Alternating Current
Questions for Short Answer
1. What is the reactance of a capacitor connected to a DC source?
ANSWER: The reactance of a capacitor is given as,
Xₖ = 1/⍵C
For a DC source, ⍵ = 0.
Hence, Xₖ = 1/0 =∞.
So the reactance of a capacitor connected to a DC source is infinite.
2. The voltage and current in a series AC circuit are given by
V =Vₒcos ⍵t and i = iₒsin ⍵t
What is the power dissipated in the circuit?
ANSWER: The average power delivered by AC source is,
P = iᵣₘₛ Vᵣₘₛ cos φ
where iᵣₘₛ =root mean square value of the current,
Vᵣₘₛ = root mean square value of the voltage,
and φ = phase difference between the current and the voltage.
In the given AC circuit,
i = iₒ sin ωt
V = Vₒ cos ωt =Vₒ sin(ωt +π/2)
Hence the phase difference between the current and the voltage is φ =π/2.
Thus the power,
P =iᵣₘₛ Vᵣₘₛ cos (π/2)
=iᵣₘₛ Vᵣₘₛ *0
=zero.
So no power is dissipated in the circuit.
3. Two alternating currents are given by
i₁ = iₒsin ⍵t and i₂ = iₒsin {⍵t+π/3}.
Will the rms values of the currents be equal or different?
ANSWER: The rms value of a current is,
iᵣₘₛ = iₒ/√2
So the rms value of a current only depends upon the peak value of the current iₒ. The given currents have the same peak value with a phase difference of π/3. Hence the rms values of the currents will be equal.
4. Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?
ANSWER: The magnitude of the voltage across an inductor is,
V =L*di/dt
=L*d{iₒ sin(ωt+φ)}/dt
=Liₒω cos(ωt+φ)
=Vₒ cos(ωt+φ)
So the peak voltage Vₒ =Lωiₒ
In an LCR circuit, impedance
Z =√{R²+(1/ωC -ωL)²}
At the resonant frequency, 1/ωC =ωL, and Z=R. So iₒ =Eₒ/R.
So the peak voltage across the inductor
Vₒ =Lωiₒ
=LωEₒ/R
=(Lω/R)Eₒ
Thus in a case where Lω >R in an LCR circuit, the factor Lω/R >1. In such a situation, Vₒ > Eₒ. So the peak voltage across an inductor may be greater than the peak voltage of the source in an LCR circuit.
5. In a circuit containing a capacitor and an AC source, the current is zero at the instant the source voltage is maximum. Is it consistent with Ohm's law?
ANSWER: No, it is not consistent with Ohm's law because Ohm's law is valid only for a resistive circuit. It is not valid if there is a reactive element (reactance, capacitive, or inductive) in the circuit.
6. An AC source is connected to a capacitor. Will the rms current increase, decrease, or remain constant if a dielectric slab is inserted into the capacitor?
ANSWER: Reactance of a capacitor,
X = 1/ωCₒ.
If a dielectric slab is inserted into the capacitor, the capacitance is given by,
C =KCₒ
where K =dielectric constant of the inserted slab,
So reactance of the capacitor is now,
X =1/ωKCₒ, where Cₒ is the capacitance without a dielectric slab. Since the dielectric constant K >1, reactance X will decrease with the insertion of the dielectric slab.
Now the rms current,
iᵣₘₛ =iₒ/√2
=Eₒ/(X√2)
With the decrease in reactance X, iᵣₘₛ will increase.
7. When the frequency of the AC source in an LCR circuit equals the resonant frequency, the reactance of the circuit is zero. Does it mean that there is no current through the inductor or the capacitor?
ANSWER: At the resonant frequency in an LCR circuit, the reactance of the circuit is zero but there is resistance in the circuit.
So the current in the circuit is
i =(Eₒ/R) sin ωt.
Thus there is a current through the inductor or capacitor. The only thing is that the potential difference across the combined inductor and capacitor will be zero because the current leads the emf by π/2 in the capacitor but it lags by π/2 in the inductor.
8. When an AC source is connected to a capacitor there is a steady-state current in the circuit. Does it mean that the charges jump from one plate to the other to complete the circuit?
ANSWER: When an AC source is connected to a capacitor there is a potential difference across the plates. The charge stored on the plates is proportional to the potential difference. Since the potential difference across the plates varies due to the AC source, the accumulated charge on the plates also varies. So in the steady state of current, charges do not jump from one plate to another but either accumulate to one plate or discharge from the other plate.
