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

𝑽𝒂 = 𝑽𝒃

𝑽𝒂 =𝟐 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕

𝟐× 𝟏 = 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕

𝑽𝒃 =𝟐 𝐬𝐢𝐧 𝟏𝟎𝟎𝒕

𝟐(𝟏 +𝟐

𝒔)

× (𝟏 +𝟐

𝒔) = 𝒔𝒊𝒏𝟏𝟎𝟎𝒕

The current through 1F capacitor is zero

Option (a)

2. The half – power bandwidth of the resonant circuit of figure can be increased by:

R2C

R1

L

(a) increasing R1

(b) decreasing R1

(c) increasing R2

(d) decreasing R2

[GATE 1989: 2 Marks]

Sol. Bandwidth of the circuit ∝𝟏

𝑸

For increasing bandwidth, Q is to be decreased

𝑸 =𝝎𝑳

𝑹𝟏=

𝟏

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 2H 2F

(a) 1

4𝜋√3𝐻𝑍

(b) 1

4𝜋𝐻𝑍

(c) 1

2𝜋√10𝐻𝑍

(d) 1

4𝜋√2𝐻𝑍

[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 =𝟏

𝝎𝑪

𝝎 =𝟏

√𝑳𝒆𝒒𝑪=

𝟏

√𝟐 × 𝟐=

𝟏

𝟐 𝒓𝒂𝒅/𝒔𝒆𝒄

𝒇 =𝟏

𝟒𝝅𝑯𝒛

Option (b)

4. In the series circuit shown in figure, for series resonance, the value of the coupling

coefficient 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 the

dot in both coils

𝑳𝒆𝒒 = 𝑳𝟏 + 𝑳𝟐 + 𝟐𝑴

𝑴 = 𝑲√𝑳𝟏𝑳𝟐 = 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

𝟖=

𝟏

𝟒= 𝟎. 𝟐𝟓

Option (a)

5. In figure, A1, A2, and A3 are ideal ammeters. If A1 reads 5A, A2 reads 12A, then A3

should read

A3

AC

A2

A1

R

C

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) = √𝑰𝟏𝟐(𝒓𝒎𝒔) + 𝑰𝟐

𝟐(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

(b) 200∠ + 900 𝑉

(c) 400∠ + 900 𝑉

(d) 400∠ − 900 𝑉

[GATE 1994: 1 Mark]

Soln. For a series LCR circuit

𝒁 = 𝑹 + 𝒋𝑿𝑳 − 𝒋𝑿𝑪

𝒔𝒊𝒏𝒄𝒆 𝑿𝑳 = 𝑿𝑪, 𝒁 = 𝑹

𝑰 =𝑽

𝑹=

𝟐𝟎𝟎

𝟏𝟎= 𝟐𝟎𝑨

The voltage across the capacitor= 𝑰. (−𝒋𝑿𝑪)

= 𝟐𝟎. (−𝒋𝟐𝟎)

= −𝟒𝟎𝟎𝒋

= 𝟒𝟎𝟎∠ − 𝟗𝟎𝟎𝑽

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

(c) C only

(d) R and L combination

[GATE 1995: 1 Mark]

Soln. 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 L C

Option (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

(b) 0.500

(c) 0.632

(d) 1.000

[GATE 1995: 1 Mark]

Soln. The total energy supplied by the source

𝑾𝑺 = 𝑽𝑰. 𝒕

= 𝑽 ∫ 𝑰𝒅𝒕 = 𝑽𝒒

𝒕

𝟎

= 𝑽𝑪𝑽

= 𝑪𝑽𝟐

Voltage across the uncharged capacitor

𝑽𝑪(𝒕) = 𝑽𝑺 (𝟏 − 𝒆−𝒕

𝑹𝑪⁄ )

Power in the capacitor = 𝑽𝑺 (𝟏 − 𝒆−𝒕

𝑹𝑪⁄ ) 𝑪𝒅

𝒅𝒕𝑽𝑺 (𝟏 − 𝒆

−𝒕𝑹𝑪⁄ )

=𝑽𝑺

𝟐

𝑹(𝒆

−𝒕𝑹𝑪⁄ − 𝒆

−𝟐𝒕𝑹𝑪⁄ )

Energy stored in the capacitor at steady state

𝑽𝑺𝟐

𝑹∫ 𝒆

−𝒕𝑹𝑪⁄ 𝒅𝒕 −

𝑽𝑺𝟐

𝑹

∞

𝟎

∫ 𝒆−𝟐𝒕

𝑹𝑪⁄

∞

𝟎

𝒅𝒕

=𝑽𝑺

𝟐

𝑹[𝑹𝑪 −

𝑹𝑪

𝟐]

=𝟏

𝟐𝑪𝑽𝑺

𝟐

𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒕𝒐𝒓𝒆𝒅 𝒊𝒏 𝒕𝒉𝒆𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒐𝒓

𝒆𝒏𝒆𝒓𝒈𝒚 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒔𝒐𝒖𝒓𝒄𝒆= 𝟎. 𝟓

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:

(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. 𝒊(𝒕) = 𝟑 + 𝟒 𝐬𝐢𝐧(𝟏𝟎𝟎𝒕 + 𝟒𝟓𝟎) + 𝟒 𝐬𝐢𝐧(𝟑𝟎𝟎𝒕 + 𝟔𝟎𝟎)

