Steady state AC Analysis

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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_2

is 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

\pm

15 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

\mu

F. 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.

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.