Question 1 |
The minimum number of equations required to analyze the circuit shown in the figure is
3
| |
4 | |
6 | |
7 |
Question 1 Explanation:
As voltage at 1 node is known
\therefore using nodal analysis only 3 equations required.
\therefore using nodal analysis only 3 equations required.
Question 2 |
A source of angular frequency 1 rad/sec has a source impedance consisting of 1 \Omega resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is
1 \Omega resistance | |
1 \Omega resistance in parallel with 1 H inductance | |
1 \Omega resistance in series with 1 F capacitor | |
1 \Omega resistance in parallel with 1 F capacitor |
Question 2 Explanation:
\begin{aligned} Z_{L}&=R_{S}-j X_{s} \\ \therefore \quad Z_{L}&=1-1 j \end{aligned}
Question 3 |
A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q=100. If each of R, L and C is doubled from its original value, the new Q of the circuit is
25
| |
50 | |
100 | |
200 |
Question 3 Explanation:
\begin{aligned} Q=& \frac{f_{o}}{B W} \\ f_{0} &=\frac{1}{2 \pi \sqrt{L C}} \\ B W &=\frac{R}{L} \\ \text { (Characteristic equation } &\left.=s^{2}+\frac{R s}{L}+\frac{1}{L C}\right)\\ \text{or}\qquad&=\frac{1}{R} \sqrt{\frac{L}{C}} \\ \text{When R, L, C are doubled,}\\ Q&=\frac{1}{2} Q=50 \end{aligned}
Question 4 |
The Laplace transform of i(t) is given by
I(s)=\frac{2}{s\left ( 1+s \right )}
At t\rightarrow \infty
The value of i(t) tends to
I(s)=\frac{2}{s\left ( 1+s \right )}
At t\rightarrow \infty
The value of i(t) tends to
0
| |
1
| |
2 | |
\infty
|
Question 4 Explanation:
\begin{aligned} \lim _{t \rightarrow \infty} i(t) &=\lim _{s \rightarrow 0} s I(s) \\ &=\operatorname{lims}_{s \rightarrow 0} \frac{2}{s(1+s)}=2 \end{aligned}
Question 5 |
The differential equation for the current i(t) in the circuit of the figure is
2\frac{d^{2}i}{dt^{2}}+2\frac{di}{dt}+i(t)=\sin t
| |
\frac{d^{2}i}{dt^{2}}+2\frac{di}{dt}+2i(t)=\cos t
| |
2\frac{d^{2}i}{dt^{2}}+2\frac{di}{dt}+i(t)=\cos t | |
\frac{d^{2}i}{dt^{2}}+2\frac{di}{dt}+2i(t)=\sin t
|
Question 5 Explanation:
Applying KVL,
\sin t=i(t) \times 2+L \frac{d i(t)}{d t}+\frac{1}{c} \int i(t) d t
\sin t=2 i(t)+2 \frac{d i(t)}{d t}+\int i(t) d t
Differentiating with respect to t
\cos (t)=\frac{2 d i(t)}{d t}+\frac{2 d^{2} i(t)}{d t^{2}}+i(t)
\sin t=i(t) \times 2+L \frac{d i(t)}{d t}+\frac{1}{c} \int i(t) d t
\sin t=2 i(t)+2 \frac{d i(t)}{d t}+\int i(t) d t
Differentiating with respect to t
\cos (t)=\frac{2 d i(t)}{d t}+\frac{2 d^{2} i(t)}{d t^{2}}+i(t)
There are 5 questions to complete.