Q2GATE 2023NAT

For the circuit shown, if $\mathrm{i}=\sin 1000 t$ , the instantaneous value of the Thevenin's equivalent voltage (in Volts) across the terminals a-b at time $\mathrm{t}=5 \mathrm{~ms}$ is ___ (Round off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

By source transformation, the circuit become Apply KVL in loop,

\begin{aligned} & (10+j 1)+4 i_{x}-(10+j 10+10-j 10) i_{x}=0 \\ & 10+\mathrm{j} 10+4 \mathrm{i}_{x}-20 \mathrm{i}_{x}=0 \\ & \Rightarrow \quad \mathrm{i}_{\mathrm{x}}=0.884 \angle 45^{\circ} \mathrm{A} \\ & \text { Now, } \quad \mathrm{V}_{\text {Th }}=\mathrm{i}_{x}(10-\mathrm{j10}) \\ & =0.884 \angle 45^{\circ} \times 14.142 \angle-45^{\circ} \\ & =12.5 \angle 0^{\circ} \mathrm{V} \\ & \therefore \quad \mathrm{V}_{\text {Th }}=12.5 \sin 1000 \mathrm{t} \\ & \text { At } \quad \mathrm{t}=5 \mathrm{msec} \\ & \mathrm{V}_{\text {Th }}=12.5 \times \sin \left(\frac{5 \times 180^{\circ}}{\pi}\right) \\ & =-11.986 \mathrm{~V} \end{aligned}

Q3GATE 2021NAT

In the given circuit, for maximum power to be delivered to $R_{L}$ , its value should be ______________ $\Omega$ . (Round off to 2 decimal places.)

🚀 Solve in Practice Mode

📖 Explanation

$=\frac{-j}{\omega C}=\frac{-j}{1000 \times 0.5 \times 10^{-3}}=-j 2 \Omega$ $Z_{\text {\in }}=2 \| j 2=\frac{j 4}{2+j 2}=\frac{j 2}{1+j 1}$ For maximum power transfer, $R_{L}=\left|Z_{\mathrm{TH}}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \Omega$

Q4GATE 2021MCQ

For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

🚀 Solve in Practice Mode

📖 Explanation

Given circuit can be resolved as shown below, $V_{T H}=15+50=65 \mathrm{~V}$

\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}

Q5GATE 2021NAT

In the given circuit, for voltage $V_y$ to be zero, the value of $\beta$ should be _________. (Round off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \frac{V_{x}-6}{1}+\frac{V_{x}}{4}+\frac{V_{x}-V_{y}}{2} & =0 \\ 4 V_{x}-24+V_{x}+2 V_{x}-2 V_{y} & =0 \\ 7 V_{x}-2 V_{y} & =24 \\ \text { If } V_{y}=0 \qquad\qquad 7 V_{x} & =24 \\ \Rightarrow\qquad\qquad V_{x} & =\frac{24}{7} \\ \Rightarrow\qquad\qquad \frac{V_{y}-V_{x}}{2}+\frac{V_{y}-\beta V_{x}}{3} & =2 \\ 3 V_{y}-3 V_{x}+2 V_{y}-2 \beta V_{x} & =12 \\ 5 V_{y}-(3+2 \beta) V_{x} & =12 \\ V_{y} & =0\\ -(3+2 \beta) \frac{24}{7} &=12 \\ (3+2 \beta) &=\frac{-7}{2} \\ 2 \beta &=\frac{-7}{2}-3=\frac{-7-6}{2}=\frac{-13}{2} \\ \beta &=\frac{-13}{4}=-3.25 \end{aligned}

Q6GATE 2021NAT

A signal generator having a source resistance of $50\:\Omega$ is set to generate a $\text{1 kHz}$ sinewave. Open circuit terminal voltage is $\text{10 V}$ peak-to-peak. Connecting a capacitor across the terminals reduces the voltage to $\text{8 V}$ peak-to-peak. The value of this capacitor is _____________ $\mu F$ . (Round off to 2 decimal places.)

