HC Verma Solutions for Class 12 Physics Chapter 39 – Alternating current

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Contents

1 HC Verma Solutions for Class 12 Physics Chapter 39 – Alternating current
2 Chapterwise HC Verma Solutions Class 12 Physics :
3 About the Author – HC Verma

HC Verma Solutions for Class 12 Physics Chapter 39 – Alternating current

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Page No 328:

Question 1:

What is the reactance of a capacitor connected to a constant DC source?

Answer:

The reactance of a capacitor is given by,

Xc=1ωCFor a DC source, ω = 0

⇒Xc=10×C=∞So, for a constant DC source, reactance of a capacitor is infinite.

Question 2:

The voltage and current in a series AC circuit are given by V = V₀cos ωt and i = i₀ sin ωt. What is the power dissipated in the circuit?

Answer:

Voltage, V = V₀cos ωt
Current, i = i₀ sin ωt or i = i₀ cos (ωt

– π2)

Power dissipated in an AC circuit is given by,

P=IrmsVrmscosϕ,
where I_rms = rms value of current
V_rms= rms value of voltage
Ï• = phase difference between current and voltage
Here, Ï• =

π/2

⇒cosϕ=cosπ2=0∴P=IrmsVrms×0=0

Question 3:

Two alternating currents are given by

i1=i0 sin ωt and i2=i0 sin ωt+π3Will the rms values of the currents be equal or different?

Answer:

The rms value of current is given by,

irms=i02Since peak value of current i₀ is same for both currents, their rms values will be same.

Question 4:

Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?

Answer:

Let a LCR circuit is connected across an AC supply with the emf E = E₀ sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be

Z=R2+(XL-XC)2Where,
R = resistance in the circuit
X_L = reactance due to inductor
X_C = reactance due to capacitor
The magnitude of the voltage across the inductor is given by

V=LdidtThe current in the circuit can be written as

I=I0sin(ωt+ϕ)Where, Ï• is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as

V=LI0cos(ωt+ϕ)Thus peak value of the voltage across the inductor is given by

V=LI0⇒V=E0Z×LTherefore, the peak voltage across the inductor is given by

V=E0Z×LAt resonance Z = R,

V=E0R×L,
If

LR>1V > E₀
Therefore if magnitude of

LR>1at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.

Question 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:

Ohm’s Law is valid for resistive circuits only. It is not valid for capacitive or inductive circuits, or a combination of both.

Question 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:

The reactance of a capacitor is given by,

Xc=1ωCAlso,

C=Kε0Ad,
where C = capacitance
K = dielectric constant
A = area of plates
d = distance between the plates.

K>1

∴The capacitance C of the capacitor will increase on inserting the dielectric slab and, consequently, the reactance X_c will decrease.
Rms current,

irms=ε02XCTherefore, rms current will decrease.

Question 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:

The condition for resonance is:

1ωC=ωLThe peak current through the circuit is given by,

i0=V0R2+1ωC-ωL2From the condition of resonance, we get:

i0=V0RThe current will flow through the all circuit elements. But since the reactance of the capacitor and inductor are equal, the potential difference across them will be equal and opposite and will cancel each other.

Question 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:

No. When an AC source is connected to a capacitor, there is a steady in the circuit to transfer change to the plates of the capacitor. This produces a potential difference between the plates. The capacitance is alternatively charged and discharged as the current reverses after each half cycle.

Page No 329:

Question 9:

A current i₁ = i₀ sin ωt passes through a resistor of resistance R. How much thermal energy is produced in one time period? A current i₂ = −i₀ sin ωt 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 for an AC circuit in one time period is given by,

H=Irms2×R×2πωFor current, i₁ = i₀ sin ωt,

Irms=i02

⇒H=i02R2×2πω=πi02RωFor current, i₂ = −i₀ sin ωt,

Irms=i02Hence, the same thermal energy will be produced due to this current.

Since, the direction of i₁ and i₂ are opposite and their magnitude is same, the net current through the resistor will become zero when both are passed together. Yes, the principle of superposition is obeyed in this case.

