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Alternating Current Test - 9

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Alternating Current Test - 9

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Question 1
4 / -1

In an a.c. circuit voltage V and current i are given by V = 100 sin 100 t volts, i = 100 sin (100t + p/3) mA. The power dissipated in the circuit is :

A
10⁴ W

B
10 W

C
2.5 W

D
5 W

Solution

V_rms = 100/√2 V
I_rms = 100/√2 mA
= 10-1/√2 A
π/3 = 30º
Average power = V_rms×I_rms× cos
= (100/√2)×(10-1√2)× cos 30º
= 10/2×(1/2)
= 10/4
= 2.5W
Question 2
4 / -1

The potential difference V and current i flowing through an a.c. circuit are given by V = 5 cos wt volt, i = 2 sin wt amp. the power dissipated in the circuit.

A
0 W

B
10 W

C
5 W

D
2.5 W

Solution

The voltage can be written as,
V=5[sinωt+(π/2)]
The angle between the voltage and the current is 90∘.
The power dissipated in the circuit is given as,
P=V_rmsI_rmscosϕ
=V_rmsI_rmscos90o
=0
Thus, the power dissipated in the circuit is 0.
Question 3
4 / -1

An a.c. circuit consists of an inductor of inductances 0.5 H and a capacitor of capacitance 8 mF in series. The current in the circuit is maximum when the angular frequency of an a.c. source is

A
500 Hz

B
2 × 10⁵ Hz

C
4000 Hz

D
5000 Hz

Solution

Given,
Inductance and capacitance are joined in series, and inductance L=0.5H,
And capacitance C=8μF=8×10⁻⁶F.
We know that,
ω=1/√(LC)
ω=1/√(8×10⁻⁶×0.5)=1/(2×10⁻³)=0.5×10³HZ=500HZ
Question 4
4 / -1

The rms value of an AC of 50 Hz is 10 amp. The time taken by an alternating current in reaching from zero to maximum value and the peak value will be ;

A
2 × 10^-2 sec and 14.14 amp.

B
1 × 10^-2 sec and 7.07 amp.

C
5 × 10^-3 sec and 7.07 amp.

D
5 × 10^-3 sec and 14.14 amp.

Solution

Time taken by alternating current to reach maximum from zero is 4T.
where T is the time period of the AC function.
T=1/f, Frequency (f)=50 Hz.
So time taken to reach maximum=T/4=1/4f=1/(4×50)=5×10⁻³ sec
I_RMS=I₀/√2, I0=Peak value of current.
Peak value of current I₀=√2×I_RMS=√2×10=14.14 A
Question 5
4 / -1

A voltage of peak value 283 V varying frequency is applied to a series L-C-R combination in which R = 3W; L = 25 mH and C = 400 mF. Then, the frequency (in Hz) of the source at which maximum power is dissipated in the above, is

A
51.5

B
50.7

C
51.1

D
50.3

Solution

Here, V₀=283V,R=3Ω,L=25×10⁻³H
C=400μF=4×10⁻⁴F
Maximum power is dissipated at resonance, for which

ν=1/[2π√(LC)]= (1×7)/[2×22(25×10⁻³×4×10⁻⁴)]
=(7×10³)/(44√10)
=50.3Hz
Question 6
4 / -1

The power factor of the circuit is 1/. The capacitance of the circuit is equal to

A
400 mF

B
300 mF

C
500 mF

D
200 mF

Solution

Here, it is given that,
power factor in the circuit, = 1/√2
Resistance, R = 10Ω
inductance, L = 0.1 H
Supply voltage, V = 2sin(100t)
peak voltage, V₀ = 2
ω = 100 rad/s
capacitance = ?
=> power factor = cos∅ = R/Z
1/√2 = 10/Z
=> By Taking the square on both the sides,
1/2 = 100/Z²
Z² = 200
=> Now, we can use below formula to calculate the capacitance of the circuit.
R2 + (Xc - XL)² = Z²
10² + (1 / cω - Lω)² = 200
(1/cω - Lω)² = 100
1/cω - Lω = 10
1/cω = Lω + 10
1/cω = 0.1 * 100 + 10
1/cω = 20
c = 1/(20xω)
c = 1/(20x100)
c = 1/2000 F
c = 10⁶ / 2000 (μ)F [∵ 1 F = 10⁶ (μ)F ]
= 500 (μ) F
Question 7
4 / -1

An ac-circuit having supply voltage E consists of a resistor of resistance 3W and an inductor of reactance 4W as shown in the figure. The voltage across the inductor at t = p/w is

A
2 volts

B
10 volts

C
zero

D
4.8 volts

Solution

Here,
X_L = 4 Ω
R = 3 Ω
Z = √(X_L² + R²)=√(4² + 3²)
= 5 Ω
E₀ = 10 V
In the LR circuit current in the ckt is given by
I = (E₀/Z)sin(ωt - Φ)
Φ = tan^-1(X_L/R)
=> Φ = tan^-1(4/3)
I = (E_0/Z)sin(ωt - Φ)
= (10/5) sin(ωt - Φ)
I at t = T/2
= 2sin(ωT/2 – Φ)
= 2sin(π - Φ)
= 2sin(tan^-1(4/3))
= 2×0.8
= 1.6 A
Potential difference across the resistor = 1.6×R
= 1.6×3
= 4.8 V
Potential difference across the inductor = 1.6XL
= 1.6×4 = 6.4 V
Question 8
4 / -1

