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Sinusoidal Steady State-MCQs
1. The value of current through the 1 Farad capacitor of figure is
0.5F
1
1
1F
2
2 sin 100 t
AC
1H
1
1
0.5F
(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
2 sin 100 t
2+s
AC
a
1F
b
1+2/s
The product of the opposite arms are equal
𝑽𝒂 = 𝑽𝒃
𝑽𝒂 =
𝑽𝒃 =
𝟐 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕
× 𝟏 = 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕
𝟐
𝟐 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕
𝟐
× (𝟏 + ) = 𝒔𝒊𝒏𝟏𝟎𝟎𝒕
𝒔
𝟐(𝟏 + 𝒔 )
𝟐
The current through 1F capacitor is zero
Option (a)
2. The half – power bandwidth of the resonant circuit of figure can be increased by:
R1
R2
C
L
(a) increasing R1
(b) decreasing R1
Sol.
(c) increasing R2
(d) decreasing R2
[GATE 1989: 2 Marks]
𝟏
Bandwidth of the circuit ∝ 𝑸
For increasing bandwidth, Q is to be decreased
𝝎𝑳
𝟏
𝑸=
=
𝑹𝟏 𝝎𝑪𝑹𝟏
If R1 → 0, R2→∞, the circuit has only L & C elements and has high selectivity.
𝑹𝟏 ↑, 𝑹𝟐 ↓, 𝑸 ↓ 𝑩𝑾 ↑
Option (a) & (d)
3. The resonant frequency of the series circuit shown in figure is
M=1H
2H
(a)
(b)
1
𝐻
4𝜋√3 𝑍
1
𝐻
4𝜋 𝑍
2H
2F
(c)
(d)
1
2𝜋√10
1
4𝜋√2
𝐻𝑍
𝐻𝑍
[GATE 1990: 2 Marks]
Sol.
The coils are connected in a series opposing way. As the current is entering the
dot of coil L1 and leaving the dot of coil L2
𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 − 𝟐𝑴
=𝟐+𝟐−𝟐
= 𝟐𝑯
At resonance XL = XC
𝝎 𝑳𝒆𝒒 =
𝝎=
𝟏
√𝑳𝒆𝒒 𝑪
=
𝟏
𝝎𝑪
𝟏
√𝟐 × 𝟐
𝒇=
=
𝟏
𝒓𝒂𝒅/𝒔𝒆𝒄
𝟐
𝟏
𝑯𝒛
𝟒𝝅
Option (b)
4. In the series circuit shown in figure, for series resonance, the value of the coupling
coefficient k will be
K
18
-j12
j2
(a) 0.25
(b) 0.5
j8
(c) 0.999
(d) 1.0
[GATE 1993: 1 Mark]
Sol.
The coils are connected in series additive manner as the current is entering the
dot in both coils
𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 + 𝟐𝑴
𝑴 = 𝑲√𝑳𝟏 𝑳𝟐 = 𝑲√𝒋𝟐. 𝒋𝟖 = 𝑲𝒋𝟒
At resonance |𝑿𝑳 | = |𝑿𝑪 |
𝑿𝑳 = 𝑿𝑳𝟏 + 𝑿𝑳𝟐 + 𝑿𝑴
= 𝒋𝟐 + 𝒋𝟖 + 𝟐𝑲𝒋𝟒
𝒋𝟏𝟎 + 𝟐𝑲𝒋𝟒
At resonance 𝑿𝑳 = 𝑿𝑪
𝒋𝟏𝟎 + 𝟐𝑲𝒋𝟒 = 𝒋𝟏𝟐
𝟐𝑲𝒋. 𝟒 = 𝒋. 𝟐
𝒋𝑲. 𝟖 = 𝒋. 𝟐
𝑲=
𝟐 𝟏
= = 𝟎. 𝟐𝟓
𝟖 𝟒
Option (a)
5. In figure, A1, A2, and A3 are ideal ammeters. If A1 reads 5A, A2 reads 12A, then A3
should read
R
A1
A3
C
A2
AC
100 sin ωt
(a) 7 A
(b) 12 A
(c) 13 A
(d) 17 A
[GATE 1993: 2 Marks]
Sol.
Since the source is a.c.
