sinsusoidal steady state
TRANSCRIPT
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Sinusoidal Steady State-MCQs
1. The value of current through the 1 Farad capacitor of figure is
AC 2 sin 100 t
1
1
1
1
2
1H
0.5F
0.5F
1F
(a) zero
(b) one
(c) two
(d) three[GATE 1987: 2 Marks]
Sol. The given circuit is a bridge circuit and can be redrawn as
AC 2+s a b
1+2/s
1F2 sin 100 t
The product of the opposite arms are equal
=
= = = + ( +) = The current through 1F capacitor is zero
Option (a)
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2. The half power bandwidth of the resonant circuit of figure can be increased by:
R 2C
R 1
L
(a) increasing R 1 (b) decreasing R 1
(c) increasing R 2 (d) decreasing R 2
[GATE 1989: 2 Marks]
Sol. Bandwidth of the circuit
For increasing bandwidth, Q is to be decreased
= = If R 1 0, R 2, the circuit has only L & C elements and has high selectivity. , ,
Option (a) & (d)
3. The resonant frequency of the series circuit shown in figure is
M=1H
2H 2H 2F
(a) (b)
(c) (d)
[GATE 1990: 2 Marks]
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Sol. The coils are connected in a series opposing way. As the current is entering thedot of coil L 1 and leaving the dot of coil L 2
= + = + =
At resonance X L = X C
=
= = = / =
Option (b)
4. In the series circuit shown in figure, for series resonance, the value of the couplingcoefficient k will be
18 -j12 j2 j8
K
(a) 0.25(b) 0.5
(c) 0.999(d) 1.0
[GATE 1993: 1 Mark]
Sol. The coils are connected in series additive manner as the current is entering thedot in both coils
= + + = = . =
At resonance | | = | |
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= + + = + + +
At resonance = + = . = .
. = .
= = = .
Option (a)
5. In figure, A 1, A 2, and A 3 are ideal ammeters. If A 1 reads 5A, A 2 reads 12A, then A 3 should read
A 3
AC
A 2
A 1R
C
100 sin t
(a) 7 A(b) 12 A
(c) 13 A(d) 17 A
[GATE 1993: 2 Marks]
Sol. Since the source is a.c.
= + = + = =
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Option (c)
6. A series LCR circuit consisting of = 10 ,| | = 20 | | = 20 is connectedacross an a.c. supply of 200 V rms. The rms voltage across the capacitor is(a) 200 90 (b) 200+90 (c)
400+90 (d) 400 90 [GATE 1994: 1 Mark]
Soln. For a series LCR circuit
= + = , =
= = = The voltage across the capacitor = . = . = =
Option (d)
7. A DC voltage source is connected across a series R-L-C circuit. Under steady stateconditions, the applied DC voltage drops entirely across the
(a) R only(b) L only
(c) C only(d) R and L combination
[GATE 1995: 1 Mark]
Soln. Under steady state condition, inductor behaves as a short circuit and capacitoras an open circuit.
The applied voltage drops entirely across the capacitor
R L C
Option (C)
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8. Consider a DC voltage source connected to a series R-C circuit. When the steady statereaches, the ratio of the energy stored in the capacitor to the total energy supplied by thevoltage source, is equal to(a) 0.362(b) 0.500
(c) 0.632(d) 1.000
[GATE 1995: 1 Mark]
Soln. The total energy supplied by the source
= .
= =
= = Voltage across the uncharged capacitor
= Power in the capacitor =
= Energy stored in the capacitor at steady state
= [ ] =
= .
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Option (b)
9. The current, i(t), through a 10- resistor in series with an inductance, is given by
i(t) = 3 + 4 sin (100 t + 450
) + 4 sin (300 t + 600
) amperes.The RMS value of the current and the power dissipated in the circuit are:
(a) 41A, 410 W, respectively(b) 35 A, 350 W, respectively
(c) 5 A, 250 W, respectively(d) 11 A, 1210 W respectively
[GATE 1995: 1 Mark]
Soln. = + ++ +
=
+ + =
Power dissipated = = =
Option (c)
10. In the circuit of Figure assume that the diodes are ideal and the meter is an averageindicating ammeter. The ammeter will read
A
4 sin ( t)Volts
D2
10 k
10 k
+
-
(a) 0.4 2 mA(b) 0.4 mA
(c) .8 (d) .
[GATE 1996: 1 Mark]
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Soln. For positive half cycle of input D 1 is conducting and D 2 is off
The current through ammeter is the average value of positive half sine wave i.e.
= =
=
.
Option (d)
11. A series RLC circuit has a resonance frequency of 1 KHz and a quality factor Q = 100. Ifeach of R, L and C is doubled from its original value, the new Q of the circuit is
(a) 25
(b) 50
(c) 100
(d) 200[GATE 2003: 1 Mark]
Soln.
= =
=
=
When R, L and C are doubled = Option (b)
12. An input voltage = 10 2cos +10+10 5cos 2 + 10 V is applied to aseries combination of R = 1 and an inductance L = 1H. The resulting steady statecurrent i(t) in ampere is(a) 10cos +55+10cos 2 +10 +2 (b) 10cos +55+10 cos 2 +55 (c) 10cos 35+10cos 2 +10 2
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(d) 10cos 35+10 cos 2 35 [GATE 2003: 2 Marks]
Soln.
= ++ +
= ++ + = , = Steady state current = +
= ++ + ++
= ++ + ++
= + + + = + + Option (C)
13. The circuit shown in the figure, with = , = , = 3 F has input voltage V(t) =Sin2t. The resulting current i(t) is
i(t)
V(t)R L C
(a) 5sin2 +53.1 (b) 5sin2 53.1 (c)
25sin2 +53.1 (d) 25sin2 53.1
[GATE 2003: 1 Mark]
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Soln. Admittance
= + +
= + +
= + = +
= . = + = / = + . Option (a)
14. For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is
= 2sin10. The output voltage V 0 (t) is equal toR
vi(t) v0(t)
(a) sin10 45 (b) sin10 +45 (c)
sin10 53 (d) sin10 +53
[GATE 2004: 1 Mark]
Soln. = = / =
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= +
= + = +
= + =
Option (a)
15. Consider the following statements S 1 and S 2 S1: At the resonant frequency the impedance of a series R-L-C circuit is zeroS2: In a parallel G-L-C circuit, increasing the conductance G results in increase in its Qfactor.Which one of the following is correct?
(a) S1 is FALSE and S 2 is TRUE(b) Both S 1 and S 2 are TRUE
(c) S1 is TRUE and S 2 is FALSE(d) Both S 1 and S 2 are FALSE
[GATE 2004: 2 Marks]
Soln. The impedance of a series RLC circuit at resonant frequency is
= + . = . At resonance
So, Z = R purely resistive
In a parallel RLC circuit,
3dB bandwidth =
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= = =
is the resonant frequency.As conductance (As conductance G increases R reduces so Q reduces)Option (d)
16. An AC source of RMS voltage 20 V with internal impedance = 1+2 feeds aload of impedance = 7+4 in the figure below. The reactive power consumed bythe load is
DC
2 0 0 0 V
ZS = (1+2j)
Z L =
( 7 + 4 j )
(a) 8 VAR(b) 16 VAR
(c) 28 VAR(d) 32 VAR
[GATE 2009: 2 Marks]
Soln. The current I in the given circuit is
=
+ =
+
= + /
= . = .
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Reactive power = = = 16 VAR
Option (b)