When 100 V DC is applied across a solenoid a current of 1A flows in it. When 100 V AC is applied across the same coil, the current drops to 0.5 A. If the frequency of the AC source is 50 Hz, the impedance and inductance of the solenoid are :

A
100 W, 0.93 H

B
200W, 1.0 H

C
10W, 0.86 H

D
200W, 0.55 H

Solution

Impedance= V_ac/I_ac =100V/0.5A =200Ω
R= V_dc/I_dc =100/1=100Ω
Lω =√ (impedance²−R²)=√(200²−100²)=173.2
L=173.2/ ω= 173.2/2π50=0.55H
Question 9
4 / -1

An inductive circuit contains a resistance of 10 ohm and an inductance of 2.0 henry. If an ac voltage of 120 volt and frequency of 60 Hz is applied to this circuit, the current in the circuit would be nearly :

A
0.32 amp

B
0.16 amp

C
048 amp

D
0.80 amp

Solution

i= E/√[R²+(2πfL)²]
=120/√[100+(753.6)2]
=120/753.66
=0.1592≅0.16Amp
Question 10
4 / -1

In the circuit shown if the emf of source at an instant is 5V, the potential difference across capacitor at the same instant is 4V. The potential difference across R at that instant may be

A
3V

B
√41v

C

D
None

Solution
Question 11
4 / -1

Let f = 50 Hz, and C = 100 mF in an AC circuit containing a capacitor only. If the peak value of the current in the circuit is 1.57 A at t = 0. The expression for the instantaneous voltage across the capacitor will be

A
E = 50 sin 100 pt

B
E = 100 sin (50 pt)

C
E = 50 sin (100 pt – p/2)

D
E = 50 sin (100 pt + p/2)

Solution

Peak value of voltage
V₀=i₀X_C =i₀/2πvC
=>1.57/(2x3.14x50x100x10^-6)
Hence if equation of current i = i_o sin ω t then in capacitive circuit voltage is
V=V₀ (ωt - π/2)
=>50[sin2π x50t- (π/2)]
=50[100πt-(π/2)]
Question 12
4 / -1

In a series CR circuit shown in figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8V. Then the voltage across R, and the phase difference between current and the applied voltage will respectively be

A
6V, tan^-1

B
3V, tan^-1

C
6V, tan^-1

D
None

Solution

V_R=√ [V₂−V_C²]=√ [(10)²−(8)²]=6V
tanϕ= X_C/X_R = V_C/V_R =8/6=4/3
Question 13
4 / -1

The phase difference between current and voltage in an AC circuit is p/4 radian. If the frequency of AC is 50 Hz, then the phase difference is equivalent to the time difference :

A
0.78 s

B
15.7 ms

C
0.25 s

D
2.5 ms

Solution

The period of a whole cycle at 50Hz is 1/50 = 0.02 Seconds.(This is the time it would take a generator rotor to complete one rotation).That is a full circle 2π radians (or 360 Degrees).
The phase difference (angle between curent and voltage) is π/4 part of this circle.That corresponds to 0.02 Seconds * (π/4) / (2 π) = 0.02 Seconds * 1/8 = 0.0025 Seconds.Or 2.5 milliseconds.So, the current vector leads (or lags) the voltage vector by 2.5 milli seconds.
Question 14
4 / -1

The given figure represents the phasor diagram of a series LCR circuit connected to an ac source. At the instant t' when the source voltage is given by V = V₀coswt, the current in the circuit will be

A
I = I₀ cos(wtφ + p/6)

B
I = I₀ cos(wtφ – p/6)

C
I = I₀ cos(wtφ + p/3)

D
I = I₀ cos(wtφ – p/3)
Question 15
4 / -1

Power factor of an L-R series circuit is 0.6 and that of a C_R series circuit is 0.5. If the element (L, C, and R) of the two circuits are joined in series the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

A
6/5

B
5/6

C

D

Solution

cosϕ₁=0.6=5/3
tanϕ₁=4/3= X_C/R₂…(1)
cosϕ₁=0.5=1/2
tanϕ₂=√3= R₂/X_C…(2)
From (1) and (2)3√3/4=R₁/R₂
Question 16
4 / -1

The direct current which Would give the same heating effect in an equal constant resistance as the current shown in figure, i.e. the r.m.s. current, is

A
zero

B

C
2A

D
Question 17
4 / -1

The effective value of current i = 2 sin 100p t + 2 sin (100pt + 30º) is

A

B

C
4

D
None

Solution

cosθ=sin(90^o−θ)
sinα+sinβ=2sin (α+β/2)cos (α−β/2)
i=2sin100πt+2cos(100πt+30^o)
=2sin100πt+2sin(90^o−(100πt+30^o))
=2sin100πt+2sin(60^o−100πt)
=2 x 2 x sin [{(100−100)πt+60^o }/2] x cox[(100+100)πt−60^o/2]
=4 x sin30o∗cos(100πt−30^o)
=4x(1/2) x cos(100πt−30^o)
=2cos(100πt−30^o)
therefore, I_o=2A
I_rms=2 / √2= √2A
Question 18
4 / -1

