Q2GATE 2025MCQ

Let $i_{C}, i_{L}$ , and $i_{R}$ be the currents flowing through the capacitor, inductor, and resistor, respectively, in the circuit given below. The AC admittances are given in Siemens (S). Which one of the following is TRUE?

🚀 Solve in Practice Mode

📖 Explanation

$\mathrm{i}_{\mathrm{C}}=\mathrm{V}\left(\mathrm{jB} \mathrm{B}_{\mathrm{C}}\right)=\left[1 \angle 90^{\circ}\right]\$ [\mathrm{j} 0.25]=0.25 \angle+180^{\circ} \mathrm{A}\mathrm{i}{\mathrm{L}}=\mathrm{V}\left(-\mathrm{j} \mathrm{B}{\mathrm{L}}\right)=\left[1 \angle 90^{\circ}\right]$[-\mathrm{j} 0.1]=0.1 \angle 0^{\circ} \mathrm{A}\mathrm{i}_{\mathrm{R}}=\mathrm{V}(0.2)=\left[1 \angle 90^{\circ}\right]$[0.2]=0.2 \angle 90^{\circ} \mathrm{A}$

Q3GATE 2024NAT

In the given circuit, the current $I_{x}$ (in $\mathrm{mA}$ ) is _____

🚀 Solve in Practice Mode

📖 Explanation

$V_{1}$ and $V_{2}$ are super node. KCL at $V_{1}$ and $V_{2}$ .

\begin{aligned} & \frac{V_{1}}{10^{3}}+\frac{V_{2}}{1 \times 10^{3}}+2 \times 10^{-3}=5 \times 10^{-3} \\ & I_{0}=\frac{V_{1}}{10^{3}} \\ & V_{2}-V_{1}=10^{3} I_{0}=10^{3} \times \frac{V_{1}}{10^{3}}=V_{1} \end{aligned}

\begin{aligned} & V_{2}=2 V_{1} \\ & V_{1}=\frac{V_{2}}{2} \end{aligned}

Put in equation (i)

\begin{aligned} \frac{1}{2} V_{2}+V_{2} & =3 \\ V_{2} & =2 \mathrm{~V} \\ I_{x} & =\frac{V_{2}}{10^{3}}=2 \mathrm{~mA} \\ I_{x} & =2 \mathrm{~mA} \end{aligned}

Q4GATE 2023NAT

In the circuit shown below, the current $i$ flowing through $200 \Omega$ resistor is ____ $m A$ (rounded off to two decimal places).

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📖 Explanation

By applying source transformation, Apply nodal at node $V_{A}$ ,

\begin{aligned} \frac{V_{A}-2}{2 \mathrm{k} \Omega}+\frac{V_{A}}{2 \mathrm{k} \Omega}+\frac{V_{A}+1}{1.2 \mathrm{k} \Omega} & =1 \mathrm{~mA} \\ V_{A}\left[\frac{1}{2}+\frac{1}{2}+\frac{1}{1.2}\right] & =1+\frac{2}{2}-\left(\frac{1}{1.2}\right) \\ V_{A} & =0.636 \mathrm{~V} \end{aligned}

The current through $200 \Omega$ resistor,

\begin{aligned} & i=\frac{V_{A}+1}{1.2 \mathrm{k} \Omega}=\frac{0.636+1}{1.2} \\ & i=1.36 \mathrm{~mA} \end{aligned}

Q5GATE 2022MCQ

The current $I$ in the circuit shown is ________

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📖 Explanation

Applying Nodal equation at Node-A

\begin{aligned} \frac{V_A}{2k}+\frac{V_A-5}{2k}&=10^{-3}\\ \Rightarrow 2V_A-5&=2k \times 10^{-3}\\ V_A&=3.5V\\ Again,&\\ I&=\frac{5-V_A}{2k}\\ &=\frac{5-3.5}{2k}\\ &=0.75 \times 10^{-3}A \end{aligned}

Q6GATE 2022MCQ

Consider the circuit shown in the figure. The current $I$ flowing through the $10\Omega$ resistor is _________.

