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## Network Theory MCQ Set 1

1. To check for the Reciprocity Theorem we consider ______ of response to excitation.
a) ratio
c) product
d) subtraction

Answer: a [Reason:] For the Reciprocity Theorem to satisfy the ratio of response to the excitation of the circuit should be equal to the ratio of response to excitation after the source is replaced.

2. For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be?
a) different
b) same
c) before source is replaced is greater than after the source is replaced
d) before source is replaced is less than after the source is replaced

Answer: b [Reason:] For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be same and if that condition satisfies the reciprocity theorem is valid for the given circuit.

3. The circuit which satisfies Reciprocity Theorem is called?
a) Short circuit
b) Open circuit
c) Linear circuit
d) Non-linear circuit

Answer: c [Reason:] The circuit which satisfies Reciprocity Theorem is called linear circuit. A linear circuit is an electronic circuit in which, for a sinusoidal input voltage of frequency f, any steady-state output of the circuit (the current through any component, or the voltage between any two points) is also sinusoidal with frequency f.

4. Find the current through the 2Ω(c-d) resistor in the circuit shown below.

a) 0.143
b) 1.43
c) 14.3
d) 143

Answer: b [Reason:] Total resistance in the circuit= 2+[3||(2+2│├|2) ]=3.5Ω. The current drawn by the circuit (It)=20/3.5=5.71Ω. The current drawn by 2Ω resistor = 1.43A.

5. The current drawn by 2Ω resistor (a-b) after the source is replaced in the question 4?
a) 143
b) 14.3
c) 1.43
d) 0.143

Answer: c [Reason:] The circuit after the source is replaced is Total resistance = 3.23Ω. The current drawn by the circuit (It)=20/3.23=6.19A. The current in branch a-b is 1.43A.

6. The circuit in the question 4 satisfies Reciprocity Theorem.
a) True
b) False

Answer: a [Reason:] The ratio of response to excitation before the source is replaced is equal to 0.0715. And the ratio of response to excitation before the source is replaced is equal to 0.0715. So, the circuit satisfies the Reciprocity theorem.

7. Find the current through 3Ω resistor in the circuit shown below.

a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] The 6Ω resistor is parallel to 3Ω resistor and the resultant is in series with 2Ω resistor. Total current from source = 12/(2+(6│|3) )=3A. Current through 3Ω resistor=3× 6/(6+3)=2A.

8. Find the current through 2Ω resistor after source is replaced in the question 7.
a) 4
b) 3
c) 2
d) 1

Answer: c [Reason:] The circuit after source is replaced is Total current from the source (It)=12/(3+(6│|2) )=2.67A. Current through 2Ω resistor=2.67× 6/(6+2)=2A.

9. The circuit shown in the question 7 satisfies the reciprocity theorem.
a) False
b) True

Answer: b [Reason:] The ratio of response to excitation before the source is replaced is equal to 0.167. And the ratio of response to excitation before the source is replaced is equal to 0.167. So, the circuit satisfies the Reciprocity theorem.

10. While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of?
a) voltage to voltage
b) current to current
c) voltage to current
d) None of the above

Answer: c [Reason:] While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of voltage to current or current to voltage.

## Network Theory MCQ Set 2

1. If we apply a sinusoidal input to RL circuit, the current in the circuit is __________ and the voltage across the elements is _______________
a) square, square
b) square, sinusoid
c) sinusoid, square
d) sinusoid, sinusoid

Answer: d [Reason:] If we apply a sinusoidal input to RL circuit, the current in the circuit is square and the voltage across the elements is sinusoid. In the analysis of the RL series circuit, we can find the impedance, current, phase angle and voltage drops.

2. The circuit shown below consists of a 1kΩ resistor connected in series with a 50mH coil, a 10V rms, 10 KHz signal is applied. Find impedance Z in rectangular form.

a) (1000+j0.05) Ω
b) (100+j0.5) Ω
c) (1000+j3140) Ω
d) (100+j3140) Ω

Answer: c [Reason:] Inductive Reactance XL= ωL = 2πfL = (6.28)(104)(50×10-3) =3140Ω. In rectangular form, total impedance Z = (1000+j3140) Ω.

