Question 1 |
Consider the circuit shown in the figure with input V(t) in volts. The sinusoidal
steady state current I(t) flowing through the circuit is shown graphically (where t is
in seconds). The circuit element Z can be ________.
a capacitor of 1 F | |
an inductor of 1 H | |
a capacitor of \sqrt{3} F | |
an inductor of \sqrt{3} H |
Question 1 Explanation:
i(t)=\frac{V(t)}{1+Z} where,
V(t)= \sin t
\because Given i(t) is lagging (from plot)
\begin{aligned} I_{max}&=\frac{V_{max}}{Z}=\frac{1}{Z}\\ \Rightarrow Z&=1/\sqrt{2}\\ Z&=(1+j\omega L)\Rightarrow L=1\\ as\; \omega &=1(\sin \omega t)=\sin t \end{aligned}
\because Given i(t) is lagging (from plot)
\begin{aligned} I_{max}&=\frac{V_{max}}{Z}=\frac{1}{Z}\\ \Rightarrow Z&=1/\sqrt{2}\\ Z&=(1+j\omega L)\Rightarrow L=1\\ as\; \omega &=1(\sin \omega t)=\sin t \end{aligned}
Question 2 |
For the circuit shown, the locus of the impedance Z(j\omega ) is plotted as \omega increases
from zero to infinity. The values of R_1 and R_2 are:
R_1=2k\Omega ,R_2=3k\Omega | |
R_1=5k\Omega ,R_2=2k\Omega | |
R_1=5k\Omega ,R_2=2.5k\Omega | |
R_1=2k\Omega ,R_2=5k\Omega |
Question 2 Explanation:
\begin{aligned}
Z(j\omega )&=R_1+\frac{R_2\cdot 1/j\omega c}{R_2+ 1/j\omega c}\\
&=R_1+\frac{R_2}{\frac{1}{2}+jR_2\omega c}\\
Z(j\omega )_{\omega =0}&=R_1+R_2\\
\Rightarrow R_1+R_2&=5k\Omega \\
Z|j\omega |_{\omega \rightarrow \infty }&=R_1\\
\Rightarrow R_1&=2k\Omega \\
R_2&=3k\Omega
\end{aligned}
Question 3 |
In the circuit shown, if v(t)=2sin(1000t) volts, R=1k\Omega and R=C=1\mu F, then the steady-state current i(t), in milliamperes (mA), is
sin(1000t) + cos(1000t) | |
2sin(1000t) + 2cos(1000t) | |
3sin(1000t) + cos(1000t) | |
sin(1000t) + 3cos(1000t) |
Question 3 Explanation:
\begin{aligned} \text{Here,}X_{C}&=\frac{1}{\omega C}=\frac{1}{10^{3} \times 10^{-6}}=\frac{1}{10^{-3}}\\ x_{C} &=10^{3} \Omega \\ R &=10^{3} \Omega \quad \text { (Given) } \\ v(t) &=2 \sin 1000 t \vee=2 \angle 0^{\circ} \mathrm{V} \end{aligned}
Redrawing the given network, we get,
As the bridge is balanced, it can be redrawn as
\begin{aligned} \therefore \quad Y_{e q} &=Y_{1}+Y_{2} \\ &=\frac{3}{2} R+\frac{1}{-2 j X_{C}} \\ &=\frac{3}{2} \times 10^{-3}+j \frac{1}{2} \times 10^{-3} \\ \therefore \quad i(t) &=(t) \times Y_{\mathrm{eq}}=2 \angle 0^{\circ}\left[\frac{3}{2}+j \frac{1}{2}\right] \mathrm{m} \mathrm{A} \\ &=(3+j 1) \mathrm{mA} \\ &=3 \sin (1000 t)+\cos (1000 t) \mathrm{mA} \end{aligned}
Question 4 |
For the circuit given in the figure, the voltage V_{C} (in volts) across the capacitor is
1.25\sqrt{2}sin(5t-0.25\pi ) | |
1.25\sqrt{2}sin(5t-0.125\pi ) | |
2.5\sqrt{2}sin(5t-0.25\pi ) | |
2.5\sqrt{2}sin(5t-0.125\pi ) |
Question 4 Explanation:
\frac{1}{\omega C}=\frac{1}{5 \times 10^{-6}}=200 \mathrm{k} \Omega
\begin{aligned} V_{C} &=\frac{5 \angle 0^{\circ}}{200-j 200} \times(-j 200) \vee \\ &=\frac{5 \angle 0^{\circ} \times 1 \angle-90^{\circ}}{\sqrt{2} \angle-45^{\circ}} \mathrm{V} \\ &=\frac{5}{\sqrt{2}} \angle-45^{\circ} \mathrm{V}=2.5 \sqrt{2} \sin \left(5 t-\frac{\pi}{4}\right) \mathrm{V} \\ &=2.5 \sqrt{2} \sin (5 t-0.25 \pi) \mathrm{V} \end{aligned}
\begin{aligned} V_{C} &=\frac{5 \angle 0^{\circ}}{200-j 200} \times(-j 200) \vee \\ &=\frac{5 \angle 0^{\circ} \times 1 \angle-90^{\circ}}{\sqrt{2} \angle-45^{\circ}} \mathrm{V} \\ &=\frac{5}{\sqrt{2}} \angle-45^{\circ} \mathrm{V}=2.5 \sqrt{2} \sin \left(5 t-\frac{\pi}{4}\right) \mathrm{V} \\ &=2.5 \sqrt{2} \sin (5 t-0.25 \pi) \mathrm{V} \end{aligned}
Question 5 |
In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V.
The ratio \frac{amplitude \; of \; voltage \; across \; the \; capacitor} {amplitude \; of \; voltage \; across \; the \; resistor} is
The ratio \frac{amplitude \; of \; voltage \; across \; the \; capacitor} {amplitude \; of \; voltage \; across \; the \; resistor} is
0.1 | |
0.2 | |
0.3 | |
0.4 |
Question 5 Explanation:
Given that, V and J have same phase. So, the circuit is in resonance.
At resonance,
\begin{aligned} &V_{C}=Q V_{R}\\ &\text { So, } \frac{\text { Amplitude of } V_{C}}{\text { Amplitude of } V_{R}}=Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{5} \sqrt{\frac{5}{5}}=0.2 \end{aligned}
There are 5 questions to complete.