9. A current i₁ =iₒsin ⍵t passes through a resistor of resistance R. How much thermal energy is produced in one time period? A current i₂ = -iₒsin⍵t passes through the resistor. How much thermal energy is produced in one time period? If i₁ and i₂ both pass through the resistor simultaneously, how much thermal energy is produced? Is the principle of superposition obeyed in this case?
ANSWER: The thermal energy produced in one time period is,
H =iᵣₘₛ²RT
=(iₒ/√2)²R(2π/⍵)
=πiₒ²R/⍵
For the current i₂ = -iₒsin ⍵t, thermal energy produced in one time period,
H' =iᵣₘₛ²RT
=(-iₒ/√2)²R*(2π/⍵)
=πiₒ²R/⍵
So H = H'. It means, that in both conditions, the thermal energy produced is the same.
When i₁ and i₂ both pass through the resistor simultaneously, the net current in the resistor is zero. Thus the thermal energy produced in the resistor is zero. So the principle of superposition is obeyed in determining the net current in the resistor.
10. Is energy produced when a transformer steps up the voltage?
ANSWER: No. When a transformer steps up the voltage, the resulting current decreases. Thus following the principle of conservation of energy, no energy is produced when stepping up the voltage.
11. A transformer is designed to convert an AC voltage of 220 V to an AC voltage of 12 V. If the input terminals are connected to a DC voltage of 220 V, the transformer usually burns. Explain.
ANSWER: The coils of the transformers have inductance only and their resistance is very low. When it is connected to a DC source, initially the induced emf resists the current from increasing but it tries to reach a steady state. The steady-state current is given as,
i = E/R =220/R
Since R is very low here, the value of i will be a large one. Since the heat produced is proportional to the square of i, it is sufficient to burn the coil of the transformer.
12. Can you have an AC series circuit in which there is a phase difference of 180° between the emf and the current? 120°?
ANSWER: In a resistive circuit the current is in phase with the emf. For a capacitive circuit, the current leads the emf by π/2, and for an inductive circuit, the current lags the emf by π/2. For an LCR circuit the phase difference φ is in a range π/2 >φ > -π/2 because
tan φ =(1/⍵C -⍵L)/R.
So φ will never be 180° or 120°.
13. A resistance is connected to an AC source. If a capacitor is included in the series circuit, will the average power absorbed by the resistance increase or decrease? If an inductor of small inductance is also included in the series circuit, will the average power absorbed increase or decrease further?
ANSWER: When a capacitor is included in the series circuit, the impedance of the circuit,
Z =√(R²+Xc²). Where Xc is the reactance of the capacitor.
Since the impedance increases, rms current in the circuit will decrease. The average power absorbed by a resistor is given by,
P =iᵣₘₛ²RT.
Hence the power absorbed by the resistor will decrease.
Now if an inductor of small inductance is also included in the series circuit, the impedance now is,
Z =√{R²+(Xc -XL)²}
Thus with the further inclusion of the inductor in the circuit, impedance reduces a little. So the rms current increases. The average power absorbed by the resistor will increase further.
14. Can a hot-wire ammeter be used to measure a direct current having a constant value? Do we have to change the graduations?
ANSWER: A hot-wire ammeter measures the rms value of an AC current. Since rms current of AC circuit is equivalent to direct current i where both produce the same amount of joule heating in the same time interval. So the hot wire ammeter can be used to measure a direct current having a constant value.
We need not to change the graduations.
Objective-I
1. A capacitor acts as an infinite resistance for
(a) DC
(b) AC
(c) DC as well as AC
(d) neither AC nor DC
ANSWER: (a).
EXPLANATION: For a DC, angular frequency ⍵ =0. Hence, the reactance of the capacitor,
X =1/⍵C
=∞.
Hence the capacitor acts as an infinite resistance for DC.
2. An AC source producing emf
Ɛ =Ɛₒ[cos(100π s⁻¹)t +cos(500π s⁻¹)t]
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be
i =i₁ cos[(100π s⁻¹)t+φ₁]+i₂cos[(500π s⁻¹)t+φ₂]
(a) i₁ > i₂
(b) i₁ = i₂
(c) i₁ < i₂
(d) the information is insufficient to find the relation between i₁ and i₂.
ANSWER: (c).
EXPLANATION: In a steady state, the charge on a capacitor is,
Q =CƐ
=CƐₒ[cos(100π s⁻¹)t+cos(500π s⁻¹)t]
Hence the current,
i =dQ/dt
=-100πCƐₒsin(100π s⁻¹)t-500πCƐₒsin(500π s⁻¹)t]
=100πCƐₒcos{(100π s⁻¹)t+π/2}+500πCƐₒcos{(500π s⁻¹)t+π/2}
Comparing with the given current,
i₁ =100πCƐₒ
i₂ =500πCƐₒ
Hence i₁ < i₂.