3. Find the current I (mA) in the circuit shown above.
a) 3.03
b) 30.3
c) 303
d) 0.303

Answer: a [Reason:] Total impedance Z = (1000+j3140) Ω. Magnitude = 3295.4Ω Current I=Vs/Z = 10/3295.4=3.03mA.

4. Find the phase angle θ in the circuit shown above.
a) 62.33⁰
b) 72.33⁰
c) 82.33⁰
d) 92.33⁰

Answer: b [Reason:] Phase angle θ = tan-1⁡(XL/R). The values of XL, R are XL = 3140Ω and R = 1000Ω. On substituting the values in the equation, phase angle θ =tan-1⁡(3140/1000)=72.33⁰.

5. In the circuit shown above find the voltage across resistance.
a) 0.303
b) 303
c) 3.03
d) 30.3

Answer: c [Reason:] Voltage across resistance = IR. The values of I = 3.03 mA and R = 10000Ω. On substituting the values in the equation, the voltage across resistance = 3.03 x 10-3×1000 = 3.03V.

6. In the circuit shown above find voltage across inductive reactance.
a) 9.5
b) 10
c) 9
d) 10.5

Answer: a [Reason:] Voltage across inductor = IXL. The values of I = 3.03 mA and XL = 10000Ω. On substituting the values in the equation, the voltage across inductor =3.03×10-3×1000 = 9.51V.

7. Determine the source voltage if voltage across resistance is 70V and the voltage across inductor is 20V as shown in the figure.

a) 71
b) 72
c) 73
d) 74

Answer: c [Reason:] If voltage across resistance is 70V and the voltage across inductor is 20V, source voltage Vs=√(702+202 ) = 72.8≅73V.

8. Find the phase angle in the circuit shown above.
a) 15
b) 16
c) 17
d) 18

Answer: b [Reason:] The phase angle is the angle between the source voltage and the current. Phase angle θ=tan-1(VL/VR). The values of VL = 20V and VR = 70V. On substituting the values in the equation, phase angle in the circuit =tan-1(20/70)=15.94o≅ 16o.

9. An AC voltage source supplies a 500Hz, 10V rms signal to a 2kΩ resistor in series with a 0.1µF capacitor as shown in figure. Find the total impedance.

a) 3750.6Ω
b) 3760.6Ω
c) 3780.6Ω
d) 3790.6Ω

Answer: b [Reason:] The capacitive reactance XC = 1/2πfC =1/(6.28×500×0.1×10-6) )=3184.7Ω. In rectangular form, X = (2000-j3184.7)Ω. Magnitude = 3760.6Ω.

10. Determine the phase angle in the circuit shown above.
a) 58
b) 68
c) -58
d) -68

Answer: c [Reason:] The phase angle in the circuit is phase angle θ =tan-1⁡(-XC/R) =tan-1⁡((-3184.7)/2000)=-57.87o≅-58o.

11. Find the current I (mA) in the circuit shown above.
a) 2.66
b) 3.66
c) 4.66
d) 5.66

Answer: a [Reason:] The term current is the ratio of voltage to the impedance. The current I (mA) in the circuit is current I = VS / Z = 10/3760.6 = 2.66mA

12. Find the voltage across the capacitor in the circuit shown above.
a) 7
b) 7.5
c) 8
d) 8.5

Answer: d [Reason:] The voltage across the capacitor in the circuit is capacitor voltage = 2.66×10-3×3184.7 =8.47V.

13. Determine the voltage across the resistor in the circuit shown above.
a) 3
b) 4
c) 5
d) 6

Answer: c [Reason:] The voltage across the resistor in the circuit resistive voltage = 2.66×10-3×3184.7 =5.32V.

14. In the circuit shown below determine the total impedance.

a) 161
b) 162
c) 163
d) 164

Answer: b [Reason:] Reactance across capacitor = 1/(6.28×50×10×10-6) =318.5Ω. Reactance across inductor = 6.28×0.5×50=157Ω. In rectangular form, Z = (10+j157-j318.5) Ω = (10-j161.5)Ω. Magnitude=161.8Ω.

