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## Network Theory MCQ Set 1

1. If a resistor Rx is connected between nodes X and Y, Ry between X and Y, Rz between Y and Z to form a delta connection, then after transformation to star, the resistor at node X is?
a) RxRy/( Rx+Ry+Rz)
b) RxRz/(Rx+Ry+Rz)
c) RzRy/(Rx+Ry+Rz)
d) (Rx+Ry)/(Rx+Ry+Rz)

Answer: a [Reason:] After transformation to star, the resistor at node X is RxRy/( Rx+Ry+Rz) and this resistance lies between Rx, Ry in star connection.

2. In the question above the resistance at node Y is?
a) RzRy/(Rx+Ry+Rz)
b) RzRx/(Rx+Ry+Rz)
c) RxRy/(Rx+Ry+Rz)
d) (Rz+Ry)/(Rx+Ry+Rz)

Answer: b [Reason:] After transformation to star, the resistor at node Y is RzRx/(Rx+Ry+Rz) and this resistance lies between Rx, Rz in star connection.

3. In the question above, the resistance at node Z is?
a) RyRx/(Rx+Ry+Rz)
b) RyRx/(Rx+Ry+Rz)
c) RzRy/(Rx+Ry+Rz)
d) (Rz+Rx)/(Rx+Ry+Rz)

Answer: c [Reason:] After transformation to star, the resistor at node Y is RzRy/(Rx+Ry+Rz) and this resistance lies between Rz, Ry in star connection.

4. If the resistors of star connected system are R1, R2, R3 then the resistance between 1 and 2 in delta connected system will be?
a) (R1R2+ R2R3+ R3R1)/R3
b) (R1R2+ R2R3+ R3R1)/R1
c) (R1R2+ R2R3+ R3R1)/R2
d) (R1R2+ R2R3+ R3R1)/(R1+R2)

Answer: a [Reason:] After transformation to delta, the resistance between 1 and 2 in delta connected system will be (R1R2+ R2R3+ R3R1)/R3 and this resistance lies between R1, R2 in delta connection.

5. If the resistors of star connected system are R1, R2, R3 then the resistance between 2 and 3 in delta connected system will be?
a) (R1R2+ R2R3+ R3R1)/R3
b) (R1R2+ R2R3+ R3R1)/R2
c) (R1R2+ R2R3+ R3R1)/R1
d) (R1R2+ R2R3+ R3R1)/(R3+R2)

Answer: c [Reason:] After transformation to delta, the resistance between 2 and 3 in delta connected system will be (R1R2+ R2R3+ R3R1)/R1 and this resistance lies between R3, R2 in delta connection.

6. If the resistors of star connected system are R1, R2, R3 then the resistance between 3 and 1 in delta connected system will be?
a) (R1R2+ R2R3+ R3R1)/R1
b) ( R1R2+ R2R3+ R3R1)/R3
c) (R1R2+ R2R3+ R3R1)/R2
d) (R1R2+ R2R3+ R3R1)/(R3+R1)

Answer: c [Reason:] After transformation to delta, the resistance between 2 and 3 in delta connected system will be (R1R2+ R2R3+ R3R1)/R2 and this resistance lies between R1, R3 in delta connection.

7. Find the equivalent resistance at node A in the delta connected circuit shown in the figure below.

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

Answer: d [Reason:] Performing delta to star transformation we obtain the equivalent resistance at node A is R=(11×12)/(11+12+13)=4Ω.

8. Find the equivalent resistance at node C in the delta connected circuit shown in the figure in the question 7.
a) 3.66
b) 4.66
c) 5.66
d) 6.66

Answer: b [Reason:] Performing delta to star transformation we obtain the equivalent resistance at node A is R=(12×13)/(11+12+13)=4.66Ω.

9. Find the equivalent resistance between node 1 and node 3 in the star connected circuit shown below.

a) 30
b) 31
c) 32
d) 33

Answer: c [Reason:] The equivalent resistance between node 1 and node 3 in the star connected circuit is R=(10×10+10×11+11×10)/10=32Ω.

10. Find the equivalent resistance between node 1 and node 2 in the star connected circuit shown in the question 9.
a) 2
b) 29
c) 30
d) 31

Answer: b [Reason:] The equivalent resistance between node 1 and node 3 in the star connected circuit is R=(10×10+10×11+11×10)/11=29Ω.

## Network Theory MCQ Set 2

1. Impedance is a complex quantity having the real part as _______ and the imaginary part as ______
a) resistance, resistance
b) resistance, reactance
c) reactance, resistance
d) reactance, reactance

Answer: b [Reason:] Almost all electric circuits offer impedance to the flow of current. Impedance is a complex quantity having the real part as resistance and the imaginary part as reactance.