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \frac{V_{0}(j \omega)}{V_{i}(j \omega)} &=\frac{1}{1+j \omega R C} \\ \frac{8}{10} \angle \theta &=\frac{1}{1+j \omega R C} \\ 1+j \omega R C &=1.25 \angle \theta \\ \omega R C &=\sqrt{1.25^{2}-1^{2}} \\ \omega R C &=0.75 \\ C &=\frac{0.75}{2000 \pi \times 50}=2.38 \mu\mathrm{F} \end{aligned}

Q7GATE 2020MCQ

A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value at which maximum power is transferred, and the corresponding load voltage and current.

🚀 Solve in Practice Mode

📖 Explanation

Maximum power transistor of VI product is maximum. If draw the the curve, it intersect (10, 4) that will give maximum power. The terminal voltage is 10 V (Load voltage) and current is 4 A (Load current). Load resistance is $\frac{10}{4}=2.5\Omega$ .

Q8GATE 2020MCQ

$x_R\; and \; x_A$ are, respectively, the rms and average values of $x(t) = x(t - T)$ , and similarly, $y_R\; and \; y_A$ are, respectively, the rms and average values of $y(t) = kx(t). k, \;T$ are independent of t. Which of the following is true?

🚀 Solve in Practice Mode

📖 Explanation

Given that,

\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}

Q9GATE 2020NAT

The Thevenin equivalent voltage, $V_{TH}$ , in V (rounded off to 2 decimal places) of the network shown below, is _______ .

🚀 Solve in Practice Mode

📖 Explanation

Only voltage source 4V is there and current source 5A is open circuited From the above circuit, $V_{TH1}=4V$ Case-II: Only current source 5A is there and voltage source 4V is short circuited. From the above circuit, $V_{TH2}=2 \times 5=10V$ By applying superposition theorem, $V_{TH}=V_{TH1}+V_{TH2}=10+4=14V$

Q10GATE 2019NAT

The current flowing in the circuit shown below in amperes is _____

🚀 Solve in Practice Mode

📖 Explanation

By Millman'e theorem, $E=\frac{\frac{200}{50}+\frac{160}{40}-\frac{100}{25}-\frac{80}{20}}{\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}}=0V$ $\frac{1}{R}=\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}$ Simplified circuit, $\therefore \;\;I=0A$

Q11GATE 2017NAT

For the given 2-port network, the value of transfer impedance $Z_{21}$ in ohms is_______

🚀 Solve in Practice Mode

📖 Explanation

Where, $R_A=1\Omega$ $R_B=1\Omega$ $R_C=\frac{1}{2}\Omega$ After rearrangement consider the following circuit, From the circuit diagram, we get, $Z_{21}=\frac{V_2}{I_1}=3\Omega$

Q12GATE 2017MCQ

For the network given in figure below, the Thevenin's voltage $V_{ab}$ is

🚀 Solve in Practice Mode

📖 Explanation

Consider the following circuit, After rearrangement, we get, From circuit using KCL, $\frac{V_{ab}+30}{15}+\frac{V_{ab}-8}{5}=0$ $V_{ab}+30+3V_{ab}-24=0$ $V_{ab}=-1.5V$

Q13GATE 2017NAT

In the circuit shown below, the maximum power transferred to the resistor R is _______ W.

🚀 Solve in Practice Mode

📖 Explanation

To get $R_{th}$ and $V_{th}$ , consider the following steps. Case-1: For $R_{th}$ For $V_{th}$ Applying KCL at node, $\frac{V_{th}-5}{5}+\frac{V_{th}+16}{5}=0$ $2V_{th}=-11$ $V_{th}=-5.5 V$ Maximum power transferred, $P_{max}=\frac{V_{th}^2}{4R_{th}}= \frac{5.5^2}{4 \times 2.5}=3.025W$

Q14GATE 2017MCQ

In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is

🚀 Solve in Practice Mode

📖 Explanation

The frequency at which the load is resistive and it is equal to $0.5\Omega$ i.e. The load is resistive means, the imaginary part of load is equal to zero and real part is equal to $0.5\Omega$ . $Z_{load}=\frac{1 \times \frac{1}{Cs}}{1 + \frac{1}{Cs}}+Ls$ $\;\;=\frac{1}{1+Cs}+Ls$ $\;\; =\frac{(1-Cs)}{1-C^2s^2}+Ls$ Put, $s=j\omega$ ; $Z_{load}=\frac{1-j\omega C}{1+C^2\omega ^2}+j\omega L$ $\;\;\;=\frac{1}{1+C^2\omega ^2}+j\omega \left ( L-\frac{C}{(1+C^2\omega ^2)} \right )$ Real part of the $Z_{load}=\frac{1}{1+C^2\omega ^2}=0.5$ Putting, $\omega =100 r/s$ we get, $C=10mF$

Q15GATE 2015NAT

The circuit s hown in the figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by $v(t)=12\sin t$ . If the circuit is in steady state, then the rms value of the current (in Ampere) flowing in the circuit is ______ .