So option (c) is correct.
3. The peak voltage in a 220 V AC source is
(a) 220 V
(b) about 160 V
(c) about 310 V
(d) 440 V.
ANSWER: (c).
EXPLANATION: vᵣₘₛ =220 V.
We know that rms voltage vᵣₘₛ =vₒ/√2.
Hence peak voltage,
vₒ =√2*vᵣₘₛ
=√2*220 ≈310 V.
Hence option (c) is correct.
4. An AC source is rated 220 V, 50 Hz. The average voltage is calculated in a time interval of 0.01 s. It
(a) must be zero
(b) may be zero
(c) is never zero
(d) is (220√2) V
ANSWER: (b).
EXPLANATION: The frequency is given,
f = 50 Hz.
Time period, T =1/f =1/50 s =0.02 s.
So to complete one cycle time taken is 0.02 s. 
Diagram for Q-4
We have been given a time interval of 0.01 s which is half-cycle time. If this time interval starts from O, then we have time from O to C (above diagram). Clearly, in this interval, the source emf is positive everywhere. Thus the average voltage in this interval will certainly be positive (non-zero). So option (a) is not correct.
Now consider that the time starts a quarter cycle after O, at A. Then the time interval ends at B. The source emf in this interval decreases from Vₒ at A to zero at C in a time interval of 0.005 s. Now the direction of emf reverses and increases from zero at C to Vₒ at B in the next 0.005 s. If we calculate the average voltage in 0.01 s from A to B, it will be zero. So option (c) is also not correct.
Since the average voltage in 0.01 s depends upon the start of the time, it has no fixed value. Option (d) is also not correct.
The average voltage may be zero is the only correct option, i.e. (b).
5. The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is
(a) 20 Hz
(b) 50 Hz
(c) 200 Hz
(d) 500 Hz.
ANSWER: (b).
EXPLANATION: The magnetic field energy in an inductor is,
E =½Li².
So it will be maximum when 'i' has a peak value and minimum when i =0. We can see in the diagram of Problem-4 (assuming it is drawn for current) that it takes a quarter cycle time for the current to change from maximum to minimum i.e. from A to C.
Thus in this case T/4 =5.0 ms
→T = 20 ms =20/1000 s
→T =1/50 s.
Thus the frequency of the source,
f = 1/T
=1/(1/50) s⁻¹
=50 Hz.
Option (b) is correct.
6. Which of the following plots may represent the reactance of a series LC combination? 
Figure for Q-6
ANSWER: (d).
EXPLANATION: The reactance of an LC circuit is,
X =⍵L -1/⍵C
→X =2πfL -1/2πfC ----(i)
If 2πfL =1/2πfC, then X =0
i.e. for f =1/{2π√(LC)}, X =0.
Also from (i) it is clear that the plot between X and f is not a straight line. Thus the given plots 'a' and 'b' are not correct. The non-straight line plot 'c' has never a zero value, thus it is also not correct. Only plot 'd' fulfills the conditions, hence option (d) is correct.
7. A series AC circuit has a resistance of 4 Ω and a reactance of 3 Ω. The impedance of the circuit is
(a) 5 Ω
(b) 7 Ω
(c) 12/7 Ω
(d) 7/12 Ω.
ANSWER: (a).
EXPLANATION: Given, R =4 Ω, X =3 Ω
Hence the impedance of the circuit
Z =√(R²+X²)
=√(4²+3²) Ω
=√25 Ω
=5 Ω.
Option (a) is correct.
8. Transformers are used
(a) in DC circuits only
(b) in AC circuits only
(c) in both DC and AC circuits
(d) neither in DC nor in AC circuits.
ANSWER: (b).
EXPLANATION: The transformer works on the principle that due to a changing emf in the primary coil, a changing magnetic flux is generated. This changing magnetic flux induces emf in the secondary coil. The emf in the secondary coil depends upon the ratio of the number of turns of the secondary coil to the primary coil.
Since the emf changes only in AC circuits, the transformers can be used only in AC circuits.
Option (b) is correct.
9. An alternating current is given by
i =i₁ cos ⍵t +i₂ sin ⍵t.
The rms current is given by
(a) (i₁+i₂)/√2
(b) |i₁+i₂|/√2
(c) √{(i₁²+i₂²)/2}
(d) √{(i₁²+i₂²)/√2}
ANSWER: (c).
