Q1GATE 2024NAT

In the network shown below, maximum power is to be transferred to the load $R_{L}$ The value of $R_{L}$ (in $\Omega$ ) is _______

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📖 Explanation

$I=\frac{V-V_{x}}{2}+\frac{V}{5}=\frac{V}{2}-\frac{V_{x}}{2}+\frac{V}{5}$ But, $V_{x}=\frac{V}{5} \times 3$ $\therefore \quad I=\frac{V}{2}-\frac{1}{2} \times \frac{3 V}{5}+\frac{V}{5}=\frac{V}{2}-\frac{3 V}{10}+\frac{V}{5}$ $I=\frac{5 V-3 V+2 V}{10}=\frac{4 V}{10}$ $\therefore \quad \frac{V}{I}=\frac{10}{4}=2.5 \Omega$

Q2GATE 2020MCQ

In the circuit shown below, the Thevenin voltage $V_{TH}$ is

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Q3GATE 2019MCQ

Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure). Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1?

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📖 Explanation

According to reciprocity theorem, In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged. $\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}$

Q4GATE 2017NAT

Consider the circuit shown in the figure. The Thevenin equivalent resistance (in $\Omega$ ) across P - Q is _________.

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📖 Explanation

The equivalent circuit to calculate the Thevenin equivalent resistance ( $R_{Th}$ ) is as follows: lt can be further reduced as follows: By applying KVL in the Loop L

\begin{aligned} V_{x} &=3 i_{0}+\left(1-i_{0}\right)=2 i_{0}+1 \\ \text { Also, } \quad V_{x} &=i_{0}(1 \Omega) \end{aligned}

\begin{aligned} \text { So, } 2 i_{0}+1&=i_{0} \\ i_{0} &=-1 \mathrm{A} \\ V_{x} &=-1 \mathrm{V}\\ \text { So, } \quad R_{\mathrm{Th}}&=\frac{V_{x}}{1 \mathrm{A}}=-1 \Omega \end{aligned}

Q5GATE 2016NAT

In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is __________

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📖 Explanation

For maximum power transfer,

\begin{array}{l} R=R_{T h} \\ V_{0}=5 \times \frac{2 \mathrm{k} \Omega}{5 \mathrm{k} \Omega}=2 \mathrm{V} \end{array}

From output loop, $V_{T H}=100 \times 2 \times \frac{40 \mathrm{k} \Omega}{50 \mathrm{k} \Omega}$

\begin{array}{l} V_{\mathrm{TH}}=160 \mathrm{V}\\ \text { and } R_{\mathrm{TH}}=10 \mathrm{k} \Omega|| 40 \mathrm{k} \Omega=\frac{10 \times 40}{50}=8 \mathrm{k} \Omega \\ \therefore \text { Maximum power }=\frac{V^{2}_{\mathrm{Th}}}{4 R_{\mathrm{Th}}}=\frac{160 \times 160}{4 \times 8000}=0.8 \mathrm{W} \end{array}

Q6GATE 2016MCQ

In the circuit shown below, $V_{S}$ is a constant voltage source and $I_{L}$ is a constant current load. The value of $I_{L}$ that maximizes the power absorbed by the constant current load is

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📖 Explanation

In maximum power transformation, half of the voltage drops across source resistance, remaining half across the load. $\therefore$ voltage across source (R)

\begin{aligned} I_{L} R &=\frac{V_{s}}{2} \\ I_{L} &=\frac{V_{s}}{2 R} \end{aligned}

Q7GATE 2015NAT

For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a-b is ________.

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📖 Explanation

\begin{aligned} V_{Th}&=V_{6\;\omega}\\ \frac{V_{A}-12}{3}+\frac{V_{A}}{6} &=V_{6 \Omega} \\ V_{A}\left[\frac{1}{3}+\frac{1}{6}\right] &=1+4 \\ V_{A} \frac{3}{6} &=5 \\ V_{T h} &=V_{6 \Omega}=V_{A}=10 \mathrm{V} \end{aligned}

Q8GATE 2015NAT

In the circuit shown, the Norton equivalent resistance (in $\Omega$ ) across terminals a-b is _______.

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📖 Explanation

\begin{aligned} I&=\frac{V_{0}}{4} \\ I_{1}&=\frac{V_{0}}{2}\\ \text{Applying KCL}\\ \frac{V_{0}-4 I}{2}+\frac{V_{0}}{2}+\frac{V_{0}}{4}&=I_{0}\\ \text{From here}\\ V_{0} \cdot \frac{3}{4} &=I_{0} \\ R_{N} &=\frac{V_{0}}{I_{0}}=\frac{4}{3}=1.33 \Omega \end{aligned}

Q9GATE 2015NAT

In the given circuit, the maximum power (in Watts) that can be transferred to the load $R_{L}$ is_______.

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📖 Explanation

For maximum power transfer

\begin{aligned} R_{L} &=\left|Z_{T h}\right|=|2 \| j 2| \\ &=\left|\frac{2 \times j 2}{2+j 2}\right|=1.414 \Omega \\ V_{T h} &=\frac{8 \angle 90^{\circ}}{2+j 2}=2.828 \angle 45^{\circ} \end{aligned}

\begin{aligned} I &=\frac{2.828 \angle 45^{\circ}}{1.414 \angle 45^{\circ}+1.414}=1.08 \angle 22.5^{\circ} \\ \text { Power } &=I^{2} R=(1.08)^{2} \times \sqrt{2}=1.649 \mathrm{W} \end{aligned}

Q10GATE 2014NAT

In the circuit shown in the figure, the angular frequency $\omega$ (in rad/s), at which the Norton equivalent impedance as seen from terminals $b-b'$ is purely resistive, is _____.