2. The voltage function v(t) in the circuit shown below is?

a) v(t) = Vm e-tjω
b) v(t) = Vmetjω
c) v(t) = etjω
d) v(t) = e-tjω

Answer: b [Reason:] The voltage function v(t) in the circuit is a complex function and is given by v(t) = Vm etjω=Vm(cosωt+jsinωt).

3. The current i(t) in the circuit shown above is?
a) i(t)=(Vm/(R-jωL))etjω
b) i(t)=(Vm(R+jωL)) etjω
c) i(t)=(Vm(R-jωL)) etjω
d) i(t)=(Vm/(R+jωL)) etjω

Answer: d [Reason:] i(t)=Im etjω. Vm etjω=RIm etjω+L d/dt (Im etjω ). Vmetjω=RIm etjω+L Im(jω)etjω. Im= Vm/(R+jωL). i(t)=(Vm/(R+jωL)) etjω.

4. The impedance of the circuit shown below is?
a) R + jωL
b) R – jωL
c) R + 1/jωL
d) R – 1/jωL

Answer: a [Reason:] Impedance is defined as he ratio of the voltage to current function. The impedance of the circuit Z= Vmetjω/((Vm/(R+jωL)) etjω )=R+jωL.

5. The magnitude of the impedance of the circuit shown above.
a) √( R+ωL)
b) √(R-ωL)
c)√(R2+(ωL)2 )
d) √(R2-(ωL)2 )

Answer: c [Reason:] Complex impedance is the total opposition offered by the circuit elements to ac current and can be displayed on the complex plane. The magnitude of the impedance of the circuit is √(R2+(ωL)2 ).

6. The phase angle between current and voltage in the circuit shown above is?
a) tan-1⁡ωL/R
b) tan-1⁡ωR/
c) tan-1⁡R/ωL
d) tan-1⁡L/ωR

Answer: a [Reason:] The angle between impedance and reactance is the phase angle between the current and voltage applied to the circuit. θ=tan-1⁡ωL/R

7. The voltage function v(t) in the circuit shown below is?

a) v(t) = e-tjω
b) v(t) = etjω
c) v(t) = Vmetjω
d) v(t) = Vme-tjω

Answer: c [Reason:] If we consider the RC series circuit and apply the complex function v(t) to the circuit then voltage function is given by v(t)= Vmetjω.

8. The impedance of the circuit shown above is?
a) R + jωC
b) R – jωC
c) R + 1/jωC
d) R – 1/jωC

Answer: c [Reason:] The impedance of the circuit shown above is R + 1/jωC. Here the impedance Z consists resistance which is real part and capacitive reactance which is imaginary part of the impedance.

9. The magnitude of the impedance of the circuit shown above is?
a) √(R+1/ωC)
b) √(R-1/ωC)
c) √(R2+(1/ωC)2 )
d) √(R2-(1/ωC)2 )

Answer: c [Reason:] The magnitude of the impedance of the circuit shown above is √(R2+(1/ωC)2 ). Here the impedance is the vector sum of the resistance and the capacitive reactance.

10. The angle between resistance and impedance in the circuit shown above.
a) tan-11/ωRC
b) tan-1⁡C/ωR
c) tan-1⁡R/ωC
d) tan-1⁡ωRC

Answer: a [Reason:] The angle between resistance and impedance in the circuit shown above tan-1⁡1/ωRC. The angle between resistance and impedance is the phase angle between the applied voltage and current in the circuit.

## Network Theory MCQ Set 3

1. For the circuit shown below, find the voltage across the capacitor C1 at the time the switch is closed.

a) 0
b) V/4
c) V/2
d) V

Answer: d [Reason:] We use two different cicuits to illustrate how an impulse function can be created with a switching operation. The capacitor is charged to an initial voltage of V at the time the switch is closed.

2. The charge on capacitor C2 in the circuit shown in question 1 is?
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] In the circuit, the initial charge on C2 is zero. So, charge on capacitor C2 = zero. Capacitor circuit and series inductor circuit are two different cicuits to illustrate how an impulse function can be created with a switching operation.

3. The current in the circuit shown in question 1 is?
a) (V/R)/(s+1/Ce)
b) (V/R)/(s+1/RCe)
c) (V/R)/(s-1/RCe)
d) (V/R)/(s-1/Ce)

Answer: b [Reason:] The current flowing through the circuit is given by I= (V/s)/(R+1/sC1+1/sC2) = (V/R)/(s+1/RCe) where the equivalent capacitance C1C2/(C1+C2) and is replaced by Ce.

4. By taking the inverse transform of the current in the question 3, the value of the current is?
a) V/R et/Ce
b) V/R et/RCe
c) V/R e-t/RCe
d) V/R e-t/Ce

Answer: c [Reason:] By taking the inverse transform of the current, we obtain i = V/R e-t/RCe which indicates that as R decreases, the initial current increases and the time constant decreases.

