2t) =3 L (t4 )-2L (t3 )+4L (e-3t )-2L (sin5t) +3L (cos2t) = 72/s5 ) -12/s4 +4/(s+3)-10/(s2+25)+3s/(s2+4).

6. Find the Laplace transform of eatsinbt.
a) b/((s-a)2+b2 )
b) b/((s+a)2+b2 )
c) b/((s+a)2-b2 )
d) b/((s-a)2-b2 )

Answer: a [Reason:] The Laplace transform of sinbt is L(sinbt)=b/(s2+b2 ). So the Laplace transform of eatsinbt is L(exp(at) sinbt)=b/((s-a)2+b2 ).

7. Find the Laplace transform of (t + 2)2 et.
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)

Answer: c [Reason:] The Laplace transform of t2+2t+4 is L(t2+2t+4)=2/(s)3 +2/(s)2+4/s. So the Laplace transform of (t + 2)2 et is L((t + 2)2 et) = 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/s2
c) 1/s3
d) 1/s4

Answer: b [Reason:] We know .

9.Find the Laplace transform of the function f (t) = tsin2t.
a) 4s/(s2+4)2
b) -4s/(s2+4)2
c) -4s/(s2-4)2
d) 4s/(s2-4)2

Answer: a [Reason:] The Laplace transform of the function of sin2t is L(sin2t)=2/(s2+4). So the Laplace transform of the function f (t) = tsin2t is L(tsin2t)=-d/ds [2/(s2+4)]= 4s/(s2+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+eas)
d) 1/s(1-eas)

Answer: b [Reason:] As u (t) = 1 for t >= 0 and u (t) = 0 for t < 0, the Laplace transform of [u (t) – u (t – a)] is L[u (t)– u (t – a) ] = 1/s-e(-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

Answer: d [Reason:] Capacitive reactance = 1/(6.28×5×103×0.2×10-6)=159.2Ω.Current in the resistance branch IR=VS/R=20/100 = 0.2A.Current in capacitive branch IC=VS/XC = 20/159.2 = 0.126A.Total current IT = (IR+j IC) A = (0.2+j0.13) A. In polar form, IT = 0.24∠33⁰ A.

2. Find the phase angle in the circuit shown above.
a) 31⁰
b) 32⁰
c) 33⁰
d) 34⁰

Answer: c [Reason:] We obtained IT = 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

Answer: b [Reason:] Z = VS/IT = 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

Answer: a [Reason:] Since the voltage across each element is the same as the applied voltage, the current in resistive branch is IR= VS/R = 20∠0⁰/50∠0⁰=0.4A. Current in the inductive branch is IL= VS/XL= 20∠0⁰/30∠90⁰= 0.66∠-90⁰. Total current is IT = 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.

Answer: c [Reason:] The current lags behind the voltage by 58.8⁰. Total impedance Z = VS/IT = 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

Answer: b [Reason:] First the inductive reactance is calculated. XL=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. ZT = 10+ (20)(j31.42)/(20+j31.42)=24.23+j9.06=25.87∠20.5⁰

7. Find IT 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

Answer: d [Reason:] The current lags behind the applied voltage by 20.5⁰. Total current IT = VS/ZT = 20/25.87∠20.5o=0.77∠-20.5o.

8. Find the phase angle θ in the circuit shown above.
a) 20.5⁰
b) 20⁰
c) 19.5⁰
d) 19⁰

Answer: a [Reason:] As IT = 0.77∠-20.5o, 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

Answer: c [Reason:] Capacitive reactance XC = 1/2πfC = 1/(6.28×50×100×10-6)=31.83Ω. Capacitive susceptance BC = 1/XC = 1/31.83 = 0.031S. Conductance G=1/R = 1/50 = 0.02S. Total admittance Y=√(G2+Bc2 )=√(0.022+0.0312 )=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

Answer: c [Reason:] Phase angle θ=tan-1(R/Xc). Resistance R = 50Ω and capacitive reactance Xc = 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

Answer: b [Reason:] The resonant frequency of the circuit is fr = 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

Answer: a [Reason:] The quality factor of the circuit is Q = XL/R = 2πfrL/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

Answer: c [Reason:] The bandwidth of the circuit is BW = fr/Q. we obtained fr = 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/VL
b) VL/V
c) V x VL
d) VL

Answer: b [Reason:] The ratio of voltage across inductor to the voltage applied at resonance can be defined as magnification. The magnification in resonance considering the voltage across inductor is Q = VL/V.

5. Considering the voltage across the capacitor, the magnification in resonance is?
a) VC
b) V x VC
c) VC/V
d) V/VC

Answer: c [Reason:] The ratio of voltage across capacitor to the voltage applied at resonance can be defined as magnification. Considering the voltage across the capacitor, the magnification in resonance is Q = VC/V.

6. The value of ωr in parallel resonant circuit is?

a) 1/(2√LC)
b) 1/√LC
c) 1/(π√LC)
d) 1/(2π√LC)

Answer: b [Reason:] Basically parallel resonance occurs when XL = XL. The frequency at which the resonance occurs is called the resonant frequency. The value of ωr in parallel resonant circuit is ωr = 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

Answer: a [Reason:] The condition for resonance occurs when XL = XL. The expression of resonant frequency for parallel resonant circuit is fr = 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

Answer: d [Reason:] The expression for resonant frequency is fr = 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

Answer: b [Reason:] If the value of Q of the circuit is high, then small bandwidth because bandwidth is inversely proportional to the quality factor.

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

Answer: d [Reason:] If in a circuit, if Q value is decreased then bandwidth increases and the bandwidth do not decrease.

## 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

Answer: a [Reason:] The phase difference between voltage and current in case of resistor is in phase. The amplitudes of the waveform may differ according to the value of resistance.

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

Answer: c [Reason:] In case of inductor, the voltage and current are out of phase. So, the voltage leads the current by 90⁰.

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

Answer: d [Reason:] For inductor, the voltage and current are out of phase. So, the current lags behind the current by 90⁰.

4. The value of inductance reactance is?
a) R
b) ωL
c) 1/ ωL
d) ωC

Answer: b [Reason:] The value of inductance reactance is ωL. Hence a pure inductor has an impedance whose value is ωL.

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

Answer: d [Reason:] In a capacitor, there exists a phase difference between current and voltage. The current in pure capacitor leads the voltage by 90⁰.

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

Answer: c [Reason:] In a capacitor, there exists a phase difference between current and voltage. In case of pure capacitor, the voltage lags behind the voltage by 90⁰.

7. The impedance value of a pure capacitor is?
a) ωC
b) 1/ ωC
c) ωL
d) R

Answer: b [Reason:] The impedance value of a pure capacitor is 1/ ωC. And the impedance of the capacitor is called capacitive reactance.

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

Answer: c [Reason:] As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. First sine wave leads reference angle by 15⁰ from the 0⁰ reference.

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