Question 1 |
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
R=1 \Omega, L=1 \mu \mathrm{H} and C=1 \mu \mathrm{F} | |
R=0.1 \Omega, L=1 \mu \mathrm{H} and C=1 \mu \mathrm{F} | |
R=0.01 \Omega, L=1 \mu \mathrm{H} and C=1 \mu \mathrm{F} | |
R=0.001 \Omega, L=1 \mu \mathrm{H} and C=1 \mu \mathrm{F} |
Question 1 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
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
Question 2 |
The transfer function \frac{V_{2}\left ( s \right )}{V_{1}\left ( s \right )} of the circuit shown below is

\frac{0.5s+1}{s+1}
| |
\frac{3s+6}{s+2}
| |
\frac{s+2}{s+1}
| |
\frac{s+1}{s+2}
|
Question 2 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}
Question 3 |
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

The value of the load resistance R_{L} is
\frac{R}{4} | |
\frac{R}{2} | |
R | |
2R |
Question 3 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}
Question 4 |
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

The component values are
L = 5 H,R = 0.5 \Omega,C = 0.1 F | |
L = 0.1H,R = 0.5 \Omega,C = 5 F | |
L = 5 H,R = 2 \Omega,C = 0.1 F | |
L = 0.1H,R = 2 \Omega,C = 5 F |
Question 4 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}
Question 5 |
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


4 | |
1 | |
1/2 | |
1/4 |
Question 5 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}
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}
There are 5 questions to complete.