Q1GATE 2023MCQ

A series $R L C$ circuit has a quality factor $Q$ of 1000 at a center frequency of $10^{6} \mathrm{rad} / \mathrm{s}$ . The possible values of $R, L$ and $C$ are

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📖 Explanation

Given: $Q=1000$ and $\omega_{0}=10^{6} \mathrm{rad} / \mathrm{sec}$ We know, for series $R L C$ circuit,

\begin{aligned} Q & =\frac{\omega_{0} L}{R} \\ \omega_{0} & =\sqrt{\frac{1}{L C}} \\ Q & =\frac{1}{\sqrt{L C}} \times \frac{1}{R}=\frac{1}{R} \sqrt{\frac{L}{C}} \end{aligned}

So, $L=1 \mu \mathrm{H}, C=1 \mu \mathrm{F}$ and $R=0.001$

Q2GATE 2013MCQ

The transfer function $\frac{V_{2}\left ( s \right )}{V_{1}\left ( s \right )}$ of the circuit shown below is

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📖 Explanation

\begin{aligned} \frac{V_{2}(s)}{V_{1}(s)}&=\frac{10 \times 10^{3}+\frac{1}{100 \times 10^{-6} s}}{10 \times 10^{3}+\frac{1}{100 \times 10^{-6} s}+\frac{1}{100 \times 10^{-6} s}} \\ \frac{V_{2}(s)}{V_{1}(s)}&=\frac{s \times 10^{4}+10^{4}}{s \times 10^{4}+10^{4}+10^{4}}=\frac{10^{4}(1+s)}{10^{4}(s+2)} \\ \frac{v_{2}(s)}{V_{1}(s)}&=\frac{s+1}{s+2} \end{aligned}

Q3GATE 2009MCQ

If the transfer function of the following network is $\frac{V_o(s)}{V_i(s)}=\frac{1}{2+sCR}$ The value of the load resistance $R_{L}$ is

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📖 Explanation

\begin{aligned} Z &=\frac{R_{L}}{1+S R_{L} C} \\ H(s) &=\frac{Z}{Z+R}=\frac{R_{L}}{\left(R_{L}+R\right)+S R R_{L} C} \\ \text{if}\quad R &=R_{L} \\ H(s) &=\frac{1}{2+s R C} \end{aligned}

Q4GATE 2008MCQ

The driving point impedance of the following network is given by $Z(s)=\frac{0.2s}{s^{2}+0.1s+2}$ The component values are

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📖 Explanation

\begin{aligned} Z(s) &=\frac{0.2 s}{s^{2}+0.1 s+2} \\ r(s) &=\frac{s^{2}+0.1 s+2}{0.2 s} \\ &=\frac{s}{0.2}+\frac{1}{2}+\frac{2}{0.2 s} \\ &=5 s+0.5+\frac{10}{s}\\ \text{Comparing with}\\ n(s) &=C s+\frac{1}{R}+\frac{1}{L s} \\ C &=5 F, R=\frac{1}{0.5}=2 \Omega \\ L &=\frac{1}{10}=0.1 \mathrm{H} \end{aligned}

Q5GATE 2007MCQ

The RC circuit shown in the figure is

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📖 Explanation

At $\omega \rightarrow \infty$ , Capacitor $\rightarrow$ short circuited Circuit looks like, at $\omega \rightarrow 0$ , Capacitor $\rightarrow$ open circuited Circuit looks like So frequency response of the circuit will be, So the circuit is Band pass filter

Q6GATE 2007MCQ

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be $B_1$ and that of Filter 2 be $B_2$ . The value $\frac{B_{1}}{B_{2}}$ is

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📖 Explanation

Bandwidth of series RLC circuit is R/L Bandwidth of filter 1; $B_{1}=\frac{R}{L_{1}}$ Bandwidth of filter 2; $B_{2}=\frac{R}{L_{2}}$ $=\frac{R}{L_{1} / 4}=\frac{4 R}{L_{1}}$ So, $\frac{B_{1}}{B_{2}}=\frac{1}{4}$

Q7GATE 2006MCQ

A negative resistance $R_{neg}$ is connected to a passive network N having driving point impedance $Z_{1}$ (s) as shown below. For $Z_{2}$ (s) to be positive real,

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📖 Explanation

For $Z_{2}(s)$ to be positive real,

\begin{aligned} \operatorname{Re}(Z(s)) & \geq\left|R_{\operatorname{nog}}\right| \\ \Rightarrow \quad\left|R_{\operatorname{neg}}\right| & \leq \operatorname{Re}\left\{Z_{1}(j \omega)\right\}, \text { for all } \omega \end{aligned}

Q8GATE 2006MCQ

The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by

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📖 Explanation

RC impedance function has (1) $1^{st}$ critical frequency due to pole (2) Last critical frequency due to pole

Q9GATE 2005MCQ

The first and the last critical frequency of an RC-driving point impedance function must respectively be

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Q10GATE 2003MCQ

The driving point impedance Z(s) of a network has the pole-zero locations as shown in the figure. If Z(0) = 3, then Z(s) is

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📖 Explanation

\begin{array}{l} Z(s)=\frac{K(s-z)}{\left(s-p_{1}\right)\left(s-p_{2}\right)}=\frac{K(s+3)}{(s+1+j)(s+1-j)} \\ Z(s)=\frac{K(s+3)}{(s+1)^{2}-j 2}=\frac{K(s+3)}{(s+1)^{2}+1} \\ Z(0)|_{\omega=0}=0 \\ \Rightarrow \frac{3 K}{2}=3 \\ \Rightarrow \quad K=2 \\ \therefore \quad Z(s)=\frac{2(s+3)}{s^{2}+2 s+2} \end{array}

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