15. Find the current in the circuit shown above.
a) 0.1
b) 0.2
c) 0.3
d) 0.4

Answer: c [Reason:] The term current is the ratio of voltage to the impedance. The current in the circuit is current I=VS/Z = 50/161.8 = 0.3A.

## Network Theory MCQ Set 3

1. The circuit is said to be in resonance if the current is ____ with the applied voltage.
a) in phase
b) out of phase
c) 45⁰ out of phase
d) 90⁰ out of phase

Answer: a [Reason:] The circuit is said to be in resonance if the current is in phase with the applied voltage and not if the current is out of phase with the applied voltage. The study of resonance is very useful particularly in the area of communications.

2. In a series resonance circuit, series resonance occurs when?
a) XL = 1
b) XC = 1
c) XL = XC
d) XL = -XC

Answer: c [Reason:] In a series resonance circuit, series resonance occurs when capacitive reactance is equal to the inductive reactance that is XL = XC.

3. As XL = XC in a series resonance circuit, the impedance is_________
a) purely capacitive
b) purely inductive
c) purely resistive
d) capacitive and inductive

Answer: c [Reason:] As XL = XC in a series resonance circuit, the impedance is purely resistive. In a series RLC circuit the current lags behind or leads the applied voltage depending on the values of XL and XC.

4. At resonant frequency, the voltage across capacitor is _______ the voltage across inductor.
a) greater than
b) less than
c) greater than or equal to
d) equal to

Answer: d [Reason:] At resonant frequency, the voltage across capacitor is equal to the voltage across inductor. If one of the parameters of the series RLC circuit is varied in such a way that the current in the circuit is in phase with the applied voltage, then the circuit is said to be in resonance.

5. In series RLC circuit, the voltage across capacitor and inductor are ______ with each other.
a) in phase
b) 180⁰ out of phase
c) 90⁰ out of phase
d) 45⁰ out of phase

Answer: b [Reason:] In series RLC circuit, the voltage across capacitor and inductor are 180⁰ out of phase with each other. The frequency at which the resonance occurs is called resonant frequency.

6. The voltage across the LC combination in a series RLC circuit is?
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] Since the voltage across capacitor and inductor are 180⁰ out of phase with each other, the voltage across the LC combination in a series RLC circuit is 0V.

7. The expression of resonant frequency in a series resonant circuit is?
a) 1/(2π√C)
b) 1/( 2π√L)
c) 2π√LC
d) 1/(2π√LC)

Answer: d [Reason:] The expression of resonant frequency is resonant frequency = 1/(2π√LC). In a series RLC circuit resonance may be produced by varying the frequency, keeping L and C constant.

8. For the circuit shown in figure determine the capacitive reactance at resonance.

a) 15
b) 20
c) 25
d) 30

Answer: c [Reason:] We know at resonance, capacitive reactance is equal to the inductive reactance that is XL = XC. Given inductive reactance XL = 25. On substituting in the equation we get XC = 25Ω.

9. The value of the impedance at resonance in the circuit shown in the question 9 is?
a) 25
b) 50
c) 75
d) 100

Answer: b [Reason:] We know that at resonance the value of impedance at resonance. So Z = R. Given R = 50Ω. On substituting in the equation we get Z = 50Ω.

10. Determine the resonant frequency (kHz) for the circuit shown below.

a) 2.25
b) 22.5
c) 225
d) 2250

Answer: a [Reason:] The expression of resonant frequency is resonant frequency fr = 1/(2π√LC). Given L = 0.5mH and C = 10uF. On substituting in the equation we get resonant frequency fr = 1/(2π√(10×10-6)×0.5×10-3)) = 2.25kHz.

