# GATE EC 2004

 Question 1
Consider the network graph shown in the figure. Which one of the following is NOT a 'tree' of this graph ? A a B b C c D d
Network Theory   Graph Theory and State Equations
Question 1 Explanation:
It is forming a closed loop. So it can't be a tree.
 Question 2
The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is A $L_{1}+L_{2}+M$ B $L_{1}+L_{2}-M$ C $L_{1}+L_{2}+2M$ D $L_{1}+L_{2}-2M$
Network Theory   Basics of Network Analysis
Question 2 Explanation: If current enters the dotte terminals of coil 1 then a voltage is developed across coil 2 whose higher potential is at dotted terminals.
\begin{aligned} V &=\frac{-M d I}{d t}+\frac{L_{1} d I}{d t}-\frac{M d I}{d t}+L_{2} \frac{d I}{d t} \\ &=\left(L_{1}+L_{2}-2 M\right) \frac{d I}{d t}\\ V&=L_{e q} \frac{d I}{d t} \end{aligned}

 Question 3
The circuit shown in the figure, with $R=\frac{1}{3}\Omega ,L=\frac{1}{4}H$ and C = 3 F has input voltage v(t) = sin2t. The resulting current i(t) is A 5 sin(2t + 53.1$^{\circ}$) B 5 sin(2t - 53.1$^{\circ}$) C 25 sin(2t + 53.1$^{\circ}$) D 25 sin(2t - 53.1$^{\circ}$)
Network Theory   Sinusoidal Steady State Analysis
Question 3 Explanation:
\begin{aligned} i(t)=& V(t), Y \\ Y=& V(t)\left[\frac{1}{R_{1}}+\frac{1}{j \omega L}+j \omega C\right] \\ =& \sin 2 t\left[3+\frac{4}{2 j}+j \times 2 \times 3\right] \\ =& \sin 2 t[3-2 j+6 j]=\sin 2 t[3+4 j] \\ =& 5 \sin 2 t \angle \tan ^{-1} \frac{4}{3}=5 \sin \left(2 t+53.1^{\circ}\right) \end{aligned}
 Question 4
For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is $v_{i}(t)=\sqrt{2}sin10^{3}t$. The output voltage$v_{o}(t)$ is equal to A $sin(10^{3}t-45^{\circ})$ B $sin(10^{3}t+45^{\circ})$ C $sin(10^{3}t-53^{\circ})$ D $sin(10^{3}t+53^{\circ})$
Network Theory   Sinusoidal Steady State Analysis
Question 4 Explanation:
\begin{aligned} v_{0}(t)&=\frac{\frac{1}{j \omega C}}{R+\frac{1}{j \omega C}}\; v_{i}(t)=\frac{1}{1+j \omega C R} \sqrt{2} \sin 10^{3} t \\ &=\frac{1}{1+j \times 10^{3} \times 10^{-3}} \sqrt{2} \sin 10^{3} t \\ v_{0}(t) &=\sin \left(10^{3} t-45^{\circ}\right) \end{aligned}
 Question 5
For the R - L circuit shown in the figure, the input voltage $v_{i}$ (t) = u(t). The current i(t) is  A a B b C c D d
Network Theory   Transient Analysis
Question 5 Explanation:
\begin{aligned} I(s)&=\frac{V(s)}{s+2}=\frac{1}{s(s+2)} \\ I(s)&=\frac{1}{s(s+2)}=\frac{1}{2}\left[\frac{1}{s}-\frac{1}{s+2}\right] \\ i(t)&=\frac{1}{2}\left(1-e^{-2 t}\right) \\ \text { At } \quad t&=0, i(t)=0 \\ t&=\infty, i(t)=0.5 \\ t&=\frac{1}{2}, i(t)=0.31 \end{aligned} Graph (c) satisfies all conditions .

There are 5 questions to complete.