Q2GATE 2022NAT

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)

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📖 Explanation

We have, overall Q-factor of given circuit is, $Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48$

Q3GATE 2022NAT

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)

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📖 Explanation

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

Q4GATE 2019NAT

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 ______

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📖 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}

Q5GATE 2018MCQ

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?

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📖 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 $P_1,P_3 \lt 0$

Q6GATE 2018NAT

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).

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📖 Explanation

Given that, $v(t)=5-10 \cos (\omega t+60^{\circ})$ $i(t)=5+C \cos (\omega t-0^{\circ})$ $P_{req}=0$ $0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$ $-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$ $X=10$

Q7GATE 2016MCQ

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

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Q8GATE 2016NAT

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___________.

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📖 Explanation

The average power consumed by the load = $P=V_1I_1 \cos \phi$ $\;\;=\frac{100}{\sqrt{2}}\frac{10}{\sqrt{2}} \cos 60^{\circ}=250W$

Q9GATE 2016NAT

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 ______.

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📖 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.

Q10GATE 2015NAT

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 ______.

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Q11GATE 2014NAT

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 _____.

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📖 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, $V_{X_L}=IX_L$ $X_L=\frac{V_{X_L}}{I}$ $X_L=\frac{173.32}{10}=17.332\Omega$

Q12GATE 2012MCQ

The average power delivered to an impedance $(4-j3)\Omega$ by a current 5cos(100 $\pi$ t+100)A is

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📖 Explanation

$i=5 \cos (100 \pi t +100^{\circ})A$ $\;\;=5\angle 100^{\circ}A$ $Z=(4-j3)\Omega$ $\;\;=5\angle -36.87^{\circ}\Omega$ $V=iZ=25\angle 63.13^{\circ}V$ The average power is $P=\frac{1}{2}V_mI_m \cos \phi$ $\;\;=\frac{1}{2} \times 25 \times 5 \times \cos (36.87^{\circ})$ $\;\;=\frac{1}{2} \times 25 \times 5 \times \frac{4}{5}=50W$ Alternate Method: $P=|I_{rms}|^2R$ $P=\left ( \frac{5}{\sqrt{2}} \right )^2 4=50W$

Q13GATE 2012MCQ

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

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📖 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, $P_L=\left ( \frac{V_3}{|Z|} \right )^2R_L$ $\;\;=\left ( \frac{136}{11.11} \right )^2 \times 5$ $\;\;=749.1\approx 750W$

Q14GATE 2011MCQ

The r.m.s value of the current i(t) in the circuit shown below is

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📖 Explanation

$V_s=1 \sin t\equiv V_m \sin \omega t$ $V_m=1 V$ and $\omega =1$ rad/sec Impedance of the branch containing inductor and capacitor $Z=j(X_L-X_C)$ $\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )$ $\;\;=j\left ( 1 \times 1 -\frac{1}{1 \times 1} \right )=0$ So, this branch is short circuit and the whole current flow through it $i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t$ rms value of the current $=\frac{1}{\sqrt{2}}A$

Q15GATE 2011MCQ

An RLC circuit with relevant data is given below. The current $I_C$ in the figure above is

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📖 Explanation

Using KCL, $-I_s+I_{RL}+I_C=0$ $\Rightarrow \; I_C=I_s-I_{RL}$ $\;\;\;=\sqrt{2}\angle \frac{\pi}{4}-\sqrt{2}\angle -\frac{\pi}{4}$ $\;\;\;=2\angle 90^{\circ}=+j2A$

Q16GATE 2011MCQ

The voltage applied to a circuit is $100 \sqrt{2} cos (100 \pi t)$ volts and the circuit draws a current of $10 \sqrt{2} sin(100 \pi t+\pi/4)$ amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is

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📖 Explanation

$V(t)=100\sqrt{2} \cos (100 \pi t)$ Voltage represented in phasor form: $V_{ph}=V_{rms}\angle \phi$ $V_{ph}=\frac{100\sqrt{2}}{\sqrt{2}}\angle 0^{\circ}$ $i(t)=10\sqrt{2} \sin \left ( 100 \pi t +\frac{\pi}{4} \right )$ $i(t)=10\sqrt{2} \cos \left ( 100 \pi t +\frac{\pi}{4}-\frac{\pi}{2} \right )$ $I_{ph}=\frac{10\sqrt{2}}{\sqrt{2}}\angle\left ( \frac{\pi}{4}-\frac{\pi}{2} \right )$ $\;\;=10\angle \left ( -\frac{\pi}{4} \right )A$

