Question 1 |
The value of parameters of the circuit shown in the figure are
\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega, \mathrm{L}=10 \mathrm{mH}, \mathrm{C}=100 \mu \mathrm{F}
For time t \lt 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t=0, the value of the voltage across the inductor \left(\mathrm{V}_{\mathrm{L}}\right) at t=0^{+}in Volts is _____ (Round off to 1 decimal place).
\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{R}_{3}=3 \Omega, \mathrm{L}=10 \mathrm{mH}, \mathrm{C}=100 \mu \mathrm{F}
For time t \lt 0, the circuit is at steady state with the switch 'K' in closed condition. If the switch is opened at t=0, the value of the voltage across the inductor \left(\mathrm{V}_{\mathrm{L}}\right) at t=0^{+}in Volts is _____ (Round off to 1 decimal place).
6.5 | |
8 | |
12.5 | |
16 |
Question 1 Explanation:
Case (i) \mathbf{t} \lt 1
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Using current division,
\begin{aligned} \mathrm{i}_{\mathrm{L}}\left(0^{-}\right) & =\frac{3}{3+2} \times 10=6 \mathrm{~A} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right) & =5 \times 2=12 \mathrm{~V} \end{aligned}
Case (ii) \mathbf{t} \gt 0 :
Switch is opened.
\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}
and \quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}
Redraw the circuit \mathrm{t}=0^{+}:
Apply KVL in loop,
\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}
At steady state, capacitor behaves as open circuit and inductor behaves as short circuit.
Redraw the circuit:
Using current division,
\begin{aligned} \mathrm{i}_{\mathrm{L}}\left(0^{-}\right) & =\frac{3}{3+2} \times 10=6 \mathrm{~A} \\ \therefore \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right) & =5 \times 2=12 \mathrm{~V} \end{aligned}
Case (ii) \mathbf{t} \gt 0 :
Switch is opened.
\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}
and \quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}
Redraw the circuit \mathrm{t}=0^{+}:
Apply KVL in loop,
\begin{array}{rlrl} 12+8-12-\mathrm{V}_{\mathrm{L}} & =0 \\ \Rightarrow & \mathrm{V}_{\mathrm{L}} & =8 \mathrm{~V} \end{array}
Question 2 |
The network shown below has a resonant frequency of 150 kHz and a bandwidth of
600 Hz. The Q-factor of the network is __________. (round off to nearest integer)
250 | |
100 | |
150 | |
450 |
Question 2 Explanation:
Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250
Question 3 |
An inductor having a Q-factor of 60 is connected in series with a capacitor having a Q-factor of 240. The overall Q-factor of the circuit is ________. (round off to nearest
integer)
12 | |
24 | |
48 | |
96 |
Question 3 Explanation:
We have, overall Q-factor of given circuit is,
Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48
Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48
Question 4 |
A 0.1\mu F capacitor charged to 100 V is discharged through a 1 k\Omega resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______
0.25 | |
0.65 | |
0.45 | |
0.85 |
Question 4 Explanation:
\begin{aligned} v_c(t)&=V_0e^{-t/\tau } \\ V_0&=100V \\ \tau &=RC=(10^3)(10^{-7}) \\ &=10^{-4}sec \\ \therefore \;v_c(t)&=100e^{-10^4 t }V \end{aligned}
Let the time required by the voltage across the capacitor to drop to 1 V is t_1,
\begin{aligned} \therefore \; v_c(t_1)&=100e^{-10^4t_1} \\ \text{But, } v_c(t_1)&=0 \\ \text{So, }0&=100e^{-10^4t_1} \\ t_1&=0.46msec \end{aligned}
Question 5 |
The voltage across the circuit in the figure, and the current through it, are given by the
following expressions:
v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V
i(t) = 5 + X cos(\omega t) A
where \omega =100 \pi radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).
v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V
i(t) = 5 + X cos(\omega t) A
where \omega =100 \pi radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).
5 | |
10 | |
15 | |
20 |
Question 5 Explanation:
Given that,
v(t)=5-10 \cos (\omega t+60^{\circ})
i(t)=5+C \cos (\omega t-0^{\circ})
P_{req}=0
0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
X=10
v(t)=5-10 \cos (\omega t+60^{\circ})
i(t)=5+C \cos (\omega t-0^{\circ})
P_{req}=0
0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]
X=10
There are 5 questions to complete.