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Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q1: The value of parameters of the circuit shown in the figure are
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)For time t < 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t = 0, the value of the voltage across the inductor (VL) at t = 0+in Volts is _____ (Round off to 1 decimal place).       (2023)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 6.5
(b) 8
(c) 12.5
(d) 16
Ans:
(b)
Sol: Case (i) t < 1
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Using current division,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Case (ii) t > 0 :
Switch is opened.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Redraw the circuit t = 0+:
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Apply KVL in loop, Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q2: The network shown below has a resonant frequency of 150 kHz and a bandwidth of 600 Hz. The Q-factor of the network is __________. (round off to nearest integer)       (2022)
(a) 250
(b) 100
(c) 150
(d) 450
Ans:
(a)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q3: An inductor having a Q-factor of 60 is connected in series with a capacitor having a Q-factor of 240. The overall Q-factor of the circuit is ________. (round off to nearest integer)      (2022)
(a) 12
(b) 24
(c) 48
(d) 96
Ans: 
(c)
Sol: We have, overall Q-factor of given circuit is,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q4: A 0.1 μF capacitor charged to 100 V is discharged through a 1 kΩ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______       (2019)
(a) 0.25
(b) 0.65
(c) 0.45
(d) 0.85
Ans: 
(c)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Let the time required by the voltage across the capacitor to drop to 1 V is t1,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q5: The voltage across the circuit in the figure, and the current through it, are given by the following expressions:
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)where ω = 100π radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).      (2018)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 5
(b) 10
(c) 15
(d) 20
Ans:
(b)
Sol: Given that,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q6: In the figure, the voltages are
v1(t) = 100cos(ωt),
v2(t) = 100cos(ωt + π/18) and
v3(t) = 100cos(ωt + π/36).
The circuit is in sinusoidal steady state, and R<<ωL. P1, P2 and P3 are the average power outputs. Which one of the following statements is true?       (2018)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(a) 𝑃1=𝑃2=𝑃3=0P1 = P2 = P3 = 0
(b) 𝑃1<0,𝑃2>0,𝑃3>0P< 0, P> 0, P> 0
(c) 𝑃1<0,𝑃2>0,𝑃3<0P< 0, P> 0, P< 0  
(d) P> 0, P< 0, P> 0
Ans:
(c)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)V2 leads V1 and V3,
So, V2 is a source, V1 and V3 are absorbing.  
Hence, P2 > 0 and P1, P3 < 0

Q7: The voltage (V) and current (A) across a load are as follows.
v(t) = 100sin(ωt),
i(t) = 10sin(ωt − 60°) + 2sin(3ωt) + 5sin(5ωt).  
The average power consumed by the load, in W, is___________.        (SET-2 (2016))
(a) 100
(b) 150
(c) 200
(d) 250
Ans: 
(d)
Sol: The average power consumed by the load = Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q8: A resistance and a coil are connected in series and supplied from a single phase, 100 V, 50 Hz ac source as shown in the figure below. The rms values of plausible voltages across the resistance (VR) and coil (VC) respectively, in volts, are      (SET-2  (2016))
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 65, 35
(b) 50, 50
(c) 60, 90
(d) 60, 80
Ans:
(d)
Sol: It is a series RL circuit and
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q9: In the circuit shown below, the supply voltage is 10sin (1000t) volts. The peak value of the steady state current through the 1Ω  resistor, in amperes, is ______.     (SET-1 (2016))
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 0.5
(b) 1
(c) 2
(d) 3
Ans:
(b)
Sol: If we observe the parallel LC combination, we get that at ω = 1000 rad/sec the parallel LC is at resonance, thus it is open circuited.
The circuit given in question can be redrawn as
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)So, Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)So, peak value is 1 Amp.

Q10: A symmetrical square wave of 50% duty cycle has amplitude of ±15 V and time period of 0.4 π ms. This square wave is applied across a series RLC circuit with R = 5 Ω, L = 10 mH, and C = 4 μF. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ______.       (SET-2 (2015))
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 100
(b) 140
(c) 192
(d) 212
Ans:
(c)

Q11: The total power dissipated in the circuit, shown in the figure, is 1 kW.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)The voltmeter, across the load, reads 200 V. The value of XL is _____.      (SET-2 (2014)
(a) 8.5
(b) 17.3
(c) 22.4
(d) 28.6
Ans:
(b)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