Question 10:

Is energy produced when a transformer steps up the voltage?

Answer:

When a transformer steps up the voltage, the voltage increases but current decreases. Neglecting any loss of energy, the power remains constant and, hence, energy is not produced. It remains constant.

Question 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:

A transformer is ideally an inductive coil. For an inductor connected across a DC voltage,

V-Ldidt=0⇒V=Ldidt⇒∫di=VL∫dt⇒i=VtLFor a DC source, the current across the inductor will increase with time and can reach a very large value, which can burn the transformer.

Question 12:

Can you have an AC series circuit in which there is a phase difference of (a) 180° (b) 120° between the emf and the current?

Answer:

Let us consider an AC series LCR circuit of angular frequency ω. The impedance of the circuit is given by,

Z=R2+ωL-1ωC2The phase difference between V and I is given by,

tanϕ=ωL-1ωCZFrom the above formula, we can clearly see that

ϕ∈-π2,π2So, we cannot have a phase difference of 180° or 120°.

Question 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 a series circuit, the impedance of the circuit,

Z=R2+XC2The power absorbed by a resistor is given by,

P=Irms2RSince impedance increases due to introduction of a capacitor, the rms value of current I_rms will decrease and, hence, the power absorbed by the resistor will decrease.

When a small inductance is introduced in the circuit, the impedance of the circuit,

Z=R2+XC-XL2Since the impedance now decreases a little, the rms value of current will increase and, hence, the power absorbed by the resistor will increase.

Question 14:

Can a hot-wire ammeter be used to measure a direct current of constant value? Do we have to change the graduations?

Answer:

A hot-wire ammeter measures the rms value of current for an alternating current. So, it can be used to measure the direct current of constant value because that constant value will be equal to the rms value of current. As, the rms value of the current is same as the direct current thus we need not change the graduations.

Question 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) DC

For a DC source, ω = 0. So, the reactance of the capacitance is given by,

XC=1ωC=10=∞

Question 2:

An AC source producing emf
ε = ε₀ [cos (100 π s⁻¹)t + cos (500 π 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 [(100 π s⁻¹)t + φ₁) + i₂ cos [(500π s⁻¹)t + Ï•₂]. So,

(a) i₁ > i₂
(b) i₁ = i₂
(c) i₁ < i₂
(d) The information is insufficient to find the relation between i₁ and i₂.

Answer:

The charge on the capacitor during steady state is given by,

Q=Cε=ε0Ccos100πs-1t+cos500πs-1tThe steady state current is, thus, given by,

i=dQdt=ε0C×100πsin100πs-1t+ε0C×500πsin500πs-1t⇒i=100Cπε0 cos100πs-1t+ϕ1+500Cπε0 cos500πs-1+ϕ2⇒i1=100Cπε0 & i2=500Cπε0∴i2>i1

Question 3:

The peak voltage of a 220 V AC source is

(a) 220 V
(b) about 160 V
(c) about 310 V
(d) 440 V

Answer:

Given:
V_rms = 220 V
The peak value of voltage is given by,

Vp=2×Vrms=1.414×220=311 V

Question 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) may be zero

Let the AC voltage be given by,

V=V0sinωtHere, ω = 2

πf = 314 rad/s
The average voltage over the given time,

Vavg=∫00.01Vdt∫00.01dt=-V0cosωtω00.01 =V0ω×0.011-cosω0.01 =V0314×0.011-cos314×0.01 =V03.141-cosπ =2V0π=140.127 VAlso, when

V=V0cosωt,

Vavg=∫00.01Vdt∫00.01dt=V0sinωtω00.01 =V0ω×0.01sinω0.01-0 =V0314×0.01sin314×0.01 =V03.14sinπ =0
From the above results, we can say that the average voltage can be zero. But it is not necessary that it should be zero or never zero.

Question 5:

The magnetic field energy in an inductor changes from maximum 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) 50 Hz

The magnetic field energy in an inductor is given by,

E=12Li2The magnetic energy will be maximum when the current will reach its peak value, i₀, and it will be minimum when the current will become zero.