EXPLANATION: The rms value of current is given as,
iᵣₘₛ =√(∫i²dt/∫dt)
The limit of integration will be from 0 to t=T=2π/ω where T is the time period.
Now i² =(i₁cos ωt +i₂sin ωt)²
=i₁²cos²ωt+i₂²sin²ωt+2i₁i₂cosωt.sinωt
=½i₁²(cos2ωt+1)+½i₂²(1-cos2ωt)+i₁i₂sin2ωt
=(i₁²+i₂²)/2+½(i₁²-i₂²)cos2ωt+i₁i₂sin2ωt
So,
∫i²dt=(i₁²+i₂²)t/2+¼(i₁²-i₂²)sin2ωt/ω-i₁i₂cos2ωt/2ω
Putting the limits for t, we get,
∫i²dt =(i₁²+i₂²)π/ω+0-i₁i₂/2ω-0-0+i₁i₂/2ω
=(i₁²+i₂²)π/ω
And, ∫dt =t
Putting limits,
∫dt =T =2π/ω
Hence iᵣₘₛ =√[{(i₁²+i₂²)π/ω}/(2π/ω)]
→iᵣₘₛ =√{(i₁²+i₂²)/2}
Option (c) is correct.
10. An alternating current having a peak value of 14 A is used to heat a metal wire. To produce the same heating effect, a constant current 'i' can be used where i is
(a) 14 A
(b) about 20 A
(c) 7 A
(d) about 10 A.
ANSWER: (d).
EXPLANATION: The constant current that produces the same heating effect that of an AC is called the rms current.
iᵣₘₛ =iₒ/√2
=14/√2 A
≈10 A.
Option (d) is correct.
11. A constant current of 2.8 A exists in a resistor. The rms current is
(a) 2.8 A
(b) about 2 A
(c) 1.4 A
(d) undefined for a direct current.
ANSWER: (a).
EXPLANATION: Since rms current is given as
iᵣₘₛ =√(∫i²dt/∫dt)
Since i is constant here,
iᵣₘₛ =√(i²∫dt/∫dt)
=√i²
=i =2.8 A.
Hence option (a) is correct.
Objective-II
1. An inductor, a resistor, and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance
(a) of the inductor increases
(b) of the resistor increases
(c) of the capacitor increases
(d) of the circuit increases.
ANSWER: (a).
EXPLANATION: The reactance is for the capacitor and inductor. Hence option (b) is not correct. Reactance for the capacitor is
Xc =1/⍵C
and for inductor
Xi =⍵L
With the increase in frequency, ⍵ increases. Thus Xi will increase and Xc will decrease. Option (a) is correct, (c) is not correct.
The reactance of the circuit,
X =⍵L-1/⍵C
When ⍵ is increased from a very low value, the numerical value of X decreases. Hence option (d) is not correct.
2. The reactance of a circuit is zero. It is possible that the circuit contains
(a) an inductor and a capacitor
(b) an inductor but no capacitor
(c) a capacitor but no inductor
(d) neither an inductor nor a capacitor.
ANSWER: (a), (d).
EXPLANATION: Reactance of a circuit,
X =Xc -Xi
=⍵L -1/⍵C
In case, ⍵L =1/⍵C
X =0.
That is even if there is a capacitor and an inductor, the reactance of the circuit may be zero. Option (a) is correct.
If there is only a capacitor or only an inductor, the reactance will have a certain value. Option (b) and (c) are not correct.
If there is neither a capacitor nor an inductor, the reactance will be zero. Option (d) is correct.
3. In an AC series circuit, the instantaneous current is zero when the instantaneous voltage is maximum. Connected to the source may be a
(a) pure inductor
(b) pure capacitor
(c) pure resistor
(d) combination of an inductor and a capacitor.
ANSWER: (a), (b), (d).
EXPLANATION: In a resistor, current and voltage are in the same phase when connected to an AC source. Hence option (c) is not correct.
In an inductor, the current lags behind the emf by 90° while in a capacitor the current leads the emf by 90°, hence the given condition may occur here. Options (a) and (b) are correct.
In a combination of inductor and capacitor, the current may lead or lag behind the emf depending upon the comparative value of their reactances. Hence the given condition may occur here. Option (d) is correct.
4. An inductor coil having some resistance is connected to an AC source. Which of the following quantities have zero average value over a cycle?
(a) current
(b) induced emf in the inductor
(c) Joule heat
(d) magnetic energy stored in the inductor.
ANSWER: (a), (b).