🚀 Solve in Practice Mode

📖 Explanation

$i(t)=\frac{8}{1}+\frac{12 \sin t}{\sqrt{1+1}}=8+6\sqrt{2} \sin t$ $i_{rms}(t)=\sqrt{8^{2+}\frac{1}{2}(6\sqrt{2} )^2}=10A$

Q16GATE 2015NAT

For the given circuit, the Thevenin equivalent is to be determined. The Thevenin voltage, $V_{Th}$ (in Volt), seen from terminal AB is __________.

🚀 Solve in Practice Mode

📖 Explanation

$2-1(i_1+i)-i=0$ $2-i_1-2i=0$ $2i + i_1=2\;\;...(i)$ $\frac{V-2}{1}+\frac{V}{1}+\frac{V-(-20i)}{2}=0$ $2(V-2)+2V+V+20i=0$ $5V+20i=4$ where, $\frac{V}{1}=i$ $\Rightarrow \; 25i=4$ $i=\frac{4}{25}A$ $i_1=2-2 \times \frac{4}{25} =1.68A$ $V_{AB}=1.68 \times 2=3.36 V$

Q17GATE 2015NAT

In a linear two-port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short-circuited. When 5 V is applied to Port 1, a current of 1.25 A flows through a 1 $\Omega$ resistance connected across Port 2. When 3 V is applied to Port 1, the current (in Ampere) through a 2 $\Omega$ resistance connected across Port 2 is _______.

🚀 Solve in Practice Mode

📖 Explanation

$V_1=10V, I_2=4 A, V_2=0\;\;Cond...(i)$ $V_1=5V, I_2=1.25 A,$ $V_2=1.25 \times 1=1.25V\;\;Cond...(ii)$ $V_1=3V, I_2=? , R=2\Omega \;\;Cond...(iii)$ as we know from ABCD parameter, $V_1=AV_2-BI_2$ $I_1=CV_2-DI_2$ From condition (i), $10=A(0)-B(4)$ $B=-\frac{10}{4}$ From condition (ii), $5=A(1.25)-\left ( -\frac{10}{4} \right ) \times 1.25$ $A=\frac{\left ( 5-\frac{12.5}{4} \right )}{1.25}=1.5$ From condition (iii), $3=1.5(2I)-\left ( -\frac{10}{4} \right )\times I$ $3=3I+2.5I$ $I=0.545 A$

Q18GATE 2014MCQ

Assuming an ideal transformer, the Thevenin's equivalent voltage and impedance as seen from the terminals x and y for the circuit in figure are

🚀 Solve in Practice Mode

📖 Explanation

Thevenin's equivalent voltage=voltage referred to secondary. We have: $\frac{\sin \omega t}{V_{Th}}=\frac{1}{2}$ $V_{Th}=2 \sin (\omega t)$ Also, Thevenin's impedance seen from the x and y terminals=voltage referred to secondary side $\therefore \;\;Z_{Th}=R_{Th}=2^2\times 1=4\Omega$ So, $V_{Th}= 2 \sin (\omega t)$ and $Z_{Th}=R_{Th}=4\Omega$

Q19GATE 2014MCQ

A non-ideal voltage source $V_{s}$ has an internal impedance of $Z_{s}$ . If a purely resistive load is to be chosen that maximizes the power transferred to the load, its value must be

🚀 Solve in Practice Mode

📖 Explanation

Thesituation of problem is shown in figure: For the transfer of maximum power from source to load, $R_L=\sqrt{R_s^{2+}X^2}=|Z|$

Q20GATE 2014NAT

The Norton's equivalent source in amperes as seen into terminals X and Y is ____.

🚀 Solve in Practice Mode

📖 Explanation

Using source transformation theoram, or we can simplify the network, Now, from the circuit, we get, $I_N=\frac{5}{5}=1A$

Network Theorems Practice Questions

Get the full Network Theorems experience — 100% free