A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Based on the readings taken by the student answer the following questions.
The value of resistance of the coil calculated by the student is

A
3W

B
4 W

C
5 W

D
8W
Question 19
4 / -1

A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Based on the readings taken by the student answer the following questions.
The power developed in the circuit when the capacitor of 2500 mF is connected in series with the coil is

A
28.8 W

B
23.04 W

C
17.28 W

D
9.6 W

Solution

V_DC=I_DCR
∴R=V_DC/I_DC=12/4=3Ω
I_AC=V_AC/Z=V_AC√(R2+XL2)
2.4=12√ [(3)²+X_L²]
solving this equation, we get,
X_L=4Ω
X_C=1/ω_C=1/(50×2500×10⁻⁶)
=8Ω
Z=√[R²+(X_C−X_L)²] =5Ω
∴I=V_DC/Z=12/5=2.4A=I_rms
P=I_rms²R=(2.4)²(3)
=17.28W
At given frequency X_C>X_L if omega is further decreased X_C will increase (as X_C∝1/ω) and X_L will increase (as X_L∝ω).
Therefore X_C−X_L and hence Z will increase. So, current will decrease.
Question 20
4 / -1

A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4A. He then replaced the 12 V DC source by a 12 V, (w = 50 rad/s) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 mF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Based on the readings taken by the student answer the following questions.
Which of the following graph roughly matches the variations of current in the circuit (with the coil and capacitor connected in the series) when the angular frequency is decreased from 50 rad/s to 25 rad/s ?

A

B

C

D

Solution

V_DC=I_DCR
∴R=V_DC/I_DC=12/4=3Ω
I_AC=V_AC/Z=V_AC√(R²+X_L²)
2.4=12√ [(3)²+X_L²]
solving this equation, we get,
X_L=4Ω
X_C=1/ω_C=1/(50×2500×10⁻⁶)
=8Ω
Z=√[R2+(XC−XL)2] =5Ω
∴I=V_DC/Z=12/5=2.4A=I_rms
P=I_rms²R=(2.4)²(3)
=17.28W
At given frequency X_C>X_L if omega is further decreased XC will increase (as X_C∝1/ω) and X_L will increase (as X_L∝ω).
Therefore X_C−X_L and hence Z will increase. So, current will decrease.
Question 21
4 / -1

Domestic power supply in India is

A
24 V D.C

B
220 V, 50 Hz

C
110 V, 60 Hz

D
416 V, 60 Hz

Solution

The standard voltage and frequency of alternating current supply in India is set to 220 V and 50Hz respectively by the government of India becausewhen power has to be transmitted from a power plant, the biggest challenge is to cut the transmission losses. For this purpose, the current value should be small and potential difference(Voltage) should be more. Also losses are minimal at 50 Hz/60 Hz frequency.
Question 22
4 / -1

For a series LCR circuit the input impedance at resonance

A
equals the resistance ωL

B
equals the resistance 1/ωC

C
equals the resistance R

D
equals the resistance R+jωL

Solution

Resonance occurs when X_L = X_C and the imaginary part of the transfer function is zero. At resonance the impedance of the circuit is equal to the resistance value as Z = R.
Question 23
4 / -1

In an inductance the current

A
is in phase

B
leads the voltage

C
lags the voltage

D
builds very fast

Solution

In an inductor, current lags behind the input voltage by a phase difference of π/2.
Current and voltage are in the same phase in the resistor whereas current leads the voltage by π/2 in a capacitor.

So, the circuit must contain an inductor only.
Question 24
4 / -1

A lamp is connected in series with a capacitor and connected to an AC source. As the capacitance value is decreased.

A
The lamp glows dimmer and dimmer

B
The lamp does not glow

C
The lamp glows brighter

D
The lamp starts turning on and off

Solution

When frequency decreases, capacitive reactance XC=1/2πνC increases and hence impedance in the circuit Z=√(R2+XC2) increases and so current I=V/Z decreases. As a result, the brightness of the bulb is reduced.
Question 25
4 / -1

You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 A, even for an instant. What is the largest root-mean-square current you can run through this bulb?

A
1.46 A

B
1.06 A

C
1.26 A

D
1.56 A

Solution

Given,
We are given the current I=1.5Am where the wire will break.
We are asked to determine the root mean square of the current Irms. The maximum current here represents the current that just after it the wire will break. The maximum value of the current is the amplitude of the current wave and it should be larger than the root mean square of the current. Using equation, we can get the I_rms in the form,
I_rms = I_max/√2
The term, 1/√2, times any factor represents the root mean square of this factor, Now, plug the value for Imax into equation 1 and get I_rms
I_rms=I_max/√2
=1.5A/√2
=1.06A.

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