Given, total power dissipated in the circuit = 1kW =1000 Watt
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Also, voltage drop across R,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage drop across load,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)∴ voltage drop across inductor,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q12: In the circuit shown, the three voltmeter readings are V= 220 V , V= 122 V , V= 136 V.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)If  RL = 5Ω, the approximate power consumption in the load is      (2012)
(a) 700 W
(b) 750 W
(c) 800 W
(d) 850 W
Ans:
(b)
Sol: Given,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Power consumed by load,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q13: In the circuit shown, the three voltmeter readings are V= 220 V , V= 122 V , V= 136 V.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)The power factor of the load is       (2012)
(a) 0.45
(b) 0.5
(c) 0.55
(d) 0.6
Ans:
(a)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q14: The average power delivered to an impedance (4 − j3)Ω by a current 5cos(100 πt+100)A is      (2012)
(a) 44.2 W
(b) 50 W
(c) 62.5 W
(d) 125 W
Ans:
(b)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)The average power is
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Alternate Method:
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q15: An RLC circuit with relevant data is given below.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)The current Iin the figure above is      (2011)
(a) -j2A
(b) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

(c) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(d) +j2A
Ans: 
(d)
Sol: Using KCL,
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q16: An RLC circuit with relevant data is given below.
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)The power dissipated in the resistor R is       (2011)
(a) 0.5 W
(b) 1 W
(c) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

(d) 2W
Ans:
(b)
Sol: Power supplied by the source = VsIcosϕ
Where, ϕ= angle between Vs and Is = π/4
Inductor and capacitor do not consume power. Therefore, power dissipated in R = Power supplied by the source
 PR = VSIS cosϕ  
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q17: The voltage applied to a circuit is 100√2𝑐𝑜𝑠(100𝜋𝑡) volts and the circuit draws a current of 10√2sin(100πt + π/4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is     (2011)
(a) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

(b) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(c) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(d) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Ans: (b)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Voltage represented in phasor form:
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q18: The r.m.s value of the current i(t) in the circuit shown below is      (2011)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 1/2 A
(b) 1/√2 A
(c) 1 A
(d) √2 A
Ans:
(b)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Impedance of the branch containing inductor and capacitor
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)So, this branch is short circuit and the whole current flow through it
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q19: An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak to peak square wave source of 150 Hz. The power in kW dissipated by the heater will be      (2006)
(a) 3.478
(b) 1.739
(c) 1.54
(d) 0.87
Ans: 
(b)
Sol: Assuming resistance of the heater = R
(i) When heater connected to 230 V, 50 Hz source, energy consumed by the heater = 2.3 units or 2.3 kWh in 1 hour
Power consumed by the heater
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)rms value of the input voltage
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(ii) When heater connected to 400 V (peak to peak) square wave source of 150Hz
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q20: The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source VIN. At 100 Hz, the R and L elements each have a voltage drop μRMS. If the frequency of the source is changed to 50 Hz, then new voltage drop across R is      (2005)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

(b) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(c) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
(d) Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Ans: (c)
Sol:  At f = 100Hz
∣VR∣ = ∣VL
as R and L are series connected, current through R and L is same, so
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q21: The RMS value of the voltage u(t)= 3 + 4 cos (3t) is      (2005)
(a) √17 V
(b) 5V
(c) 7V
(d) (3+2√2)V  
Ans:
(a)
Sol: R.M.S. value of d.c voltage Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)R.M.S. value of a.c. voltage
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Therefore, R.M.S. value of the voltage
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)
Q22: In figure, the admittance values of the elements in Siemens are YR = 0.5 + j0, YL = 0 - j1.5, YC = 0 + j0.3 respectively. The value of I as a phasor when the voltage E across the elements is 10∠0 V is      (2004)
Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)(a) 1.5 + j0.5
(b) 5 - j18
(c) 0.5 + j1.8
(d) 5 - j12
Ans:
(d)
Sol: Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE)

The document Previous Year Questions- Steady state AC Analysis | Network Theory (Electric Circuits) - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Network Theory (Electric Circuits).
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FAQs on Previous Year Questions- Steady state AC Analysis - Network Theory (Electric Circuits) - Electrical Engineering (EE)

1. What is steady-state AC analysis?
Ans. Steady-state AC analysis is a method used to analyze electrical circuits operating under sinusoidal conditions, where the voltages and currents are constant over time.
2. How is steady-state AC analysis different from transient analysis?
Ans. Steady-state AC analysis focuses on the circuit's behavior after it has reached a stable condition under sinusoidal conditions, while transient analysis deals with the circuit's response during the transition to a stable state.
3. What are the key components involved in steady-state AC analysis?
Ans. The key components involved in steady-state AC analysis include resistors, capacitors, inductors, and AC voltage sources, which are used to analyze impedance, phase angles, and power in the circuit.
4. How is impedance calculated in steady-state AC analysis?
Ans. Impedance in steady-state AC analysis is calculated using the complex representation of resistors, capacitors, and inductors, where impedance is the ratio of voltage to current in the circuit.
5. What are the applications of steady-state AC analysis in electrical engineering?
Ans. Steady-state AC analysis is commonly used in designing and analyzing electrical circuits in power systems, electronic devices, and communication systems, where understanding impedance, phase angles, and power relationships are essential.
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