## Network Theory MCQ Set 4

1. In determining short circuit impedance parameters, among V1, V2, I1, I2, which of the following are dependent variables?
a) V1 and V2
b) I1 and I2
c) V1 and I2
d) I1 and V2

Answer: b [Reason:] In determining short circuit impedance parameters, among V1, V2, I1, I2; I1 and I2 are dependent variables and V1, V2 are independent variables i.e., dependent variables depend on independent variables.

2. In determining short circuit impedance parameters, among V1, V2, I1, I2, which of the following are independent variables?
a) I1 and V2
b) V1 and I2
c) I1 and I2
d) V1 and V2

Answer: d [Reason:] In determining short circuit impedance parameters, among V1, V2, I1, I2; V1 and V2 are independent variables and I1, I2 are dependent variables. Independent variables are the variables that do not depend on any other variable.

3. Which of the following expression is true in case of short circuit parameters?
a) I1 = Y11 V1 + Y12 V2
b) I1 = Y11 I1 + Y12 V2
c) V1 = Y11 I1 + Y12 V2
d) V1 = Y11 V1 + Y12 V2

Answer: a [Reason:] The expression relating the short circuit parameters Y11, Y12 and voltages V1, V2 and current is I1, is I1 = Y11 V1 + Y12 V2.

4. Which of the following expression is true in case of short circuit parameters?
a) I2 = Y21I1 + Y22 I2
b) V2 = Y21I1 + Y22 V2
c) I2 = Y21V1 + Y22 V2
d) I2 = Y21V1 + Y22 I2

Answer: c [Reason:] The expression relating the voltages V1, V2 and current is I2 and short circuit parameters Y11, Y12 is I2 = Y21V1 + Y22V2.

5. The parameters Y11, Y12, Y21, Y22 are called?
a) Open circuit impedance parameters
c) Inverse transmission parameters
d) Transmission parameters

Answer: b [Reason:] The parameters Y11, Y12, Y21, Y22 are called short circuit admittance parameters also called network functions as they are obtained by short circuiting port 1 or port 2.

6. Find the Y – parameter Y11 in the circuit shown below.

a) 2
b) 3/2
c) 1
d) 1/2

Answer: d [Reason:] After short circuiting b-b’, the equation will be V1 = (I1) x 2. We know Y11 = I1/V1. From the equation we get I1/V1 = 2. On substituting we get Y11 = 2 mho.

7. Find the Y – parameter Y21 in the circuit shown in question 6.
a) -1/4
b) 1/4
c) 1/2
d) -1/2

Answer: a [Reason:] After short circuiting b-b’, the equation will be -I2=I1 × 2/4=I1/2 and -I2= V1/4 and on solving and substituting we get Y21 =I2/V1=-1/4 mho.

8. Find the Y – parameter Y22 in the circuit shown in question 6.
a) 3/8
b) 5/8
c) 7/8
d) 9/8

Answer: b [Reason:] On short circuiting a-a’,we get Zeq = 8/5 Ω. V2=I2× 8/5. We know Y22 = I2/V2. We got I2/V1 = 5/8. ON substituting we get Y22 = 5/8 mho.

9. Find the Y – parameter Y12 in the circuit shown in question 6.
a) 1/2
b) -1/2
c) -1/4
d) 1/4

Answer: c [Reason:] Short circuiting a-a’, -I1= 2/5 I2 and I2= 5 V2/8. On solving -I1= 2/5×5/8 V2= V2/4. We know Y12 = I1/V2. We got I1/V2 = -1/4. So the value of Y12 will be -1/4 mho.

10. Which of the following equation is true in the circuit shown in question 6?
a) I1=0.5(V1)+0.25(V2)
b) I1=0.25(V1)+0.625(V2)
c) I1=-0.25(V1)+0.625(V2)
d) I1=0.5(V1)-0.25(V2)

Answer: d [Reason:] We got the admittance parameters as Y11 = 0.5, Y12 = -0.25, Y21 = -0.25, Y22 = 0.625. So the equations in terms of admittance parameters is I1=0.5(V1)-0.25(V2) and I2=-0.25(V1)+0.625(V2).