From the above graph of alternating current, we can see that current reduces from maximum to zero in T/4 time, where T is the time period.
So, in this case, T/4 = 5 ms

⇒T=20 ms∴ Frequency, ν = 1T=120×10-3=50 Hz

Question 6:

Which of the following plots may represent the reactance of a series LC combination?

Figure

Answer:

(d)

The reactance of a series LC circuit is given by,

X=XL-XC=ωL-1ωC⇒X=2πfL-12πfCThe correct relation between the reactance and f is represented by the graph in option (d).

X=0, when

XL=XC.

Thus plot d represent the reactance of a series LC combination.

Question 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) 5 Ω

The impedance of the circuit is given by

Z=R2+X2R=4 Ω & X=3 Ω⇒Z=42+32=5 Ω

Question 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) in AC circuits only

When a DC supply is provided to a transformer, there will be no change in flux with time across the coils of the transformer. So, there will be no induced emf in the secondary coil due to changing current in the primary coil. Hence, the transformer cannot operate in DC because of the violation of its working principle.

Question 9:

An alternating current is given by i = i₁ cos ωt + i₂ sin ωt. The rms current is given by

(a)

i1+i22(b)

i1+i22(c)

i12+i222(d)

i12+i222

Answer:

(c)

i12+i222Given:
i = i₁ cos ωt + i₂ sin ωt
The rms value of current is given by,

irms=∫0Ti2dt∫0Tdti=i1cos ωt +i2 sin ωtirms=∫0Ti1cos ωt +i2 sin ωt2dt∫0Tdtirms=∫0Ti12cos2ωt +i22 sin2ωt +2i1i2sin ωt cos ωt dt∫0Tdtirms=∫0Ti12(cos 2ωt+1)2 +i22 (1-cos 2ωt) 2 +i1i2sin 2ωt dt∫0Tdt [∵cos2ωt =(cos 2ωt+1)2 , sin2ωt=(1-cos 2ωt) 2 ]We know that, T = 2π

Integrating the above expression

irms=12i12 ∫02π1dt+∫02πcos 2ωt dt + i22 ∫02π1 dt -∫02π cos 2ωt dt +i1i2∫02π sin 2ωt dt∫02πdtThe following integrals become zero

∫02πcos 2ωt dt = 0=∫02π sin 2ωt dtTherefore, it becomes

irms=i122∫02π1dt+i222∫02π1dt∫02πdtirms=i122×2π +i222×2π2π⇒irms =i12+i222

Question 10:

An alternating current of peak value 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) about 10 A

The rms value of an alternating current is equivalent to the constant current. So, the heating effect produced is actually measured in terms of the rms value, in case of alternating current. The constant current is, thus, equal to the rms value of alternating current, which is given by,

Irms=Ipeak2=142=9.9≈10 A

Question 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) 2.8 A

The constant current is equal to the rms value of current. So,
I_rms = 2.8 A

Question 1:

An inductor, a resistance 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) of the inductor increases

The reactance of an inductor is given by,

XL=ωLAnd the reactance of a capacitor is given by,

XC=1ωCHere, ω = 2πf , where f is the frequency of the source. So, when f increases, ω increases.
∴ X_L will increase and X_C will decrease.

Question 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) an inductor and a capacitor
(d) neither an inductor nor a capacitor

(a) The reactance of a circuit containing a capacitor and an inductor is given by,

X=XL-XC=ωL-1ωCIf

ωL=1ωC, X = 0. So, the circuit contains an inductor and a capacitor.

(b) For a circuit without a capacitor, reactance is given by,

X’=XL=ωL, which cannot be zero for an AC source.

X”=XC=1ωC, which also cannot be zero.

(d) For a circuit without any capacitor and inductor, reactance, X^‘ = 0 ( L = C = 0)

Page No 330:

Question 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) pure inductor
(b) pure capacitor
(d) combination of an inductor and a capacitor.

For a pure inductive circuit, voltage leads the current by

90°. So, the instantaneous current is zero when the instantaneous voltage is maximum.
Similar is the case with a purely capacitive circuit, in which, current leads the voltage by

90°.

