Question 1 |

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)

250 | |

100 | |

150 | |

450 |

Question 1 Explanation:

Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250

Question 2 |

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)

12 | |

24 | |

48 | |

96 |

Question 2 Explanation:

We have, overall Q-factor of given circuit is,

Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48

Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48

Question 3 |

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 ______

0.25 | |

0.65 | |

0.45 | |

0.85 |

Question 3 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 4 |

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).

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).

5 | |

10 | |

15 | |

20 |

Question 4 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

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 5 |

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?

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?

P_{1}=P_{2}=P_{3}=0 | |

P_{1} \lt 0,P_{2} \gt 0,P_{3} \gt 0 | |

P_{1} \lt 0,P_{2} \gt 0,P_{3} \lt 0 | |

P_{1} \gt 0,P_{2} \lt 0,P_{3} \gt 0 |

Question 5 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 6 |

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___________.

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___________.

100 | |

150 | |

200 | |

250 |

Question 6 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

P=V_1I_1 \cos \phi

\;\;=\frac{100}{\sqrt{2}}\frac{10}{\sqrt{2}} \cos 60^{\circ}=250W

Question 7 |

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

65, 35 | |

50, 50 | |

60, 90 | |

60, 80 |

Question 7 Explanation:

As per GATE Official answer key MTA (Marks to All)

Question 8 |

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 ______.

0.5 | |

1 | |

2 | |

3 |

Question 8 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.

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 9 |

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 ______.

100 | |

140 | |

192 | |

212 |

Question 10 |

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 _____.

The voltmeter, across the load, reads 200 V. The value of X_L is _____.

8.5 | |

17.3 | |

22.4 | |

28.6 |

Question 10 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

There are 10 questions to complete.