𝑰𝟑(𝒓𝒎𝒔) = √𝑰𝟐𝟏 (𝒓𝒎𝒔) + 𝑰𝟐𝟐 (𝒓𝒎𝒔)
= √𝟓𝟐 + 𝟏𝟐𝟐
= √𝟏𝟔𝟗 = 𝟏𝟑𝑨
Option (c)
6. A series LCR circuit consisting of 𝑅 = 10 Ω, |𝑋𝐿 | = 20 Ω 𝑎𝑛𝑑 |𝑋𝐶 | = 20 Ω is connected
across an a.c. supply of 200 V rms. The rms voltage across the capacitor is
(a) 200∠ − 900 𝑉
(c) 400∠ + 900 𝑉
(b) 200∠ + 900 𝑉
(d) 400∠ − 900 𝑉
[GATE 1994: 1 Mark]
Soln.
For a series LCR circuit
𝒁 = 𝑹 + 𝒋𝑿𝑳 − 𝒋𝑿𝑪
𝒔𝒊𝒏𝒄𝒆 𝑿𝑳 = 𝑿𝑪 ,
𝑽
𝑰=𝑹=
𝟐𝟎𝟎
𝟏𝟎
𝒁=𝑹
= 𝟐𝟎𝑨
The voltage across the capacitor= 𝑰. (−𝒋𝑿𝑪 )
= 𝟐𝟎. (−𝒋𝟐𝟎)
= −𝟒𝟎𝟎𝒋
= 𝟒𝟎𝟎∠ − 𝟗𝟎𝟎 𝑽
Option (d)
7. A DC voltage source is connected across a series R-L-C circuit. Under steady state
conditions, the applied DC voltage drops entirely across the
(a) R only
(b) L only
Soln.
(c) C only
(d) R and L combination
[GATE 1995: 1 Mark]
Under steady state condition, inductor behaves as a short circuit and capacitor
as an open circuit.
The applied voltage drops entirely across the capacitor
R
Option (C)
L
C
8. Consider a DC voltage source connected to a series R-C circuit. When the steady state
reaches, the ratio of the energy stored in the capacitor to the total energy supplied by the
voltage source, is equal to
(a) 0.362
(c) 0.632
(b) 0.500
(d) 1.000
[GATE 1995: 1 Mark]
Soln.
The total energy supplied by the source
𝑾𝑺 = 𝑽𝑰. 𝒕
𝒕
= 𝑽 ∫ 𝑰𝒅𝒕 = 𝑽𝒒
𝟎
= 𝑽𝑪𝑽
= 𝑪𝑽𝟐
Voltage across the uncharged capacitor
𝑽𝑪 (𝒕) = 𝑽𝑺 (𝟏 − 𝒆
−𝒕⁄
𝑹𝑪 )
Power in the capacitor = 𝑽𝑺 (𝟏 − 𝒆
=
𝑽𝟐𝑺
𝑹
(𝒆
−𝒕⁄
𝑹𝑪
−𝒕⁄
𝒅
𝑹𝑪 ) 𝑪
𝑽 (𝟏
𝒅𝒕 𝑺
−𝒆
−𝒆
−𝒕⁄
𝑹𝑪 )
−𝟐𝒕⁄
𝑹𝑪 )
Energy stored in the capacitor at steady state
∞
∞
𝑽𝟐𝑺
𝑽𝟐𝑺
−𝒕⁄
−𝟐𝒕
𝑹𝑪
∫𝒆
𝒅𝒕 −
∫ 𝒆 ⁄𝑹𝑪 𝒅𝒕
𝑹
𝑹
𝟎
=
𝑽𝟐𝑺
𝑹
[𝑹𝑪 −
𝟎
𝑹𝑪
𝟐
]
𝟏
= 𝟐 𝑪𝑽𝟐𝑺
𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒕𝒐𝒓𝒆𝒅 𝒊𝒏 𝒕𝒉𝒆𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓
= 𝟎. 𝟓
𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒔𝒐𝒖𝒓𝒄𝒆
Option (b)
9. The current, i(t), through a 10-Ω 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:
(c) 5 A, 250 W, respectively
(d) 11 A, 1210 W respectively
(a) √41A, 410 W, respectively
(b) √35 A, 350 W, respectively
[GATE 1995: 1 Mark]
Soln.