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Q7GATE 2021NAT

Consider the circuit shown in the figure. The value of $v_{0}$ (rounded off to one decimal place) is _________ V.

🚀 Solve in Practice Mode

📖 Explanation

Write KVL equation first loop,

\begin{aligned} V_{o}-4+1\left(V_{o}+2\right) &=0 \\ V_{o} &=1 \mathrm{~V} \end{aligned}

Q8GATE 2021NAT

Consider the circuit shown in the figure. The current I flowing through the $7\:\Omega$ resistor between P and Q (rounded off to one decimal place) is ________ A.

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📖 Explanation

Redraw the circuit,

\begin{aligned} 3 \Omega \;||\; 6 \Omega&=2 \Omega \\ 2 \Omega \;||\; 2 \Omega&=1 \Omega \\ \qquad I&=5 \times \frac{1}{10}=0.5 \mathrm{~A} \end{aligned}

Q9GATE 2020MCQ

The current I in the given network is

🚀 Solve in Practice Mode

📖 Explanation

$I=-[I_{1}+I_{2}]$ $I=-\left[ \frac{120\angle -90^{\circ}}{80-j35}+\frac{120\angle -30^{\circ}}{80-j35} \right]$ $I=2.38\angle 143.7^{\circ}$

Q10GATE 2018NAT

Consider the network shown below with $R_{1} = 1\Omega , R_{2} = 2\Omega \; and \; R_{3} = 3\Omega$ . The network is connected to a constant voltage source of 11V. The magnitude of the current (in amperes, accurate to two decimal places) through the source is _______.

🚀 Solve in Practice Mode

📖 Explanation

As the network is symmetric, $V_{A}= V_{B} and V_{C}= V_{D}$ So current throught $R_{2}$ resistor is zero and $V_{A}=V_{B}$ and $V_{C}$ = $V_{D}$ ,electrically the circuit can reduced as, Total resistance,

\begin{aligned} R_{T} &=2\left(R_{1} \| R_{1}\right)+\left(R_{1}\left\|R_{1}\right\| R_{3} \| R_{3}\right) \\ &=R_{1}+\left(\frac{R_{1}}{2} \| \frac{R_{3}}{2}\right) \end{aligned}

Given that, $R_{1}=1 \Omega$ and $R_{3}=3 \Omega$

\begin{aligned} \text{So,}\qquad R_{T} &=1+\left(\frac{1}{2} \| \frac{3}{2}\right) \Omega=1+\frac{3 / 2}{4}=\frac{11}{8} \Omega \\ I &=\frac{11 \mathrm{V}}{R_{T}}=\frac{11}{(11 / 8)}=8 \mathrm{A} \end{aligned}

Q11GATE 2017NAT

A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10 $\Omega$ , 5 $\Omega$ , 2 $\Omega$ are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.

🚀 Solve in Practice Mode

📖 Explanation

The connection of resistors is as shown below: Given resistor values are : $10 \Omega, 5 \Omega, 2 \Omega$ The maximum resistance possible is,

\begin{aligned} R_{T(\max )} &=10 \Omega+(5 \Omega \| 2 \Omega) \\ &=\left(10+\frac{10}{7}\right) \Omega=\frac{80}{7} \Omega \end{aligned}

The minimum resistance possible is,

\begin{aligned} R_{T(\min )} &=2 \Omega+(10 \Omega \| 5 \Omega) \\ &=\left(2+\frac{10}{3}\right) \Omega=\frac{16}{3} \Omega \\ \frac{R_{T(\max )}}{R_{T(\min )}} &=\frac{80 / 7}{16 / 3}=\frac{15}{7}=2.143 \end{aligned}

Q12GATE 2016NAT

In the figure shown, the current i (in ampere) is __________

🚀 Solve in Practice Mode

📖 Explanation

Using KCL at $V_{1}$ $\frac{V_{1}}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}}{1}=0$ or $V_{1}=4 V$ considering KVL, we get

