Question 1 |
The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage
having a frequency \omega=100 radian/s. When the inductor current i(t) is in phase with the
voltage v(t), the magnitude of the impedance Z (in \Omega) seen between the terminals a and b
is ________ (up to 2 decimal places).
25 | |
50 | |
100 | |
150 |
Question 1 Explanation:
At resonance imaginary part of Z_{eq}=0
Real of Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}
\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega
Real of Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}
\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega
Question 2 |
A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t=0 as shown in the figure. Assume i(0)=0, v(0)=0. Which one of the following
circular loci represents the plot of i(t) versus v(t) ?
A | |
B | |
C | |
D |
Question 2 Explanation:
I(s)=\frac{\frac{5}{s}}{s+\frac{1}{s}}
\;\;=\frac{5}{s^2+1}
i(t)=5 \sin t
v(t)=\frac{1}{C}\int_{0}^{t}i \; dt
v(t)=\int_{0}^{t}5 \sin t \; dt
v(t)=5[-\cos t]_0^t
v(t)=5[-\cos t+1]
v(t)=5-5 \cos t
\begin{matrix} t & i(t) & v(t)\\ 0 & 0 & 0\\ \frac{T}{4} & 5 & 5\\ \frac{T}{2}& 0 & 10\\ \frac{3T}{4} &-5 & 5\\ T& 0 & 0 \end{matrix}
Question 3 |
In the balanced 3-phase, 50 Hz, circuit shown below, the value of inductance (L) is 10 mH. The value of the capacitance (C) for which all the line currents are zero, in millifarads, is ___________.
1.32 | |
2.12 | |
3.03 | |
4.08 |
Question 3 Explanation:
Usingstar to delta conversion,
Line current will be zero when the parallel pair of induction capacitor is resonant at f=50Hz
So, 50 \times 2\pi=\frac{1}{\sqrt{LC/3}}
100 \pi=\frac{1}{\sqrt{LC/3}}
Since, L=10mH
C will be 3.03 mF.
Line current will be zero when the parallel pair of induction capacitor is resonant at f=50Hz
So, 50 \times 2\pi=\frac{1}{\sqrt{LC/3}}
100 \pi=\frac{1}{\sqrt{LC/3}}
Since, L=10mH
C will be 3.03 mF.
Question 4 |
The circuit below is excited by a sinusoidal source. The value of R, in \Omega, for which the admittance of the circuit becomes a pure conductance at all frequencies is _____________.
14.14 | |
8.62 | |
22.46 | |
12.18 |
Question 4 Explanation:
The resonant frequency for the circuit is
\omega _0=\frac{1}{\sqrt{LC}}\sqrt{\frac{R_L^2-L/C}{R_C^2-L/C}}
Since, (R_L=R_C=R)
So the circuit will have zero real part of admittance
when, R=\sqrt{\frac{L}{C}}=\sqrt{\frac{0.02}{100\mu F}}=14.14\Omega
\omega _0=\frac{1}{\sqrt{LC}}\sqrt{\frac{R_L^2-L/C}{R_C^2-L/C}}
Since, (R_L=R_C=R)
So the circuit will have zero real part of admittance
when, R=\sqrt{\frac{L}{C}}=\sqrt{\frac{0.02}{100\mu F}}=14.14\Omega
Question 5 |
An inductor is connected in parallel with a capacitor as shown in the figure.
As the frequency of current i is increased, the impedance (Z) of the network varies as
As the frequency of current i is increased, the impedance (Z) of the network varies as
A | |
B | |
C | |
D |
Question 5 Explanation:
Z=j\omega L||\frac{1}{j\omega C}
\;\;=\frac{L/C}{j\omega L+\frac{1}{j\omega C}}
\;\;=\frac{(L/C)j\omega C}{-\omega ^2 LC+1}
Z=\frac{j\omega L}{1-\omega^2 LC}
|Z|=\frac{\omega L}{1-\omega^2 LC}
For, 1\gt \omega^2 LC;
Z=+ve
For, 1 \lt \omega^2 LC;
Z=-ve
\;\;=\frac{L/C}{j\omega L+\frac{1}{j\omega C}}
\;\;=\frac{(L/C)j\omega C}{-\omega ^2 LC+1}
Z=\frac{j\omega L}{1-\omega^2 LC}
|Z|=\frac{\omega L}{1-\omega^2 LC}
For, 1\gt \omega^2 LC;
Z=+ve
For, 1 \lt \omega^2 LC;
Z=-ve
There are 5 questions to complete.