Also, in a circuit containing a combination of inductor and capacitor, the current may lead or lag the voltage by

90°, depending upon whether the voltage across the inductor or the capacitor is greater. Here too, there is a phase difference of

90°between the voltage and current. Hence, the the instantaneous current is zero when the instantaneous voltage is maximum.

Question 4:

An inductor coil of 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) Current
(b) Induced emf in the inductor

For a series L–R circuit, the AC current can be given by,

i=i0sinωt

From the graph, we can see that the average value of current over a cycle is zero.

Since it a series L–R circuit, the phase difference between current and voltage is

π2. The AC voltage can be given by,

V=V0cosωt

From the graph, we can see that the average value of voltage over a cycle is also zero.

Joule’s heat through the resistor is given by,

Havg=irms2R, which is non zero.

Similarly, magnetic energy stored in the inductor is given by,

Uavg=12Lirms2, which is also non-zero.

Question 5:

The AC voltage across a resistance can be measured using

(a) a potentiometer
(b) a hot-wire voltmeter
(c) a moving-coil galvanometer
(d) a moving-magnet galvanometer

Answer:

(b) a hot-wire voltmeter

Only a hot-wire voltmeter can be used to measure an AC voltage across a resistor.

Question 6:

To convert mechanical energy into electrical energy, one can use

(a) DC dynamo
(b) AC dynamo
(c) motor
(d) transformer

Answer:

(a) DC dynamo
(b) AC dynamo

An AC or DC dynamo can be used to convert mechanical energy to electrical energy.
A motor converts electrical energy to mechanical energy and a transformer is used to step up or down the voltage or simply to transfer electrical energy.

Question 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) may be 1000 W
(c) may be greater than 1000 W
(d) may be less than 1000 W

Answer:

(b) may be 1000 W
(d) may be less than 1000 W

The average power delivered by an AC source is given by,

Pavg=VrmsIrmscosϕGiven:
V_rms = 100 V
I_rms = 10 A

⇒Pavg=1000cosϕ

∵ϕ∈-π2,π2

⇒cosϕ∈0,1

∴0≤Pavg≤1000

Question 1:

Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.

Answer:

Frequency of alternating current, f = 50 Hz
Alternation current

iis given by,
i = i₀sinωt …(1)
Here, i₀ = peak value of current
Root mean square value of current

irmsis given by,

irms=i02 …1
On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:

i02=i0sinωt⇒12 = sinωt = sinπ4⇒π4 = ωt⇒t =π4ω = π4×2πf ∵ω=2πf = 18f = 18×50 =1400=0.0025 s =2.5 ms

Question 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:

RMS value of voltage, E_rms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage

E0is given by,
E0=Erms 2,
where E_rms = root mean square value of voltage

E0=Erms 2⇒ E0=2×220⇒ E0=311.08 V=311 V(b) Voltage

Eis given by,
E=E0sinωt,
where E₀ = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.

As E=E02, E02 = E0sin ωt⇒ωt = π4⇒t=π4ω=π4×2πf⇒ t=π8π50=1400⇒ t= 2.5 msThus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.

Question 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:

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, E_rms = 220 V
P = V²R,
where R = resistance of the bulb

∴ R=V2P=220×22060 =806.67Peak value of voltage

E0is given by,

E0=Erms2 =

220×2 = 311.08
Now, maximum current through the filament

i0is,

i0=E0R

⇒i0=311.08806.67=0.39 A

Question 4:

An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?

Answer:

Voltage across the electric bulb, E = 12 volts
Let E₀ be the peak value of voltage.
We know that heat produced by passing an alternating current

ithrough a resistor is equal to heat produced by passing a constant current

irmsthrough the same resistor. If R is the resistance of the electric bulb and T is the temperature, then

i2RT=irms2RT⇒E2R2=Erms2R2⇒E2=E022 ∵E2rms=E202⇒E02=2E2⇒E02=2×122=2×144⇒E0=2×144 =16.97=17 VThus , peak value of voltage is 17 V.