𝑰𝒓𝒎𝒔 = √𝟑𝟐 + (𝟒

√𝟐)

𝟐

+ (𝟒

√𝟐)

𝟐

= 𝟓 𝒂𝒎𝒑𝒔

Power dissipated = 𝑰𝒓𝒎𝒔𝟐 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

4 sin (ωt)Volts

D2

10 k

10 k

+

-

(a) 0.4 √2 mA

(b) 0.4 mA

(c) 0.8

𝜋𝑚𝐴

(d) 0.4

𝜋𝑚𝐴

[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

𝝅×

𝟏

𝑹=

𝟒

𝝅 × 𝟏𝟎 × 𝟏𝟎𝟑=

𝟎. 𝟒

𝝅𝒎𝑨

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

𝑹= 𝟐𝝅𝒇𝒐

𝑳

𝑹

=𝟐𝝅

𝟐𝝅√𝑳𝑪

𝑳

𝑹=

𝟏

𝑹√

𝑳

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−12)

(b) 10 cos(𝑡 + 550) + 10√3

2cos(2𝑡 + 55)

(c) 10 cos(𝑡 − 350) + 10 cos(2𝑡 + 100 − 𝑡𝑎𝑛−12)

(d) 10 cos(𝑡 − 350) + 10√3

2cos(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)

𝑹 + 𝒋𝛚𝟏𝑳+

𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎)

𝑹 + 𝒋𝛚𝟏𝑳

=𝟏𝟎√𝟐 𝐜𝐨𝐬(𝒕 + 𝟏𝟎𝟎)

𝟏 + 𝒋+

𝟏𝟎√𝟓 𝐜𝐨𝐬(𝟐𝒕 + 𝟏𝟎𝟎)

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)

13. The circuit shown in the figure, with 𝑅 =1

3Ω, 𝐿 =

1

4𝐻, 𝐶 = 3 F has input voltage V(t) =

Sin2t. The resulting current i(t) is

i(t)

V(t) R L C

(a) 5 sin(2𝑡 + 53. 10)

(b) 5 sin(2𝑡 − 53. 10)

(c) 25 sin(2𝑡 + 53. 10)

(d) 25 sin(2𝑡 − 53. 10)

[GATE 2003: 1 Mark]

Soln. Admittance

𝒀 =𝟏

𝑹+

𝟏

𝒋𝝎𝑳+ 𝒋𝝎𝑪

= 𝟑 +𝟒

𝒋𝟐+ 𝒋𝟐 × 𝟑

= 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 103𝑡. The output voltage V0 (t) is equal to

R

vi(t) v0(t)

(a) sin(103𝑡 − 450)

(b) sin(103𝑡 + 450)

(c) sin(103𝑡 − 530)

(d) sin(103𝑡 + 530)

[GATE 2004: 1 Mark]

Soln. 𝑽𝟏(𝒕) = √𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑𝒕

𝝎 = 𝟏𝟎𝟑 𝒓𝒂𝒅𝒊𝒂𝒏𝒔/𝒔𝒆𝒄

𝑹𝑪 = 𝟏𝒎 𝒔𝒆𝒄

𝑽𝟎(𝒕) =𝑽𝟏(𝒕)

(𝑹 +𝟏

𝒋𝝎𝒄)

×𝟏

𝒋𝝎𝒄

=𝑽𝟏(𝒕)

𝒋𝝎𝑹𝑪 + 𝟏=

√𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑𝒕

𝒋𝟏𝟎𝟑 × 𝟏𝟎−𝟑 + 𝟏

=√𝟐 𝐬𝐢𝐧 𝟏𝟎𝟑𝒕

𝟏 + 𝒋

= 𝐬𝐢𝐧(𝟏𝟎𝟑𝒕 − 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

(c) S1 is TRUE and S2 is FALSE

(d) Both S1 and S2 are FALSE

[GATE 2004: 2 Marks]

Soln. The impedance of a series RLC circuit at resonant frequency is

𝒁 = 𝑹 + 𝒋 (𝝎𝑳 −𝟏

𝝎.𝑪)

𝝎𝑳 =𝟏

𝝎.𝑪 At resonance

So, Z = R purely resistive

In a parallel RLC circuit,

3dB bandwidth𝝎𝟐 − 𝝎𝟏 =𝟏

𝑹𝑪

𝑸 =𝝎𝒓

𝝎𝟐 − 𝝎𝟏=

𝝎𝒓

𝟏𝑹𝑪⁄

= 𝝎𝑹𝑹𝑪

𝝎𝒓 is the resonant frequency.

As conductance 𝑮 ↑ 𝑹 ↓ 𝑸 ↓ (As conductance G increases R reduces so Q reduces)

Option (d)

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

DC

20

00V

ZS = (1+2j)

ZL =

(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

𝑰 =𝟐𝟎

𝟖 + 𝟔𝒋=

𝟏𝟎

𝟒 + 𝟑𝒋

=𝟏𝟎

√𝟒𝟐 + 𝟑𝟐∠ 𝐭𝐚𝐧−𝟏 𝟑/𝟒

=𝟏𝟎

𝟓∠𝟑𝟔. 𝟖𝟕= 𝟐∠ − 𝟑𝟔. 𝟖𝟕𝟎

Reactive power = 𝑰𝟐𝑿𝑳

= (𝟐)𝟐 × 𝟒

= 16 VAR

Option (b)