Q17GATE 2011MCQ

An RLC circuit with relevant data is given below. The power dissipated in the resistor R is

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📖 Explanation

Power supplied by the source $=V_sI_s \cos \phi$ Where, $\phi =$ angle between $V_s$ and $I_s \;=\; \pi/4$ Inductor and capacitor do not consume power. Therefore, power dissipated in R = Power supplied by the source $P_R=V_SI_S \cos \phi$ $\;\;=1 \times \sqrt{2} \times \cos \frac{\pi}{4}$ $\;\;=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 W$

Q18GATE 2006MCQ

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

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📖 Explanation

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 $=\frac{energy}{time \;eriod}=\frac{2.3kWh}{1 \; hour}$ $P_1=2.3 kW$ rms value of the input voltage $=V_{rms}=230V$ $P_1=\frac{V_{rms}^2}{R}$ $\Rightarrow \; 2.3 \times 10^3=\frac{230^2}{R}$ $\Rightarrow \; R=23\Omega$ (ii) When heater connected to 400 V (peak to peak) square wave source of 150Hz $V_{rms}$ value of the input voltage [/latex] $V_{rms}=\left[ \frac{1}{T}\int_{0}^{T}V^2 \; dt \right]^{1/2}$ $V_{rms}=\left[ \frac{1}{T}\left \{ \int_{0}^{T/2} 200^2 \; dt+ \int_{T/2}^{T} (-200)^2 \; dt\right \} \right]^{1/2}$ $V_{rms}=200V$ $P_2=\frac{V_{rms}^2}{R}=\frac{200^2}{23}\times 10^{-3}kW =1.739kW$

Q19GATE 2005MCQ

The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source $V_{IN}$ . At 100 Hz, the R and L elements each have a voltage drop $\mu _{RMS}$ . If the frequency of the source is changed to 50 Hz, then new voltage drop across R is

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📖 Explanation

At f = 100Hz $|V_R|=|V_L|$ as R and L are series connected, current through R and L is same, so

\begin{aligned}IR&=IX_L=I\omega L \\ \Rightarrow \; R&=X_L=\omega L \\ I&=\frac{V_{\in}}{\sqrt{R^{2+}X_L^2}} \\ &=\frac{V_{\in}}{\sqrt{R^{2+}R^2}}=\frac{V_{\in}}{\sqrt{2}R} \\ V_R&=u_{rms}=IR \\ V_R&=\left ( \frac{V_{\in}}{\sqrt{2}R} \right ) \times R=\frac{V_{\in}}{\sqrt{2}} \\ V_{\in}&=\sqrt{2}u_{rms}\;\;...(i) \\ f&=50 Hz \\ X_l & \propto f \\ \text{So, } X'_L&=X_L \times \frac{50}{100} \\ &=\frac{X_L}{2}=\frac{R}{2} \\ I'&=\frac{V_{\in}}{\sqrt{R^{2+}(X'_L)^2}} \\ &=\frac{V_{\in}}{\sqrt{R^{2+}\left ( \frac{R}{2} \right )^2}} \\ &=\frac{2V_{\in}}{\sqrt{5}R} \\ V'_R&=I'R \\ &=\left ( \frac{2V_{\in}}{\sqrt{5}R} \right )R \\ &=\frac{2}{\sqrt{5}}V_{\in} \\ &\text{From equation (i), }\\ V'_R&=\frac{2}{\sqrt{5}} \times (\sqrt{2}u_{rms}) \\ &=\frac{2\sqrt{2}}{\sqrt{5}}u_{rms}=\sqrt{\frac{8}{5}}u_{rms}\end{aligned}

Q20GATE 2005MCQ

The RMS value of the voltage u(t)= 3 + 4 cos (3t) is

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📖 Explanation

R.M.S. value of d.c voltage $=V_{dc}^{(rms)}=3V$ R.M.S. value of a.c. voltage $=V_{ac}^{(rms)}=\left ( \frac{4}{\sqrt{2}} \right ) V$ Therefore, R.M.S. value of the voltage $\sqrt{3^{2+}\left ( \frac{4}{\sqrt{2}} \right )^2}$ $=\sqrt{9+8}=\sqrt{17}V$

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