5. Consider the circuit shown below. On applying the Kirchhoff’s current law, the equation will be?

a) V/(2s-15)+(V-[(100/s)+30])/(3s+10)=0
b) V/(2s-15)+(V-[(100/s)+30])/(3s-10)=0
c) V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0
d) V/(2s+15)+(V-[(100/s)+30])/(3s-10)=0

Answer: c [Reason:] The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero. Applying Kirchhoff’s current law, we get V/(2s+15)+(V-[(100/s)+30])/(3s+10)=0.

6. The value of the voltage V in the circuit shown in question 5 is?
a) 40(s+7.5)/s(s+5) -12(s+7.5)/(s-5)
b) 40(s+7.5)/s(s+5) -12(s+7.5)/(s+5)
c) 40(s+7.5)/s(s+5) +12(s+7.5)/(s-5)
d) 40(s+7.5)/s(s+5) +12(s+7.5)/(s+5)

Answer: d [Reason:] Solving for V yields V=40(s+7.5)/s(s+5) +12(s+7.5)/(s+5). The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero.

7. The value of the voltage V after taking the partial fractions in the question 5 is?
a) 12+ 60/s+10/(s+5)
b) 12- 60/s+10/(s+5)
c) 12- 60/s-10/(s+5)
d) 12+ 60/s-10/(s+5)

Answer: a [Reason:] By taking the partial fractions we get V = 60/s-20/(s+5)+12+30/(s+5) and on solving the equation we get V = 12+60/s+10/(s+5).

8. Determine the voltage V after taking the inverse transform in the question 5.
a) 12δ(t)-(60-10e-5t)u(t)
b) 12δ(t)+(60+10e-5t)u(t)
c) 12δ(t)-(60+10e-5t)u(t)
d) 12δ(t)+(60-10e-5t)u(t)

Answer: b [Reason:] By taking inverse transform of V = 12+60/s+10/(s+5) we have v=12δ(t)+(60+10e-5t)u(t) volts and we have to derive the expression for the current when t > 0.

9. The current equation for the circuit shown in question 5 is?
a) I=4/s-2/(s-5)
b) I=4/s-2/(s+5)
c) I=4/s+2/(s+5)
d) I=4/s+2/(s-5)

Answer: c [Reason:] After the switch has been opened, the current in L1 is the same as the current in L1. The current equation is I=(100/s+30)/(5s+25). On solving we get I = 4/s-2/(s+5).

10. The value of the current after taking the inverse transform of the current in the question 5 is?
a) (4-2e5t )u(t)
b) (4-2e-5t)u(t)
c) (4+2e5t)u(t)
d) (4-2e-5t)u(t)

Answer: d [Reason:] By taking the inverse transform of I = 4/s-2/(s+5) gives i = (4-2e-5t)u(t). Before the switch is opened,the current in L1 is 10A and the current in L2 is 0A.

## Network Theory MCQ Set 4

1. In purely resistive circuit, energy delivered by source is ____________ by resistance.
a) dissipated in the form of heat
b) stored as electric field
c) stored as magnetic field
d) returned to source

Answer: a [Reason:] In purely resistive circuit, energy delivered by source is dissipated in the form of heat by resistance and is not stored as either electric field or magnetic field.

2. In inductor, the energy delivered by source is ____________ by inductor.
a) stored as magnetic field
b) dissipated in the form of heat
c) returned to source
d) stored as electric field

Answer: a [Reason:] In inductor, the energy delivered by source is stored as magnetic field by inductor and is not dissipated in the form of heat or stored as electric field.

3. In capacitor, the energy delivered by source is ____________ by capacitor.
a) returned to source
b) dissipated in the form of heat
c) stored as electric field
d) stored as magnetic field

Answer: c [Reason:] In capacitor, the energy delivered by source is stored as electric field by capacitor and is not stored as magnetic field or dissipated in the form of heat.

4. If there is complex impedance in a circuit, part of energy is ____________ by reactive part and part of its energy is ____________ by the resistance.
a) alternately stored and returned, alternately stored and returned
b) alternately stored and returned, dissipated
c) dissipated, alternately stored and returned
d) dissipated, dissipated

Answer: b [Reason:] If there is complex impedance in a circuit, part of energy is alternately stored and returned by reactive part and part of its energy is dissipated by the resistance. The amount of energy dissipated is determined by relative values of resistance and reactance.

5. The equation of instantaneous power is?
a) P (t) = (VmIm/2)(cos⁡(2ωt+θ)+sin⁡θ)
b) P (t) = (VmIm/2)(sin⁡(2ωt+θ)+cos⁡θ)
c) P (t) = (VmIm/2)(cos⁡(2ωt+θ)+cos⁡θ)
d) P (t) = (VmIm/2)(sin⁡(2ωt+θ)+sin⁡θ)

Answer: c [Reason:] The equation of instantaneous power is P (t) =(VmIm/2)(cos⁡(2ωt+θ)+cos⁡θ). It consists of two parts. One is a fixed part and the other is time varying which has frequency twice that of the voltage or current wave forms.

