NCERT Solutions Class 12th Physics Chapter – 7 Alternating Current
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 7 |
| Chapter Name | Alternating Current |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
NCERT Solutions Class 12th Physics Chapter – 7 Alternating Current
?Chapter – 7?
✍Alternating Current✍
?Question & Answer?
Question 1. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer: Resistance of the resistor, R = 100 Ω
Supply voltage, V = 220 V
Frequency, ν = 50 Hz
(a) The rms value of current in the circuit is given as:
I = V / R
= 220 / 100 = 2.20 A
(b) The net power consumed over a full cycle is given as:
P = VI
= 220 × 2.2 = 484 W
Question 2.
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer: (a) Peak voltage of the ac supply, V0 = 300 V
Rms voltage is given as:
(b) Therms value of current is given as:
I = 10 A
Now, peak current is given as:
Question 3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer: Inductance of inductor, L = 44 mH = 44 × 10−3 H
Supply voltage, V = 220 V
Frequency, v = 50 Hz
Angular frequency, ω = 2πv
Inductive reactance, XL = ω L = 2πvL = 2π × 50 × 44 × 10−3 Ω
Rms value of current is given as:
I = V/XL
= 220/2π×50×44×10-3
= 15.92 A
Hence, the rms value of the current in the circuit is 15.92 A.
Question 4. A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer: Given: The capacitance of the capacitor is . The RMS value of the voltage supplied is .
Question 5. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer
Answer: (a) In the case of the inductive network, we know that
the RMS current value is I = 15.92 A
the RMS voltage value is V = 220 V
Therefore, the total power taken in can be derived by the following equation :
P = VI cos Φ
Here,
Φ is the phase difference between V and I
In case of a purely inductive circuit, the difference in the phase of an alternating voltage and an alternating current is 90°,
i.e., Φ = 90°.
Therefore, P = 0
i.e., the total power absorbed by the circuit is zero.
(b) In the case of the capacitive network, we know that
The value of RMS current is given by, I = 2.49 A
The value of RMS voltage is given by, V = 110 V
Thus, the total power absorbed is derived from the following equation :
P = VI Cos Φ
For a purely capacitive circuit, the phase difference between alternating Voltage and alternating current is 90°
i.e., Φ = 90°.
Thus , P = 0
i.e., the net power absorbed by the circuit is zero.
Question 6. Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 µF and R = 10 Ω. What is the Q-value of this circuit?
Answer: Inductance, L = 2.0 H
Capacitance, C = 32 μF = 32 × 10−6 F
Resistance, R = 10 Ω
Resonant frequency is given by the relation,
Now, Q-value of the circuit is given as:
Hence, the Q-Value of this circuit is 25.
Question 7. A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer: Capacitance, C = 30μF = 30×10−6 F Inductance, L = 27 mH = 27 × 10−3 H Angular frequency is given as:
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.
Question 8. Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at a later time?
Answer:
Initially, energy is stored in the capacitor and is given by
The initial energy stored in the capacitor is shared by the inductor during oscillations of the circuit, but the total energy remains the same even at a later time.
Question 9. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer: At resonance, the frequency of supply power equals to naturan frequency of LCR circuit.
Impedence,
Question 10. A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor? [ Hint: The condition for tuning is that the natural frequency which is the frequency of free oscillations of the LC network must be of the same value as the radio wave frequency ]
Answer: Given:
A radio can tune over the range of frequencies from 800 kHz to 1200 kHz and the inductance of its LC circuit is 200 μH.
The capacitance of variable capacitor is given as,
C= 1 ω 2 L = 1 ( 2πf ) 2 ×L
Where, the frequency is f and inductance of radio is L.
By substituting the values for the lower tuning frequency, we get
C l = 1 ( 2π×800× 10 3 ) 2 ×200× 10 −6 =198 pF
By substituting the values for the upper tuning frequency, we get
C u = 1 ( 2π×1200× 10 3 ) 2 ×200× 10 −6 =88 pF
Thus, the range of capacitor is 88 pF to 198 pF.
Question 11. Figure below shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80µF, R = 40 Ω.
L = 5.0 H,
C = 80 μF,
R = 40 Ω
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the RMS potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Answer: (a) The resonance frequency is given by
The resonant frquency is 50 rad/s.
(b) At resonance,
The peak voltage,
(c) Potential drop across inductor,
Potential drop across capacitor,
Potential drop across resistor,
Potential drop across the LC combination
= (–)
at resonance =
Question 12. An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible.
Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored (i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Answer: The energy stored initially in the capacitor is given by:
Question 13.A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer: Given:
Question 14. Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady-state?
Answer: (a) It is given that the inductance of the inductor, L=0⋅50H, the resistance of the
Resistor, R=100Ω, the supply voltage, V=240V and the frequency of supply,
ν=10kHz.
The formula of angular frequency is,
ω=2πν
Substitute the value of ν.
ω=2π× 10 4 rad/s
The formula of maximum current in the circuit is,
I 0 = 2 V R 2 + ω 2 L 2
Substitute the values.