EXPLANATION: The plot of current and induced emf w.r.t. to time are sinusoidal waves with some phase difference. Hence for each half-cycle, their values will be just opposite of the next half-cycle. Thus over a cycle average value of current or induced emf will be zero. Options (a) and (b) are correct.
Since Joule heat or magnetic energy stored in an inductor is proportional to the square of rms current, they will be always positive over a cycle, not zero. Options (c) and (d) are not correct.
5. The AC voltage across a resistance can be measured using
(a) a potentiometer
(b) a hotwire voltmeter
(c) a moving coil galvanometer
(d) a moving magnet galvanometer.
ANSWER: (b).
EXPLANATION: The AC voltage across a resistance changes sinusoidally with respect to time and so is the current. The principle behind a potentiometer, a moving coil galvanometer, and a moving magnet galvanometer use a function that is directly proportional to the current. So they can not be used to measure AC voltage. Options (a), (c) and (d) are not correct.
In a hot wire voltmeter principle of Joule heating is used that is proportional to the square of the rms current. Hence it is used to measure AC voltage. Option (b) is correct.
6. To convert mechanical energy into electrical energy, one can use
(a) DC dynamo
(b) AC dynamo
(c) motor
(d) transformer.
ANSWER: (a), (b).
EXPLANATION: With the help of mechanical energy an inducting coil is rotated in a magnetic field to generate electrical energy. This is done either in a DC dynamo or AC dynamo. Options (a) and (b) are correct.
A motor works on the opposite principle. It is used to convert electrical energy into mechanical energy. Option (c) is not correct.
A transformer is used either to step up the available AC voltage or step down. It does not convert the form of energy. Option (d) is not correct.
7. An AC source rated 100 V (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source
(a) must be 1000 W
(b) maybe 1000 W
(c) maybe greater than 1000 W
(d) maybe less than 100 W.
ANSWER: (b), (d).
EXPLANATION: The average power delivered by an AC source is given as,
P =VᵣₘₛIᵣₘₛCos φ
With given data,
P =100*10*Cos φ
=1000.Cos φ
Since the value of φ has a range of (-π/2 to π/2), the cos φ remains between 0 to 1. Thus the average power supplied in this AC circuit is between 0 to 1000 W.
Options (b) and (d) are correct.
EXERCISES, Q1 to Q10
1. Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
ANSWER: Frequency of the alternating current, f =50 Hz. For an alternating current, the instantaneous current is given as,
i = iₒ sin ωt
And rms current,
iᵣₘₛ = iₒ/√2
At t = 0, i =0.
If at t =t', i =iᵣₘₛ, then
iᵣₘₛ =iₒ sin ωt'
→iₒ/√2 =iₒ sin ωt'
→sin ωt' =1/√2 =sin π/4
→ωt' = π/4
→2πft' =π/4
→t' =1/(8f)
=1/(8*50) s
=0.0025 s
=2.5 ms.
2. The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.
ANSWER: Vᵣₘₛ =220 V, f =50 Hz.
Since Vᵣₘₛ =Vₒ/√2
→Peak voltage, Vₒ =√2iᵣₘₛ
→Vₒ =√2*220 volts
=311 V.
In a cycle when E begins with zero, it reaches rms value and then to peak value. Now E begins to decrease, reaches rms value, and then to zero. So the least time to reach from Eᵣₘₛ to zero is in this decreasing phase. This time is the same when E changes from zero to Eᵣₘₛ, so we calculate this time.
Eᵣₘₛ =Eₒ sin ωt
→Eₒ/√2 =Eₒ sin ωt
→sin ωt =1/√2 =sin π/4
→ωt =π/4
→2πft =π/4
→t =1/(8f)
=1/400 s
=0.0025 s
=2.5 ms.
3. A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.
ANSWER: From the bulb's rating, it is clear that the bulb consumes a power of 60 W at rms current of 220 V. Let the resistance of the bulb =R. We know,
Power, P = V²/R, hence
60 =(Eᵣₘₛ)²/R =220²/R
→R =220²/60 Ω
=807 Ω
Maximum instantaneous current will be at a time when the voltage is maximum, i.e. at peak voltage. So,
iₒ =Eₒ/R
=√2*Eᵣₘₛ/R
=√2*220/807 A
=0.39 A.
4. An electric bulb is designed to operate at 12 volts of DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?
ANSWER: Let a constant current i pass through the bulb when connected to the DC circuit. If it gives the same brightness when connected to an AC circuit, it means that the rms value of this current is equal to the constant current i because both produce the same amount of Joule heating in a given time period. So for the AC circuit,
iᵣₘₛ = i
→iₒ/√2 =i
→iₒR/√2 =iR,
{where R =bulb's resistance,iₒ =peak current and Eₒ =peak voltage of the source}
→Eₒ/√2 =E,
→Eₒ =√2E
=√2*12 V
=17 V.
