Q41GATE 2004MCQ

For the circuit shown in the figure, the initial conditions are zero. Its transfer function $H(s)\frac{V_{c}(s)}{V_{i}(s)}$ is

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📖 Explanation

\begin{aligned} H(s) &=\frac{\frac{1}{s C}}{R+s L+\frac{1}{s C}}=\frac{1}{s^{2} L C+s C R+1} \\ &=\frac{1}{s^{2}\left(10 \times 10^{-3} \times 100 \times 10^{-6}\right)} \\ &+s\left(10 \times 10^{3} \times 100 \times 10^{-6}\right)+1 \\ H(s) &=\frac{1}{10^{-6} s^{2}+s+1}=\frac{10^{6}}{s^{2}+10^{6} s+10^{6}} \end{aligned}

Q43GATE 2003MCQ

The minimum number of equations required to analyze the circuit shown in the figure is

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📖 Explanation

As voltage at 1 node is known $\therefore$ using nodal analysis only 3 equations required.

Q44GATE 2003MCQ

An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 $\mu F$ in every cycle. The charging requires

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📖 Explanation

\begin{aligned} v_{c}(t) &=\frac{1}{C} \int_{0}^{T} i d t \\ \text{Here } T &=\frac{1}{f}=\frac{1}{500}=2 \mathrm{msec}\\ C&=2 \mu F \\ \Rightarrow \quad 3 &=\frac{i}{2 \mu F} T \\ i &=\frac{3 \times 2 \mu F}{T}=\frac{3 \times 2 \mu F}{2 m \sec }=3 m A \end{aligned}

Q45GATE 2003MCQ

Twelve 1 $\Omega$ resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

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📖 Explanation

\begin{array}{l} \qquad V_{a b}=\frac{i}{3} \times 1+\frac{i}{6} \times 1+\frac{i}{3} \times 1 \\ \Rightarrow \quad R_{\mathrm{eq}}=\frac{V_{a b}}{i}=\frac{5}{6} \Omega \end{array}

Q46GATE 2002MCQ

The dependent current source shown in Figure

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📖 Explanation

Applying KVL, $20-5 I-5\left(I+\frac{V_{1}}{5}\right)=0$ $20-10 I-20=0$ $\Rightarrow \quad I=0$ $\therefore$ Only dependent source acts. $\frac{V_{1}}{5}=4 \mathrm{A}$ Power delivered $=I^{2} R=16 \times 5=80 \mathrm{W}$