𝒊(𝒕) = 𝟑 + 𝟒 𝐬𝐢𝐧(𝟏𝟎𝟎𝒕 + 𝟒𝟓𝟎 ) + 𝟒 𝐬𝐢𝐧(𝟑𝟎𝟎𝒕 + 𝟔𝟎𝟎 )
𝟒
𝟐
𝟒
𝟐
𝑰𝒓𝒎𝒔 = √𝟑𝟐 + ( ) + ( ) = 𝟓 𝒂𝒎𝒑𝒔
√𝟐
√𝟐
Power dissipated = 𝑰𝟐𝒓𝒎𝒔 𝑹
= (𝟓)𝟐 × 𝟏𝟎
= 𝟐𝟓𝟎 𝒘𝒂𝒕𝒕𝒔
Option (c)
10. In the circuit of Figure assume that the diodes are ideal and the meter is an average
indicating ammeter. The ammeter will read
A
+
D2
4 sin (ωt)
Volts
10 k
10 k
(a) 0.4 √2 mA
(b) 0.4 mA
(c)
(d)
0.8
𝜋
0.4
𝜋
𝑚𝐴
𝑚𝐴
[GATE 1996: 1 Mark]
Soln.
For positive half cycle of input D1 is conducting and D2 is off
The current through ammeter is the average value of positive half sine wave i.e.
𝑰𝒂𝒗𝒈 =
𝑽𝒎 𝟏
𝟒
𝟎. 𝟒
× =
=
𝒎𝑨
𝟑
𝝅 𝑹 𝝅 × 𝟏𝟎 × 𝟏𝟎
𝝅
Option (d)
11. A series RLC circuit has a resonance frequency of 1 KHz and a quality factor Q = 100. If
each 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.
𝑸=
=
𝝎𝑳
𝑳
= 𝟐𝝅𝒇𝒐
𝑹
𝑹
𝟐𝝅
𝑳 𝟏 𝑳
= √
𝟐𝝅√𝑳𝑪 𝑹 𝑹 𝑪
When R, L and C are doubled 𝑸′ =
𝑸
𝟐
Option (b)
12. An input voltage 𝑣(𝑡) = 10√2 cos(𝑡 + 100 ) + 10√5 cos(2𝑡 + 100 ) V is applied to a
series combinationof R = 1 Ω and an inductance L = 1H. The resulting steady state
current i(t) in ampere is
(a) 10 cos(𝑡 + 550 ) + 10 cos(2𝑡 + 100 + 𝑡𝑎𝑛−1 2)
3
(b) 10 cos(𝑡 + 550 ) + 10√2 cos(2𝑡 + 55)
(c) 10 cos(𝑡 − 350 ) + 10 cos(2𝑡 + 100 − 𝑡𝑎𝑛−1 2)
3
(d) 10 cos(𝑡 − 350 ) + 10√2 cos(2𝑡 − 35)
[GATE 2003: 2 Marks]
Soln.
𝑽(𝒕) = 𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎 ) + 𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎 )
𝑽(𝒕) = 𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎 ) + 𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎 )
𝝎𝟏 = 𝟏, 𝝎𝟐 = 𝟐
Steady state current 𝒊(𝒕) = 𝒊𝟏 (𝒕) + 𝒊𝟐 (𝒕)
𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎 ) 𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎 )
=
+
𝑹 + 𝒋𝛚𝟏 𝑳
𝑹 + 𝒋𝛚𝟏 𝑳
=
=
𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎 ) 𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎 )
+
𝟏+𝒋
𝟏 + 𝟐𝒋
𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎 )
√𝟐∠𝟒𝟓𝟎
+
𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎 )
√𝟓∠ 𝐭𝐚𝐧−𝟏 𝟐
= 𝟏𝟎 𝐜𝐨𝐬(𝒕 − 𝟑𝟓𝟎 ) + 𝟏𝟎 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟐 − 𝐭𝐚𝐧−𝟏 𝟐)
Option (C)
1
1
13. The circuit shown in the figure, with 𝑅 = 3 Ω, 𝐿 = 4 𝐻, 𝐶 = 3 F has input voltage V(t) =
Sin2t. The resulting current i(t) is
i(t)
V(t)
(a) 5 sin(2𝑡 + 53. 10 )
(b) 5 sin(2𝑡 − 53. 10 )
R
L
C
(c) 25 sin(2𝑡 + 53. 10 )
(d) 25 sin(2𝑡 − 53. 10 )
[GATE 2003: 1 Mark]
Soln.