Q13GATE 2016NAT

In the circuit shown in the figure, the magnitude of the current (in amperes) through $R_{2}$ is ___

🚀 Solve in Practice Mode

📖 Explanation

Using KVL in the outer loop $60-5\left(0.16 V_{x}\right)-\frac{V_{x}}{5} \times 3-V_{x}=0$ $\text{or}\quad v_{x}=25 \mathrm{V}$ $\therefore$ The current flowing through $R_{2}=\frac{V_{x}}{5}=\frac{25}{5}=5 \mathrm{A}$

Q14GATE 2016MCQ

In the given circuit, each resistor has a value equal to 1 $\Omega$ . What is the equivalent resistance across the terminals a and b?

🚀 Solve in Practice Mode

📖 Explanation

$R_{eq}\Rightarrow$ By using delta to star conversion Again by star to delta conversion

\begin{aligned} R_{\mathrm{ab}} &=\{4 \| 1)+(4 \| 1)\}\|\{1 \| 4)\} \\ &=\left(\frac{4}{5}+\frac{4}{5}\right) \| \frac{4}{5}=\frac{8}{15} \Omega \end{aligned}

Q15GATE 2016NAT

An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that $R_{1}=3k \Omega$ , $R_{2}=6k \Omega$ and $R_{3}=9k \Omega$ , and that the diode is ideal. RMS current $I_{rms}$ (in mA) through the diode is ________

🚀 Solve in Practice Mode

📖 Explanation

The equivalent resistance across terminal ah ( outer loop) is

\begin{aligned} V &=\frac{I}{3} \times 3 \mathrm{k} \Omega+\frac{I}{6} \times 6 \mathrm{k} \Omega+\frac{I}{3} \times 9 \mathrm{k} \Omega \\ V &=5 I \\ \text{or}\quad \frac{V}{I} &=5 \mathrm{k} \Omega \end{aligned}

For half wave rectifier

\begin{aligned} I_{\mathrm{rms}} &=\frac{I_{m}}{(2)}=\frac{10 \sin t}{5 \mathrm{k} \Omega}=2 \sin t \mathrm{m} \mathrm{A} \\ \therefore \quad I_{\mathrm{rms}} &=\frac{I_{m}}{2}=1 \mathrm{m} \mathrm{A} \end{aligned}

Q16GATE 2015NAT

In the network shown in the figure, all resistors are identical with R = 300 $\Omega$ . The resistance $R_{ab}$ (in $\Omega$ ) of the network is ______.

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📖 Explanation

Modifying the given circuit

\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}

Q17GATE 2015NAT

In the circuit shown, the voltage $V_{X}$ (in Volts) is _______.

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📖 Explanation

\begin{aligned} \frac{V_{x}}{20}+\frac{V_{x}-V_{y}}{10}+0.5 V_{x}&=5 \mathrm{A} \\ V_{x}\left[\frac{1}{20}+\frac{1}{10}+0.5\right]&=5+\frac{V_{y}}{10} \\ 13 \mathrm{V}_{x}&=100+2 \mathrm{V}_{y} \quad\ldots(i)\\ \text { and also, } V_{y}&=0.25 \mathrm{V}_{x} \quad\dots(ii) \end{aligned}

Solving equations (i) and (ii), we have

\begin{aligned} 52 V_{y} &=100+2 V_{y} \\ 50 V_{y} &=100 \Rightarrow V_{y}=2 \mathrm{V} \\ V_{x} &=4 V_{y}=8 \mathrm{V} \end{aligned}

Q18GATE 2015MCQ

In the given circuit, the values of $V_{1} \; and \; V_{2}$ respectively are

🚀 Solve in Practice Mode

📖 Explanation

Current flowing through both the parallel $4 \Omega$ will be I. $\text { So, } V_{2}=4(I+I+2 I)+4 I \quad \text { by KVL }$ $I+I+2 I=5 \quad \text{by KCL}$