5. The peak power consumed by a resistive coil when connected to an AC source is 80 W. Find the energy consumed by the coil in 100 seconds which is many times larger than the time period of the source.
ANSWER: Let the resistance of the coil =R. Hence the peak power consumed by the coil,
Pₒ =iₒ²R
→iₒ² =Pₒ/R
→(√2iᵣₘₛ )² =80/R
→iᵣₘₛ² =40/R
Energy consumed by the resistive coil in a time period t is given as,
=iᵣₘₛ²Rt
=(40/R)*R*100 J
=4000 J
=4.0 kJ.
6. The dielectric strength of air is 3.0x10⁶ V/m. A parallel plate air capacitor has an area of 20 cm² and a plate separation of 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.
ANSWER: Given for the capacitor,
Plate area, A =20 cm² =0.002 m²,
Separation, d =0.10 mm
=1.0x10⁻⁴ m.
Dielectric strength of air,
E' =3.0x10⁶ V/m.
It means that if the electric field inside the capacitor is equal to the dielectric strength of the air, it is at the limit of safe working. Let at this electric field the potential difference across the plate is V, then
E' =V/d
→V =E'd
=3.0x10⁶*1.0x10⁻⁴ volts
=300 V.
So this should be the maximum potential difference across the capacitor i.e. equal to the peak voltage Eₒ of the AC source. Since we need to know the rms voltage corresponding to this peak voltage, we have,
Eᵣₘₛ =Eₒ/√2.
=300/√2 volts
≈210 V.
7. The current in a discharging LR circuit is given by i =iₒe^-t/𝝉 where 𝝉 is the time constant of the circuit. Calculate the rms current for the period t =0 to t =𝝉.
ANSWER: rms current is the square root of the mean square current. So mean square current =iᵣₘₛ².
→iᵣₘₛ² =∫i²dt/∫dt
=∫(iₒe^-t/𝝉)²dt/∫dt
Keeping the limit for t from 0 to 𝝉, we get
=(iₒ²/𝝉)[-𝝉e^-2t/𝝉/2]
=(iₒ²/2)[-e^-2t/𝝉]
=½iₒ²[-e⁻²+1], putting limits.
=½iₒ²(e²-1)/e²
Hence iᵣₘₛ =√{½iₒ²(e²-1)/e²}
=(iₒ/e)√{(e²-1)/2}.
8. A capacitor of capacitance 10 µF is connected to an oscillator giving an output voltage Ɛ =(10 V)sin ⍵t. Find the peak currents in the circuit for ⍵ =10 s⁻¹, 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.
ANSWER: Reactance of a capacitor connected to an AC source of emf,
Ɛ =Ɛₒ sin ωt, is
X =1/ωC
Hence the peak current,
iₒ = Ɛₒ/X
=Ɛₒ/(1/ωC)
=CƐₒω.
From the given data, Ɛₒ =10 V.
Hence iₒ =10Cω.
=10*10x10⁻⁶ω,
=1x10⁻⁴ω.
For ω =10 s⁻¹,
Peak current =1.0x10⁻⁴*10 A
=1.0x10⁻³ A.
For ω =100 s⁻¹,
Peak current =1.0x10⁻⁴*100 A
=0.01 A.
For ω =500 s⁻¹,
Peak current =1.0x10⁻⁴*500 A
=0.05 A.
For ω =1000 s⁻¹,
Peak current =1.0x10⁻⁴*1000 A
=0.1 A.
9. A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for
⍵ = 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.
ANSWER: Inductance of the coil,
L =5.0 mH =5.0x10⁻³ H.
Reactance of the inductive coil,
Xi =ωL
Hence peak current iₒ =Ɛₒ/ωL
→iₒ =10/(5.0x10⁻³ω)
=2.0x10³/ω
For ω =100 s⁻¹,
iₒ =2.0x10³/100 A
=20 A.
For ω =500 s⁻¹,
iₒ =2.0x10³/500 A
=4.0 A.
For ω =1000 s⁻¹,
iₒ =2.0x10³/1000 A
=2.0 A.
10. A coil has a resistance of 10 Ω and an inductance of 0.4 henries. It is connected to an AC source of 6.5 V, 30/π Hz. Find the average power consumed in the circuit.
ANSWER: R =10 Ω, L =0.4 H, f =30/π Hz.