Question 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:

Peak power of the resistive coil,

P0 =80 WTime, t = 100 s
RMS value of power

Prmsis given by,

Prms = P02,
where P₀ = Peak value of power

∴ Prms=P02=40 WEnergy consumed

Eis given by,
E = P_rms × t
= 40 × 100
= 4000 J = 4.0 kJ

Question 6:

The dielectric strength of air is 3.0 × 10⁶ V/m. A parallel-plate air-capacitor has area 20 cm² and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.

Answer:

Given:
Area of parallel-plate air-capacitor, A = 20 cm²
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 10⁶ V/m
E =

Vd,
where V = potential difference across the capacitor

∴V = Ed
= 3 × 10⁶× 0.1 × 10⁻³
= 3 × 10² = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage

Vrmsis given by,

Vrms=V02 =3002=212 V

Question 7:

The current in a discharging LR circuit is given by i = i₀ e^−t^/τ , where τ is the time constant of the circuit. Calculate the rms current for the period t = 0 to t = τ.

Answer:

As per the question,

i = i0e-tτWe need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:

irms2 = 1τ∫0τi02 e-2t/τ dt⇒irms2 =i02τ ∫0τ e-2t/τ dt =i02τ×-τ2 e-2t/τ0τ =-i02τ×τ2×e-2-1 =i022(1-1e2)irms=i0e e2-12

Question 8:

A capacitor of capacitance 10 μF is connected to an oscillator with output voltage ε = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s⁻¹, 100 s⁻¹, 500 s⁻¹ and 1000 s⁻¹.

Answer:

Capacitance of the capacitor, C = 10 μF = 10 × 10⁻⁶ F = 10⁻⁵ F
Output voltage of the oscillator,

ε= (10 V)sinωt
On comparing the output voltage of the oscillator with

ε=ε0sinωt, we get:
Peak voltage

ε0= 10 V
For a capacitive circuit,
Reactance,

Xc=1ωCHere,

ω= angular frequency
C = capacitor of capacitance
Peak current,

I0=

ε0Xc(a) At ω = 10 s⁻¹:
Peak current,
I₀ =

ε0Xc

= ε01/ωC=101/10×10-5 A = 1 × 10⁻³ A
(b) At ω = 100 s⁻¹:
Peak current, I₀ =

ε01/ωC

⇒I0= 101/100×10-5⇒I0 =10103=1×10-2 A =0.01 A(c) At ω = 500 s⁻¹:
Peak current, I₀ =

ε01ωC

I0 = ε01/ωC⇒I0 = 101/500×10-5⇒I0 = 10×500×10-5 = 5×10-2 A=0.05 A(d) At ω = 1000 s⁻¹:
Peak current, I₀ =

ε01ωC

⇒I0=101/1000×10-5⇒I0 =10×1000×10-5⇒I0 =10-1 A=0.1 A

Question 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 ω = 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.

Answer:

Given:
Inductance of the coil, L = 5.0 mH = 0.005 H
(a) At ω = 100 s⁻¹:
Reactance of coil

XLis given by,
X_L =

ωLHere,

ω= angular frequency

∴X_L = 100

×0.005 = 0.5

ΩPeak current, I₀ =

100.5=20 A(b) At ω = 500 s⁻¹:
Reactance, XL = 500×51000 = 2.5 Ω Peak current, I₀ =

102.5=4 A(c) ω = 1000 s⁻¹:
Reactance, X_L = 1000

×0.005= 5

ΩPeak current, I₀ =

105= 2 A

Question 10:

A coil has a resistance of 10 Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V,

30πHz. Find the average power consumed in the circuit.

Answer:

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source,

30πHzReactance of resistance-inductance circuit

Zis given by,
Z =

R2+XL2Here, R = resistance of the circuit
X_L = Reactance of the pure inductive circuit
Z =

R2+2πfL2 =

102+2×π×30π×0.42Average power consumed in the circuit (P) is given by,
P = E_rmsI_rmscos

ϕ cos

ϕ=

RZ, I_rms =

ErmsZ

∴ P =6.5×6.5Z×RZ⇒ P =6.5×6.5×10R2+ωL22⇒ P =6.5×6.5×10100+576⇒P=6.5×6.5×10676⇒P=0.625=58 W

Question 11:

A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s⁻¹)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.