I 0 = 2 ×240 ( 100 ) 2 + ( 2π× 10 4 ) 2 ( 0⋅50 ) 2 I 0 =1⋅1× 10 −2 A
Thus, the value of maximum current in the circuit is 1⋅1× 10 −2 A.
(b) The equation of voltage is,
V= V 0 cos( ωt )
The equation of current is,
I= I 0 cos( ωt−ϕ )
The equation at t=0 is,
ωt−ϕ=0 t= ϕ ω (1)
From the formula of phase angle,
tanϕ= ωL R
Substitute the values.
tanϕ= 2π× 10 4 ×0⋅5 100 =100π ϕ= 89⋅82π 180 rad
Substitute the value of ϕ in equation (1).
t= 89⋅82π 180×2π× 10 4 t=25× 10 −6 s =25 μs
Thus, from the above calculation we can conclude that maximum current is very small at high frequencies so the inductor behaves as an open circuit. In a dc circuit after steady state is achieved, ω=0, the inductor behaves like a pure conducting element.
Question 15. A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Answer:(a) ω =2πf=120π rad/s
Z2=R2+1/ω2C2
⟹Z≈48 Ω
V0=V√2=110√2
Io=Vo/Z=3.24A
(b) tanϕ=1/ωC/R=0.6635
ϕ=33.56π/180 rad
Time lag is ϕ/ω=1.55ms
Question 16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady-state.
Answer: Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R = 40 Ω
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 × 103 Hz
Angular Frequency, ω = 2 πν= 2 × π × 12 × 103
= 24π × 103 rad/s
For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:
Hence, Φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.
In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open circuit.
Question 17. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency
Answer: An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,
L = 5.0 H
C = 80 μF = 80 × 10−6 F R
= 40 Ω
Potential of the voltage source, V = 230 V
Impedance (Z) of the given parallel LCR circuit is given as:
Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum. rms current flowing through inductor L is given as:
Question 18. A circuit containing an 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is
negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle]
Answer: Here,
Inductance, L = 80 mH = 80 x 10-3H
Capacitance, C = 60 μF = 60 x 10 -6 F
Resistance, R = 0
RMS voltage, Ev = 230 V
Peak Voltage, E0 = √2 x Ev
= √2 x 230
Frequency of Ac supply, f = 50 Hz
∴ ω = 2πf = 100 π rad/s
(a) We have to find I0 = ?, Iv = ?
Therefore,
I0 = E0ωL -1ωL = 230 2100 π × 80 × 10-3 – 1100 π × 60 × 10-6
= 230 28π -10006π = 230 2-27.91 = -11.63 amp.
and,
Iv = I02 =-11.631.414 = -8.23 amp.
Negative sign appears as ωL<1ωC.
∴ e.m.f lags behind the current by 90°
(b) Rms value of potential drop across L,
V = Iv × ωL = 8.23 × 100 π × 80 × 10-3 = 206.74 volt.
Rms value of potential drop across C,
V = Iv × 1ωC = 8.23 × 1100 π × 60 × 10-6 = 436.84 volt.
As voltages across L and C are 180° out of phase, therefore, they get subtracted.
That is why applied r.m.s. voltage = 436.84 – 206.74
= 230.1 volt.
(c) Average power transferred over a complete cycle by the source to inductor is always zero because of phase difference of π/2 between voltage and current through the inductor.
(d) Average power transferred over a complete cycle by the source to the capacitor is also zero because of phase difference of π/2 between voltage and current through capacitor.
(e) Total average power absorbed by the circuit is also, therefore zero.
Question 19. Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Answer: Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Total power absorbed by the circuit = 788.44 W
Inductance of inductor, L = 80 mH = 80 × 10−3 H
Capacitance of capacitor, C = 60 μF = 60 × 10−6 F
Resistance of resistor, R = 15 Ω
Potential of voltage supply, V = 230 V
Frequency of signal, ν = 50 Hz
Angular frequency of signal, ω = 2πν= 2π × (50) = 100π rad/s
The elements are connected in series to each other. Hence, impedance of the circuit is given as:
Current flowing in the circuit,
Average power transferred to resistance is given as:
PR= I2R
= (7.25)2 × 15 = 788.44 W
Average power transferred to capacitor, PC = Average power transferred to inductor, PL = 0
Total power absorbed by the circuit:
= PR + PC + PL
= 788.44 + 0 + 0 = 788.44 W
Hence, the total power absorbed by the circuit is 788.44 W.
Question 20. A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Answer: Given: The inductance of inductor is 0.12 H, the capacitance of the capacitor is 480 nF, the resistance of resistor is 23 Ω and the applied potential difference is 230 V.
The peak voltage is given as,
V 0 =V 2
Where, the applied potential difference is V.
By substituting the given values in the above equation, we get
V 0 =230 2 V =325.22 V
(a) The angular frequency at resonance is given as,
ω= 1 LC
Where, the inductance is L, the capacitance is C and the resonance angular frequency is ω.
