Steady state AC Analysis

Question 1
A 0.1\mu F capacitor charged to 100 V is discharged through a 1 k\Omega resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______
A
0.25
B
0.65
C
0.45
D
0.85
GATE EE 2019   Electric Circuits
Question 1 Explanation: 


\begin{aligned} v_c(t)&=V_0e^{-t/\tau } \\ V_0&=100V \\ \tau &=RC=(10^3)(10^{-7}) \\ &=10^{-4}sec \\ \therefore \;v_c(t)&=100e^{-10^4 t }V \end{aligned}
Let the time required by the voltage across the capacitor to drop to 1 V is t_1,
\begin{aligned} \therefore \; v_c(t_1)&=100e^{-10^4t_1} \\ \text{But, } v_c(t_1)&=0 \\ \text{So, }0&=100e^{-10^4t_1} \\ t_1&=0.46msec \end{aligned}
Question 2
The voltage across the circuit in the figure, and the current through it, are given by the following expressions:
v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V
i(t) = 5 + X cos(\omega t) A
where \omega =100 \pi 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).
A
5
B
10
C
15
D
20
GATE EE 2018   Electric Circuits
Question 2 Explanation: 
Given that,
v(t)=5-10 \cos (\omega t+60^{\circ})
i(t)=5+C \cos (\omega t-0^{\circ})
P_{req}=0
0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
X=10
Question 3
In the figure, the voltages are
v_{1}(t)=100cos(\omega t),
v_{2} (t) = 100cos(\omega t + \pi /18) and
v_{3}(t) = 100cos(\omega t + \pi /36).
The circuit is in sinusoidal steady state, and R \lt \lt \omega L. P_{1},P_{2} and P_{3} are the average power outputs. Which one of the following statements is true?
A
P_{1}=P_{2}=P_{3}=0
B
P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0
C
P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0
D
P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0
GATE EE 2018   Electric Circuits
Question 3 Explanation: 


V_2:\frac{\pi}{18}=\frac{180^{\circ}}{18}=10^{\circ}
V_3:\frac{\pi}{36}=\frac{180^{\circ}}{36}=5^{\circ}
V_2 leads V_1 and V_3,
So, V_2is a source, V_1 and V_3 are absorbing.
Hence, P_2 \gt 0 and P_1,P_3 \lt 0
Question 4
The voltage (V) and current (A) across a load are as follows.
v(t) = 100 sin(\omega t),
i(t) = 10 sin(\omega t - 60^{\circ}) + 2 sin(3\omega t) + 5 sin(5\omega t).
The average power consumed by the load, in W, is___________.
A
100
B
150
C
200
D
250
GATE EE 2016-SET-2   Electric Circuits
Question 4 Explanation: 
The average power consumed by the load =
P=V_1I_1 \cos \phi
\;\;=\frac{100}{\sqrt{2}}\frac{10}{\sqrt{2}} \cos 60^{\circ}=250W
Question 5
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 (V_{R}) and coil (V_{C}) respectively, in volts, are
A
65, 35
B
50, 50
C
60, 90
D
60, 80
GATE EE 2016-SET-2   Electric Circuits
Question 5 Explanation: 
As per GATE Official answer key MTA (Marks to All)
Question 6
In the circuit shown below, the supply voltage is 10sin(1000t) volts. The peak value of the steady state current through the 1\Omega resistor, in amperes, is ______.
A
0.5
B
1
C
2
D
3
GATE EE 2016-SET-1   Electric Circuits
Question 6 Explanation: 
If we observe the parallel LC combination, we get that at \omega=1000 rad/sec the parallel LC is at resonance, thus it is open circuited.
The circuit given in question can be redrawn as

So, I=\frac{10 \sin 1000t}{10}=\sin 100t
So, peak value is 1 Amp.
Question 7
A symmetrical square wave of 50% duty cycle has amplitude of \pm15 V and time period of 0.4\pi ms. This square wave is applied across a series RLC circuit with R=5 \Omega, L=10 mH, and C=4 \muF. The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ______.
A
100
B
140
C
192
D
212
GATE EE 2015-SET-2   Electric Circuits
Question 8
The total power dissipated in the circuit, shown in the figure, is 1 kW.

The voltmeter, across the load, reads 200 V. The value of X_L is _____.
A
8.5
B
17.3
C
22.4
D
28.6
GATE EE 2014-SET-2   Electric Circuits
Question 8 Explanation: 


Given, total power dissipated in the circuit = 1kW =1000 Watt
\therefore \;\; 2^2 \times 1 +10^2 \times R=1000
R=\frac{998}{100}=9.98\Omega
Also, voltage drop across R,
V_R=IR
=10 \times 9.98=99.8 volt
Voltage drop across load,
V=200 volt =\sqrt{V_R^2+V_{X_L}^2}
\therefore voltage drop across inductor,
V_{X_L}=\sqrt{V^2-V_R^2}
\;\;=\sqrt{(200)^2-(99.8)^2}
\;\;=173.32 volt
Now, V_{X_L}=IX_L
X_L=\frac{V_{X_L}}{I}
X_L=\frac{173.32}{10}=17.332\Omega
Question 9
In the circuit shown, the three voltmeter readings are V_1=220 V , V_2=122 V , V_3=136 V.

If R_L=5\Omega, the approximate power consumption in the load is
A
700W
B
750W
C
800W
D
850W
GATE EE 2012   Electric Circuits
Question 9 Explanation: 
Given,
R_L= 5\Omega
\therefore \;\; \cos \phi =\frac{R_L}{|Z|}
\Rightarrow \;\;|Z|=\frac{5}{0.45}=11.11
Power consumed by load,
P_L=\left ( \frac{V_3}{|Z|} \right )^2R_L
\;\;=\left ( \frac{136}{11.11} \right )^2 \times 5
\;\;=749.1\approx 750W
Question 10
In the circuit shown, the three voltmeter readings are V_1=220 V , V_2=122 V , V_3=136 V.

The power factor of the load is
A
0.45
B
0.5
C
0.55
D
0.6
GATE EE 2012   Electric Circuits
Question 10 Explanation: 


V_1^2=V_2^2+V_3^2+2V_2V_3 \cos \phi
(220)^2=(122)^2+(136)^2+2 \times 122 \times 136 \times \cos \phi
\Rightarrow \;\; \cos \phi =0.45
There are 10 questions to complete.
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