Answer:

Given:
Peak voltage of AC source, E₀ = 12 V
Angular frequency, ω = 250

πs⁻¹
Resistance of resistor, R = 100 Ω
Energy dissipated as heat

His given by,

H=Erms2RTHere, E_rms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,

H=∫0tdH =∫E02 sin2 ωtRdt ∵Erms =E0sinωt = 144100∫010-3sin2 ωt dt = 1.44 ∫010-31-cos 2 ωt2dt =1.442∫010-3dt+∫010-3cos 2 ωt dt =0.72 10-3-sin2ωt2ω010-3 = 0.7211000-1500 π = 0.7211000-21000π = π-21000 π×0.72 = 0.0002614 = 2.61×10-4 J

Question 12:

In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε₀ = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.

Answer:

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε₀ = 50 V
Frequency of the AC source, ν = 50/

πHz
Capacitive reactance

XCis given by,

XC=1ωCHere,

ω= angular frequency of AC source
C = capacitive reactance of capacitance
∴ XC=12π×50π×25×10-6⇒ XC=10425 ΩNet reactance of the series RC circuit

R2+XC2

⇒Z =

3002+104252 =

3002+4002= 500

Ω(a) Peak value of current

I0is given by,

I0= ε0Z ⇒I0= 50500 = 0.1 A(b) Average power dissipated in the circuit

Pis given by,
P =

εrmsI_rms cosÏ•.
εrms=

ε02 and

Irms=I02

∴ P=E02×I02×RZ⇒P=50×0.1×3002×500⇒P =32=1.5 W

Question 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 correct voltage?

Answer:

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P =

V2R,
where R = resistance of electric bulb
∴ R=V2P

=110×11055=220 ΩIf L is the inductance of the coil, then total reactance of the circuit (Z) is given by,
Z =

R2+ωL2 =

2202+100πL2

∵ ω=2πfHere,

ω= angular frequency of the circuit
Now, current through the bulb,

I=VZ

∴Voltage drop across the bulb, V =

VZ×RAs per question,

110=220×2202202+100πL2110=22022202+100πL2⇒220×2=2202+100πL2⇒2202+100πL2=4402⇒48400+104π2L2=193600⇒104π2L2=193600-48400⇒L2=145200π2×104 =1.4726⇒L=1.2135≅1.2 H

Question 14:

In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ε_rms = 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 differences across the three elements is greater than the rms voltage of the source.

Answer:

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10⁻⁶ F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms = 50 V
Frequency of source, f = 50/

πHz
Reactance of the inductor (X_L) is given by,
X_L=

ωL= 2

πfL
⇒X_L = 2

×π×50π×1 = 100

ΩReactance of the capacitance

XCis given by,

XC=1ωC=

12πfC

⇒X_C =

12π×50π×20×10-6

⇒X_C =

500 Ω
(a) Impedance of an LCR circuit

Zis given by,

Z=R2+XC-XL2 ⇒Z =3002+500-1002 ⇒Z =3002+4002⇒ Z =500RMS value of current

Irmsis given by,

Irms=εrmsZ

⇒I_rms =

50500

⇒I_rms =

0.1 A(b) Potential across the capacitor

VCis given by,
V_C = I_rms × X_C

⇒V_C = 0.1 × 500 = 50 V
Potential difference across the resistor

VRis given by,
V_{R =}I_rms × R
⇒V_R = 0.1 × 300 = 30 V
Potential difference across the inductor

VLis given by,
V_L = I_rms × X_L

⇒ V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

Question 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:

Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10⁻⁶ F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage,

εrms= 50 V
RMS value of current, I_rms = 0.1 A
Peak current

I0is given by,

I0=Erms2Z=50×1.414500=0.1414 AElectrical energy stored in capacitor

UCis given by,
UC=12 CV2

⇒ UC=12×20×10-6×50×50⇒UC=25×10-3 J=25 mJMagnetic field energy stored in the coil

ULis given by,
UL=12LI02

⇒ UL=12×1×0.14142⇒UL≈ 5×10-3 J⇒UL=5 mJ

Question 16:

An inductance of 2.0 H, a capacitance of 18μF 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 maximise the current in the circuit? (b) What is the value of this maximum current?

