Question 1 |
A current impulse 5\delta \left ( t \right ), is forced through a capacitor C. The voltage V_{c}\left ( t \right ),
across the capacitor is given by
5t | |
5u(t) - C | |
\frac{5}{C}t | |
\frac{5u(t)}{C} |
Question 1 Explanation:
V_c(t)=\frac{1}{C}\int_{-\infty }^{t}i(t)dt=\frac{1}{C}\int_{-\infty }^{t}5\delta (t)dt=\frac{5}{C}u(t)
Question 2 |
Fourier Series for the waveform, f(t) shown in figure is


\frac{8}{\pi ^{2}}\left [ \sin \left ( \pi t \right )+\frac{1}{9}\sin \left ( 3\pi t \right )+\frac{1}{25}\sin \left ( 5\pi t \right )+.... \right ] | |
\frac{8}{\pi ^{2}}\left [ \sin \left ( \pi t \right )-\frac{1}{9}\cos \left ( 3\pi t \right )+\frac{1}{25}\sin \left ( 5\pi t \right )+.... \right ] | |
\frac{8}{\pi ^{2}}\left [ \cos \left ( \pi t \right )+\frac{1}{9}\cos \left ( 3\pi t \right )+\frac{1}{25}\cos \left ( 5\pi t \right )+.... \right ] | |
\frac{8}{\pi ^{2}}\left [ \cos \left ( \pi t \right )-\frac{1}{9}\sin \left ( 3\pi t \right )+\frac{1}{25}\sin \left ( 5\pi t \right )+.... \right ] |
Question 2 Explanation:
\because f(t) is an even function with half waves symmetry,
\therefore dc term as well as sine terms=0
Only the cosine terms with odd harmonics will be present.
\therefore dc term as well as sine terms=0
Only the cosine terms with odd harmonics will be present.
Question 3 |
The graph of an electrical network has N nodes and B branches. The number of
links, L, with respect to the choice of a tree, is given by
B - N + 1 | |
B + N | |
N - B + 1 | |
N - 2B -1 |
Question 3 Explanation:
Number of links =B-(N-1) =B-N+1
Question 4 |
Two in-phase, 50 Hz sinusoidal waveform of unit amplitude are fed into channel 1
and channel 2 respectively of an oscilloscope. Assuming that the voltage scale,
time scale and other settings are exactly the same for both the channels, what
would be observed if the oscilloscope is operated in X-Y mode?
A circle of unit radius | |
An ellipse | |
A parabola | |
A straight line inclined at 45^{\circ} with respect to the x-axis |
Question 4 Explanation:
\therefore \;\; \text{Phase difference}=0^{\circ} \text{also} f_x=f_y.
Question 5 |
Given a vector field \vec{F}, the divergence theorem states that
\int _S\vec{F}\cdot d\vec{S}=\int _V\vec{\triangledown }\cdot \vec{F}dV | |
\int _S\vec{F}\cdot d\vec{S}=\int _V\vec{\triangledown }\times \vec{F}dV | |
\int _S\vec{F}\times d\vec{S}=\int _V\vec{\triangledown }\cdot \vec{F}dV | |
\int _S\vec{F}\times d\vec{S}=\int _V\vec{\triangledown }\times \vec{F}dV |
There are 5 questions to complete.