Reactance of the coil X =ωL
→X =2πfL
=2π*(30/π)*0.4 Ω
=24 Ω.
Impedance of the coil,
Z =√(R²+X²)
=√(10²+24²)
=√676 Ω
Average power is given as,
P =EᵣₘₛIᵣₘₛcos φ
Now, Eᵣₘₛ =6.5 V, Iᵣₘₛ =Eᵣₘₛ/Z, and
cos φ =R/Z. Thus,
P =6.5*(6.5/√676)*10/√676 W
=(422.5/676) W
=4225/6760 W
=325/520 W, {division by 13}
=25/40 W, {division by 13}
=5/8 W.
EXERCISES, Q11 to Q19
11. A resistor of resistance 100 Ω is connected to an AC source Ɛ =(12 V) sin (250 π s⁻¹)t. Find the energy dissipated as heat during t =0 to t =1.0 ms.
ANSWER: ω =250π s⁻¹
Time period T =2π/ω s =2π/250π s
→T =0.008 s =8 ms.
Since we have to calculate the energy dissipated in 1.0 ms which is less than the time taken to complete one cycle, we can not calculate the average heat dissipated based on rms voltage. Thus we integrate.
H =∫(Ɛ²/R)dt
Limits of integration is t =0 to t =0.001 s
→H =∫{(Ɛₒsinωt)²/R} dt
=(Ɛₒ²/R)∫sin²ωt dt
=½(Ɛₒ²/R)∫(1-cos2ωt) dt
=½(Ɛₒ²/R)[t -(sin2ωt)/2ω]
Putting the limits
H =½(12²/100)[0.001-sin(2*250π*0.001)/500π] J
=0.72[0.001-(sin π/2)/500π] J
=0.72[0.001 -1/500π] J
=2.61x10⁻⁴ J.
12. In a series RC circuit with an AC source, R =300 Ω, C =25 µF, Ɛₒ =50 V, and v =50/π Hz. Find the peak current and the average power dissipated in the circuit.
ANSWER: R =300 Ω, C =25x10⁻⁶ F,
ω =2π𝝂 =2π(50/π) s⁻¹
=100 s⁻¹
Ɛₒ =50 V.
Reactance of the circuit X =1/ωC
→X =1/25x10⁻⁴ Ω
=400 Ω.
R = 300 Ω.
Hence impedance of the circuit,
Z =√(R²+X²)
=√(300²+400²)
=500 Ω.
Hence peak current, iₒ =Ɛₒ/Z
=50/500 A
=0.10 A.
The average power dissipated in the circuit is,
P =Ɛᵣₘₛiᵣₘₛcos φ
=(Ɛₒ/√2)(iₒ/√2)(R/Z)
=(50*0.10/2)*(300/500) W
=1.5 W.
13. An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets the correct voltage?
ANSWER: Let resistance of the bulb =R.
Power P =E²/R
→R =E²/P
=110²/55 Ω
=220 Ω.
Now let the inductance of the added choke coil in the circuit =L.
The reactance of the circuit,
X =ωL
=2πfL
=100πL
The impedance of the circuit,
Z=√(R²+X²)
=√{220²+(100πL)²}
So rms current through the circuit,
I =220/Z
The bulb is rated for 110 V, so the voltage drop through it in the new circuit should be 110 V. i.e.
IR =110
→(220/Z)*220 =110
→Z =220²/110
→√{220²+(100πL)²}=220²/110
→(100πL)² =(220²*220²/110²)-220²
=220²(4 -1)
=220²*3
→100πL =220√3
→L=220√3/100π =1.2 H
14. In a series LCR circuit with an AC source, R =300 Ω, C =20 µF, L =1.0 Henry, Ɛᵣₘₛ =50 V and 𝝂 =50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor, and the inductor. Note that the sum of the rms potential difference across the three elements is greater than the rms voltage of the source.
ANSWER: R =300 Ω, C =20x10⁻⁶ F,
L =1.0 H, Ɛᵣₘₛ =50 V, 𝝂 =50/π.
Capacitive reactance,
Xc =1/ωC
=1/(2π𝝂C)
=1/{2π*(50/π)*20x10⁻⁶}
=500 Ω
Inductive reactance,
Xi =ωL
=2π*(50/π)*1.0 Ω
=100 Ω.