Answer:

Given:
Inductance of inductor, L = 2.0 H
Capacitance of capacitor, C = 18 μF
Resistance of resistor, R = 10 kΩ
Voltage of AC source, E = 20 V
(a) In an LCR circuit, current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
X_L = X_C

⇒ωL=1ωC⇒ω2=1LC=12×18×10-6⇒ω2=10636⇒ω=1036⇒2πf=1036⇒f=10006×2π=26.539 Hz =27 Hz(b) At resonance, reactance is minimum.
Minimum Reactance, Z = R
Maximum current (I) is given by,
I=ER⇒I=2010×103

⇒I=2A103=2 mA

Question 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 this inductor coil is connected to a battery of emf 12 V and internal resistance 4.0 Ω, what will be the current?

Answer:

RMS value of voltage, E_rms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, I_rms = 6 A
Reactance

Ris given by,

R=EIrms⇒R=246=4 ΩLet R‘ be the total resistance of the circuit. Then,
R‘ = R + r

⇒ R’ = 4 Ω + 4 Ω
⇒R‘ = 8 Ω

Current, I =

128 A = 1.5 A

Question 18:

Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input V_i= 10 mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10 k Hz, 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

Answer:

Here,
Input voltage to the filter, Vi = 10 × 10⁻³ V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10⁻⁹ F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
Capacitive reactance

XCis given by,
XC=1ωC=12πfC

⇒XC=12π×10×103×10×10-9⇒XC=12π×10-4⇒XC =1042π=5000π Ω Net impedence of the resistance-capacitance circuit (Z) is given by,
Z=R2+XC2⇒Z=1+103+5000/π2⇒ Z=106+5000/π2 Current (I₀) is given by,

I0=ViZ⇒I0=10×10-3106+5000/π2Output across the capacitor

V0is given by,

V0=102105+50/π2×500π ⇒V0 =1.6124 V=1.6 mV(b)When frequency, f = 1 MHz =

1×10⁶ Hz
Capacitive reactance

XCis given by,

Xc=1ωC⇒XC=12πfC⇒XC=12π×106×10-9×10⇒XC =12π×10-2⇒XC=1002π⇒XC=500π ΩTotal impedence Z =R2+XC2⇒Z =1032+50/π2Current (I0) =V1Z⇒I0 =10×10-3106+50/π2Output voltage V0 =I0XC⇒V0=10-2105+50/π2×50π⇒V0=0.16 mV(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
Capacitive reactance

XCis given by,
Xc=1ωC⇒XC=12πfC⇒XC=12π×107×10×10-9⇒XC=5π ΩImpedence Z =R2+Xc2⇒Z=1032+5/π2Current I0=V1Z⇒I0=10×10-3106+5/π2V0=I0XC⇒V0=10-2106+5/π2×5π⇒Vo=16 μV

Page No 331:

Question 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 works on the principle of electromagnetic induction, which is only possible in case of AC.
Hence, when DC (zero frequency) is supplied to it, the primary coil blocks the current supplied to it. Thus, the induced current in the secondary coil is zero. So, the output voltage will be zero.

Chapterwise HC Verma Solutions Class 12 Physics :

About the Author – HC Verma

HC Verma, the author of many popular and well-renowned Physics books, was born on 8 April 1952. Passing out from one of the most prestigious colleges of the country, IIT Kanpur, he worked as an experimental physicist in the Department of Nuclear Physics.

His most famous works which he is known for include the two-volume Concepts of Physics. He also worked for the social upliftment of the economically weaker children through his organization named Shiksha Sopan. He is also the recipient of the Padma Shri, which is considered India’s fourth-highest civilian award. He received the same because of his contribution and valuable work in the field of Physics.