6. The time varying part in the equation of instantaneous power has frequency ________________ that of the frequency of voltage or current wave forms.
a) equal to
b) twice
c) thrice
d) four times

Answer: b [Reason:] The time varying part in the equation of instantaneous power has frequency twice that of voltage or current wave forms and the other part is a fixed part.

7. Instantaneous power is negative, when the polarities of voltage and current are of __________
a) opposite sign
b) same sign
c) voltage is zero
d) current is zero

Answer: a [Reason:] Instantaneous power is negative, when voltage and current have opposite sign that is if voltage is positive, the current is negative and if current is positive, the voltage is negative.

8. In P (t) equation, if θ=0, then P (t) =?
a) (VmIm/2)(1+cos⁡ωt)
b) (VmIm/2)(cos⁡ωt)
c) (VmIm/2)(cos⁡2ωt)
d) (VmIm)(1+cos⁡2ωt)

Answer: d [Reason:] In P (t) equation, if θ=0⁰, then P (t) =(VmIm/2)(1+cos⁡2ωt). The power wave has a frequency twice that of the voltage or current. Here the average value of power is VmIm/2.

9. The average value of power if θ=0⁰ is?
a) VmIm/2
b) VmIm/2
c) VmIm/4
d) VmIm/8

Answer: b [Reason:] The average value of power if θ=0⁰ is VmIm/2. So, average power = VmIm/2 at θ=0⁰. When phase angle is increased the negative portion of the power cycle increases and lesser power is dissipated.

10. At θ=π/2, positive portion is __________ negative portion in power cycle.
a) greater than
b) less than
c) equal to
d) greater than or equal to

Answer: c [Reason:] At θ=π/2, the area under positive portion is equal to the area under negative portion in power cycle. At this instant the power dissipated in the circuit is zero.

## Network Theory MCQ Set 5

1.In the circuit shown below, find the Z-parameter Z11.

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

Answer: c [Reason:] The Z –parameter Z11 is V1/I1, port 2 is open circuited. V1 = (1+2)I1 => V1/I1 = 3 and on substituting, we get Z11 = 3Ω.

2. In the circuit shown in question 1, find the Z-parameter Z12.
a) 4
b) 3
c) 2
d) 1

Answer: c [Reason:] The Z-parameter Z12 is V2/I1 |I2=0. On open circuiting port 2 we obtain the equation, V1 = (2) I2 => V1/I1 = 2. On substituting we get Z12 = 2Ω.

3. In the circuit shown in question 1, find the Z-parameter Z21.
a) 2
b) 4
c) 1
d) 3

Answer: a [Reason:] The Z-parameter Z21 is V2/I1 |I2=0. On open circuiting port 2, we get V2 = (2)I1 => V2/I1 = 2. On substituting we get Z21 = 2Ω.

4. In the circuit shown in question 1, find the Z-parameter Z22.
a) 3
b)2
c) 4
d) 1

Answer: a [Reason:] The Z-parameter Z21 is V2/I2 |I1=0. This parameter is obtained by open circuiting port 1. So we get V2 = (2 + 1)I2 => V2 = 3(I2) => V2/I2 = 3. On substituting Z21 = 3Ω.

5. In the circuit shown below, find the Z-parameter Z11.

a) 10
b) 15
c) 20
d) 25

Answer: b [Reason:] The Z –parameter Z11 is V1/I1, port 2 is open circuited. V1 = (10 + 5)I1 => V1/I1 = 15 and on substituting, we get Z11 = 2.5Ω.

6. In the circuit shown in question 5, find the Z-parameter Z12.
a) 15
b) 10
c) 5
d) 1

Answer: c [Reason:] The Z-parameter Z12 is V2/I1 |I2=0. On open circuiting port 2 we obtain the equation, V1 = (5) I2 => V1/I1 = 5. On substituting we get Z12 = 5Ω.

7. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z11.
a) 8
b) 18
c) 28
d) 38

Answer: b [Reason:] The Z-parameter Z11 is Z11 = Z11x + Z11y and Z11x = 3, Z11y = 15. On substituting we get Z11 = 3 +15 = 18Ω.

8. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z12.
a) 4
b) 5
c) 6
d) 7

Answer: d [Reason:] The Z-parameter Z12 is Z12 = Z12x + Z12y and we have Z12x = 2, Z12y. On substituting we get Z12 = 2 + 5 = 7Ω.

9. From the circuits shown in question 1 in question 5, find the combined Z-parameter Z21.
a) 7
b) 6
c) 5
d) 4