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📖 Explanation

Finding $Z_{N}$

\begin{aligned} Z_{b b}^{\prime} &=\frac{1 \times j 0.5 \omega}{1+j 0.5 \omega}+\frac{1}{j \omega}=\frac{j \omega}{2+j \omega}+\frac{1}{j \omega} \\ \text { or } Z_{b b}^{\prime} &=\frac{2-\omega^{2}+j \omega}{2 j \omega-\omega^{2}}&\ldots(i) \end{aligned}

Rationalizing equation (i), we get,

\begin{aligned} Z_{b b}^{\prime} &=\frac{\left(2-\omega^{2}\right)+j \omega}{2 j \omega-\omega^{2}} \times \frac{-\omega^{2}-j 2 \omega}{-\omega^{2}-j 2 \omega} \\ &=-\frac{2 \omega^{2}+\omega^{4}+2 \omega^{2}}{\omega^{4}+2 \omega^{2}}+j \frac{\left(\omega^{3}-4 \omega\right)}{\omega^{4}+2 \omega^{2}} \end{aligned}

In order to have a purely resistive impedance $Z_{\text {bb }}^{\prime}$ , the imaginary part of equation (ii) will be equaled to zero.

\begin{aligned} &\therefore \frac{-4 \omega+\omega^{3}}{\omega^{4}+2 \omega^{2}}=0\\ &\text{or}\quad \omega^{3}=4 \omega \\ &\text{or}\quad \omega=\sqrt{4}=2 \mathrm{rad} / \mathrm{sec} \end{aligned}

Q11GATE 2014MCQ

Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance

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Q12GATE 2014NAT

In the figure shown, the value of the current I (in Amperes) is_____.

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📖 Explanation

Using super position theorem, when 5 V source acting alone, we get $I_{1}=\frac{V}{R_{\mathrm{eq}}}=\frac{5}{10+5+5}=\frac{1}{4} \mathrm{A}\quad \ldots(i)$ When 1 A source acting alone, we get $I_{2}=\frac{1 \times 5}{5+10+5}=\frac{5}{20}=\frac{1}{4} \mathrm{A}\quad \ldots(ii)$ Therefore, $I=I_{1}+I_{2}=\frac{1}{2} A=0.5 \mathrm{A}$

Q13GATE 2013MCQ

In the circuit shown below, if the source voltage $V_{s}= 100\angle 53.13^{\circ}V$ then the Thevenin?s equivalent voltage in Volts as seen by the load resistance $R_{L}$ is

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📖 Explanation

To find $V_{Th}$ , open circuit the load voltage $R_{L}$ then

\begin{aligned} I_{2} &=0 \\ j 40 I_{2} &=0 \\ V_{L_{1}} &=\frac{V_{s} \cdot(j 4)}{3+j 4}=\frac{100 \angle 53.13^{\circ}}{5 \angle 53.13^{\circ}} \times 4 \angle 90^{\circ} \\ V_{L_{1}} &=80 \angle 90^{\circ} \\ V_{\mathrm{Th}} &=10 V_{L_{4}}+I_{2} j 6+I_{2} 3 \\ V_{\mathrm{Th}} &=10 \times 80 \angle 90^{\circ}+0 \times j 6+0 \times 3 \\ V_{\mathrm{Th}} &=800 \angle 90^{\circ} \mathrm{V} \end{aligned}

Q14GATE 2013MCQ

A source $v_{s}\left ( t \right )=V\cos 100\pi t$ has an internal impedance of $\left ( 4+j3 \right )\Omega$ . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in $\Omega$ should be

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📖 Explanation

For pure resistive load to extract the maximum power. $R_{L}=\left|Z_{s}\right|=\sqrt{R_{s}^{2}+X_{s}^{2}}=\sqrt{4^{2}+3^{2}}=5 \Omega$

Q15GATE 2012MCQ

The impedance looking into nodes 1 and 2 in the given circuit is

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📖 Explanation

To find Thevenin impedance across node 1 and 2. Connect a 1 V source and find the current through voltage source. Then $Z_{T h}=\frac{1}{I}$ By applying KCL at node B and A

\begin{aligned} i_{A B}+99 i_{b} &=I \\ i_{b} &=i_{A}+i_{A B}\\ \Rightarrow \quad i_{b}-i_{A}+99i_{b}&=1 \\ \Rightarrow \quad 100i_{b}-i_{A}&=I \ldots(i) \end{aligned}

By applying KVL in outer loop

\begin{aligned} 10 \times 10^{3} i_{b}&=1\\ i_{b} &= 10^{-4}A\\ \text{and }10 \times 10^{3} i_{b}&=10^{1} \mathrm{A} \\ \Rightarrow \quad i_{A}&=-100 i_{b} \\ \end{aligned}

$\therefore \quad$ From equation (i)

\begin{aligned} 100 i_{n}+100 i_{n}&=1 \\ \Rightarrow \qquad I &=200 i_{b} \\ &=200 \times 10^{-4}=0.02 \\ \therefore \quad Z_{T h} &=\frac{1}{I}=\frac{1}{0.02}=50 \Omega \end{aligned}

Q16GATE 2012MCQ

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is

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📖 Explanation

Redrawing the diagram Current through R will be $i=\frac{10-3}{2+R}=\left(\frac{7}{2+R}\right) \mathrm{A}$ Current through 3V source is $i_{1}=i-\frac{3}{-j 1}=i-3 j$ So power delivered to circuit B by circuit A is