The impedance of the circuit,
Z =√{R²+(Xc-Xi)²}
=√{300²+(500 -100)²} Ω
=√{300²+400²} Ω
=500 Ω
Given Ɛᵣₘₛ =50 V
(i) rms current =Ɛᵣₘₛ/Z
=50/500 A
=0.10 A
(ii) rms potential difference across capacitor
=iᵣₘₛ*Xc
=0.10*500 V
=50 V.
rms potential difference across the resistor =iᵣₘₛR
=0.10*300 V
=30 V.
rms potential difference across the inductor =iᵣₘₛXi
=0.10*100 V
=10 V.
Sum of the rms potential differences across the three elements,
=50 V +30 V +10 V
=90 V
It is greater than the rms voltage of the source which is 50 V.
15. Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.
ANSWER: Average electric field energy stored in a capacitor,
=½CV²
=½*20x10⁻⁶*50² J
=0.025 J
=25 mJ.
Average magnetic field energy stored in the coil,
=½Liᵣₘₛ²
=½*1.0*(0.10)² J
=0.005 J
=5 mJ.
16. An inductance of 2.0 H. a capacitance of 18 µF and a resistance of 10 kΩ are connected to an AC source of 20 V with adjustable frequency. (a) What frequency should be chosen to maximize the current in the circuit? (b) What is the value of this maximum current?
ANSWER: (a) The impedance of the circuit,
Z =√{R²+(Xc -Xi)²},
Since the current in the circuit,
i =E/Z,
To maximize the current we need to keep Z at a minimum. Minimum Z will be when reactance is zero i.e. Xc =Xi.
→1/ωC =ωL
→ω² =1/LC
→(2πf)² =1/(2.0*18x10⁻⁶)
→2πf =√(2.78x10⁴)
→f =167/2π
≈27 Hz.
(b) Since the maximum current reactance of the circuit is zero and it has only resistance in effect. Hence the maximum current,
Iₘ =E/R
=20/(1.0x10⁴) A
=2.0x10⁻³ A
=2.0 mA.
17. An inductor coil, a capacitor, and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0 A is observed. If the inductor coil is connected to a battery of emf 12 V and internal resistance of 4.0 Ω, what will be the current?
ANSWER: Suppose that the inductor coil has resistance R. When the frequency of the source is varied to get maximum current, the reactance of the circuit is zero. Only resistance R is effective.
R =Eᵣₘₛ/Iᵣₘₛ
=24/6 =4 Ω
When this coil is connected to a battery of V =12 volts and internal resistance r =4 Ω, the total resistance of the circuit,
=R+r =4 +4 =8 Ω.
Hence current =12/8 A =1.5 A.
18. Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input Vi =10 mV is applied at the left end and the output Vₒ is received at the right end. Find the output voltage for 𝝂 =10 kHz, 100 kHz, 1.0 MHz, and 10.0 MHz. Note that as the frequency is increased the output decreases and hence the name low-pass filter. 
Figure for Q-18
ANSWER: R =1000 Ω. C =10x10⁻⁹ F.
The reactance of the circuit X =1/ωC.
→X =1.0x10⁸/ω.
The impedance of the circuit,
Z =√(R²+X²)
=√(1000²+10¹⁶/ω²)
=(10³/ω)√(ω²+10¹⁰)
Hence current in the circuit,
i =Vᵢ/Z
=10x10⁻³/{(10³/ω)√(ω²+10¹⁰)}
=1.0x10⁻⁵ω/√(ω²+10¹⁰)
Potential difference across the capacitor,
Vₒ =i*X
={1.0x10⁻⁵ω/√(ω²+10¹⁰)}*1x10⁸/⍵
=1000/√(ω²+10¹⁰)
=1000/√(4π²𝝂²+10¹⁰)
For 𝝂 =10 kHz =10x10³ Hz
Vₒ =1000/√(39.48*10⁸+10¹⁰)
=0.0085 volts
=8.5 mV.
For 𝝂 =100 kHz =1.0x10⁵ Hz
Vₒ =1000/√(39.48x10¹⁰+10¹⁰)
=0.0016 volts
=1.6 mV.
For 𝝂 =1.0 MHz =1.0x10⁶ Hz
Vₒ =1000/√(39.48x10¹²+10¹⁰) volts
=0.00016 V
=0.16 mV.
For 𝝂 =10.0 MHz =1.0x10⁷ Hz
Vₒ =1000/√(39.48x10¹⁴+10¹⁰) volts
=1.6x10⁻⁵ V
=16 µV.
19. A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?
ANSWER: A transformer can not work when connected to a DC supply, because the DC supply has a constant emf and a constant steady-state current in the primary coil. There will be no change of flux in the primary coil that is necessary to induce an emf in the secondary coil. Thus the voltage across the secondary coil will be zero.
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