Admittance
𝒀=
𝟏
𝟏
+
+ 𝒋𝝎𝑪
𝑹 𝒋𝝎𝑳
=𝟑+
𝟒
+ 𝒋𝟐 × 𝟑
𝒋𝟐
= 𝟑 − 𝒋𝟐 + 𝒋𝟔
= 𝟑 + 𝒋𝟒
𝒊(𝒕) = 𝑽(𝒕). 𝒚
= (𝟑 + 𝟒𝒋) 𝐬𝐢𝐧 𝟐𝒕
= 𝟓 𝐬𝐢𝐧 𝟐𝒕 ∠ 𝐭𝐚𝐧−𝟏 ( 𝟒/𝟑)
= 𝟓 𝐬𝐢𝐧(𝟐𝒕 + 𝟓𝟑. 𝟏𝟎 )
Option (a)
14. For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is
𝑉𝑖 (𝑡) = √2 sin 103 𝑡. The output voltage V0 (t) is equal to
R
vi(t)
v0(t)
(a) sin(103 𝑡 − 450 )
(b) sin(103 𝑡 + 450 )
Soln.
(c) sin(103 𝑡 − 530 )
(d) sin(103 𝑡 + 530 )
[GATE 2004: 1 Mark]
𝑽𝟏 (𝒕) = √𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑 𝒕
𝝎 = 𝟏𝟎𝟑 𝒓𝒂𝒅𝒊𝒂𝒏𝒔/𝒔𝒆𝒄
𝑹𝑪 = 𝟏𝒎 𝒔𝒆𝒄
𝑽𝟎 (𝒕) =
𝑽𝟏 (𝒕)
𝟏
(𝑹 + 𝒋𝝎𝒄)
×
𝟏
𝒋𝝎𝒄
𝑽𝟏 (𝒕)
√𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑 𝒕
=
=
𝒋𝝎𝑹𝑪 + 𝟏 𝒋𝟏𝟎𝟑 × 𝟏𝟎−𝟑 + 𝟏
=
√𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑 𝒕
𝟏+𝒋
= 𝐬𝐢𝐧(𝟏𝟎𝟑 𝒕 − 𝟒𝟓𝟎 )
Option (a)
15. Consider the following statements S1 and S2
S1: At the resonant frequency the impedance of a series R-L-C circuit is zero
S2: In a parallel G-L-C circuit, increasing the conductance G results in increase in its Q
factor.
Which one of the following is correct?
(a) S1 is FALSE and S2 is TRUE
(b) Both S1 and S2 are TRUE
Soln.
(c) S1 is TRUE and S2 is FALSE
(d) Both S1 and S2 are FALSE
[GATE 2004: 2 Marks]
The impedance of a series RLC circuit at resonant frequency is
𝟏
𝒁 = 𝑹 + 𝒋 (𝝎𝑳 − 𝝎.𝑪)
𝟏
𝝎𝑳 = 𝝎.𝑪
At resonance
So, Z = R purely resistive
In a parallel RLC circuit,
𝟏
3dB bandwidth𝝎𝟐 − 𝝎𝟏 = 𝑹𝑪
𝑸=
𝝎𝒓
𝝎𝒓
=
= 𝝎𝑹 𝑹𝑪
𝝎𝟐 − 𝝎𝟏 𝟏⁄
𝑹𝑪
𝝎𝒓 is the resonant frequency.
As conductance 𝑮 ↑ 𝑹 ↓ 𝑸 ↓ (As conductance G increases R reduces so Q reduces)
Option (d)
ZL = (7+4j)
20 00V
16. An AC source of RMS voltage 20 V with internal impedance 𝑍𝑠 = (1 + 2𝑗)Ω feeds a
load of impedance 𝑍𝐿 = (7 + 4𝑗)Ω in the figure below. The reactive power consumed by
the load is
ZS = (1+2j)
DC
(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
𝑰=
=
=
𝟐𝟎
𝟏𝟎
=
𝟖 + 𝟔𝒋 𝟒 + 𝟑𝒋
𝟏𝟎
√𝟒𝟐 + 𝟑𝟐 ∠ 𝐭𝐚𝐧−𝟏 𝟑/𝟒
𝟏𝟎
= 𝟐∠ − 𝟑𝟔. 𝟖𝟕𝟎
𝟓∠𝟑𝟔. 𝟖𝟕
Reactive power = 𝑰𝟐 𝑿𝑳
= (𝟐)𝟐 × 𝟒
= 16 VAR
Option (b)