## Network Theory MCQ Set 5

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

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

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

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

3. The value of ‘c’ in complementary function of ‘i’ is?
a) c = -V/√(R2+(ωL)2) cos⁡(θ+tan-1(ωL/R))
b) c = -V/√(R2+(ωL)2) cos⁡(θ-tan-1(ωL/R))
c) c = V/√(R2+(ωL)2) cos⁡(θ+tan-1(ωL/R))
d) c = V/√(R2+(ωL)2) cos⁡(θ-tan-1(ωL/R))

Answer: b [Reason:] Since the inductor does not allow sudden changes in currents, at t = 0, i = 0. So, c = -V/√(R2+(ωL)2) cos⁡(θ-tan-1(ωL/R)).

4. The complete solution of the current in the sinusoidal response of R-L circuit is?
a) i = e-t(R/L)[V/√(R2+(ωL)2) cos⁡(θ-tan-1)⁡(ωL/R))]+V/√(R2+(ωL)2) cos⁡(ωt+θ-tan-1)⁡(ωL/R))
b) i = e-t(R/L)[-V/√(R2+(ωL)2) cos⁡(θ-tan-1)(ωL/R))]-V/√(R2+(ωL)2) cos⁡(ωt+θ-tan-1)⁡(ωL/R))
c) i = e-t(R/L)[V/√(R2+(ωL)2) cos⁡(θ-tan-1)⁡(ωL/R))]-V/√(R2+(ωL)2) cos⁡(ωt+θ-tan-1)⁡(ωL/R))
d) i = e-t(R/L)[-V/√(R2+(ωL)2) cos⁡(θ-tan-1)⁡(ωL/R))]+V/√(R2+(ωL)2) cos⁡(ωt+θ-tan-1)⁡(ωL/R))

Answer: d [Reason:] The complete solution for the current becomes i = e-t(R/L)[-V/√(R2+(ωL)2) cos⁡(θ-tan-1)⁡(ωL/R))]+V/√(R2+(ωL)2)cos⁡(ωt+θ-tan-1)⁡(ωL/R)).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is?

a) ic = ce-100t
b) ic = ce100t
c) ic = ce-200t
d) ic = ce200t

Answer: c [Reason:] By applying Kirchhoff’s voltage law to the circuit, we have 20i+0.1di/dt=100cos⁡(103 t+π/2) => (D+200)i=1000cos⁡(1000t+π/2). The complementary function is ic = ce-200t.

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.
a) ip = 0.98cos⁡(1000t+π/2-78.6o)
b) ip = 0.98cos⁡(1000t-π/2-78.6o)
c) ip = 0.98cos⁡(1000t-π/2+78.6o)
d) ip = 0.98cos⁡(1000t+π/2+78.6o)

Answer: a [Reason:] Assuming particular integral as ip = A cos (ωt + θ) + B sin(ωt + θ). We get ip = V/√(R2+(ωL)2 ) cos⁡(ωt+θ-tan-1(ωL/R)) where ω = 1000 rad/sec, V = 100V, θ = π/2, L = 0.1H, R = 20Ω. On substituting, we get ip = 0.98cos⁡(1000t+π/2-78.6o).

7. The complete solution of ‘i’ from the information provided in the question 5.
a) i = ce-200t + 0.98cos⁡(1000t-π/2-78.6o)
b) i = ce-200t + 0.98cos⁡(1000t+π/2-78.6o)
c) i = ce-200t + 0.98cos⁡(1000t+π/2+78.6o)
d) i = ce-200t + 0.98cos⁡(1000t-π/2+78.6o)

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 = ce-200t + 0.98cos⁡(1000t+π/2-78.6o).

8. The current flowing through the circuit at t = 0 in the circuit shown in the question 5 is?
a) 1
b) 2
c) 3
d) 0

Answer: d [Reason:] At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

9. The value of c in the complementary function of ‘i’ in the question 5 is?
a) c = -0.98cos⁡(π/2-78.6o)
b) c = -0.98cos⁡(π/2+78.6o)
c) c = 0.98cos⁡(π/2+78.6o)
d) c = 0.98cos⁡(π/2-78.6o)