By substituting the given values in the above equation, we get
ω= 1 0.12×480× 10 −9 = 1 5.76× 10 −8 = 10 4 2.4 =4166.67 rad/s
The resonance frequency is given as,
v= ω 2π
By substituting the given values in the above equation, we get
v= 4166.67 2( 3.14 ) =663.48 Hz
The maximum current flowing through the circuit is given as,
I 0 = V 0 R
Where, the resistance of the circuit is R.
By substituting the given values in the above equation, we get
I 0 = 325.22 23 =14.14 A
Thus, the maximum current of the circuit is 14.14 A.
(b) The maximum average power absorbed by the circuit is given as,
P m = 1 2 ( I 0 ) 2 R
By substituting the given values in the above equation, we get
P m = 1 2 ( 14.14 ) 2 ×23 =2299.3 W
Thus, the maximum average power absorbed by the circuit is 2299.3 W.
(c) Since, the power transferred to the circuit is half the power at resonant frequency.
The change in angular frequency given as,
Δω= R 2L
By substituting the given values in the above equation, we get
Δω= 23 2×( 0.12 ) =95.83 rad/s
The change in frequency is given as,
Δv= 1 2π Δω
By substituting the given values in the above equation, we get
Δv= 95.83 2π =15.26 Hz
The values of new frequencies are given as,
v 1 =v+Δv(1)
v 2 =v−Δv(2)
By substituting the given values in the above equations (1) and (2), we get
v 1 =663.48+15.26 =678.74 Hz
v 2 =663.48−15.26 =648.22 Hz
The current amplitude at these two frequencies is given as,
I ′ = 1 2 ( I 0 )
By substituting the given values in the above equation, we get
I ′ = 14.14 2 =10 A
Thus, the frequencies are 648.22 Hz, 678.74 Hz and the current amplitude is 10 A.
(d) The quality factor of the circuit is given as,
Q= ωL R
By substituting the given values in the above equation, we get
Q= 4166.67×0.12 23 =21.74
Thus, the Q-factor of the given circuit is 21.74.
Question 21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 µF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Answer: Inductance, L = 3.0 H Capacitance, C = 27 μF = 27 × 10−6 F
Resistance, R = 7.4 Ω At resonance, angular frequency of the source for the given LCR series circuit is given as:
To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changing , we need to reduce R to half i.e., Resistance
Question 22. Answer the following questions:
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series
elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer: (a) Yes. The applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. The same is not true for rms voltage, because voltages across different elements may not be in phase.
(b) When the circuit is broken, the high induced voltage is used to charge the capacitor, thus avoiding sparks, etc.
(c) For dc signal, the impedance across the inductor is negligible and the impedance of the capacitor is very high (infinite). So the dc signal appears across the capacitor. For high-frequency ac, the impedance across the inductor is high and that of a capacitor is low. So, the ac signal appears across the inductor.
(d) For a steady-state dc, L has no effect, even if it is increased by an iron core. For ac, the lamp will shine dimly because of the additional impedance of the choke. It will dim further when the iron core is inserted which increases the choke’s impedance.
(e) A choke coil reduces the voltage across the tube without wasting power. A resistor would waste power as heat.
Question 23. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer: Input voltage to the transformer, V1 = 2300 V
Primary windings in the step-down transformer, n1= 4000 turns
Output voltage, V2 = 230 V
Let n2 be the number of turns in the secondary
n2 = (n1 x V2)/V1
= (4000 x 230)/2300 = 400 turns
So, the number of turns in the secondary windings is 400.
Question 24. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms–2).
Answer: h = 300 m
Volume of water flowing = 100 m3s–1
Density of water, ρ = 1000 kg/m3
Electric power, P= hρgAv = hρgβ
here β = Av (Volume of water flowing per second, β= 100 m3s–1 )
⇒ P = hρgβ = 300 x 1000 x 9.8 x 100 = 29.4 x 107W
Efficiency of the turbine generator = 60%
Electric power available for the plant = (29.4 x 107 x 60)/100
= 176.4 x 106 W
Question 25. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V.
The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant.
Answer: Power required , P = 800 kW = 800 x 103 W
Power = Voltage x current
⇒ 800 x 103 = 4000 x I
Therefore, RMS current in the line, I = (800 x 103 )/4000
I = 200 A
Total resistance of the two wire line, R = 2 x 15 x 0.5 = 15 Ω
(a) Line Power loss = I2 R= (200)2 x 15 = 60 x 104 W = 600 kW
(b) Power supply to the plant= 800 kW + 600 kW= 1400 kW
(c) Voltage drop on the line = I R = 200 x 15 = 3000 V
Transmission voltage = 4000 V + 3000 V= 7000 V
The step-up transformer at the plant is 440 V – 7000 V
Question 26. Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption
any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
Answer: Therefore, RMS current in the line, I = (800 x 103 )/40,000
I = 20 A
Total resistance of the two wire line, R = 2 x 15 x 0.5 = 15 Ω
(a) Line Power loss = I2 R= (20)2 x 15 = 60 x 102 W = 6 kW
(b) Power supply to the plant= 800 kW + 6 kW= 806 kW
(c) Voltage drop on the line = I R = 20 x 15 = 300 V
The step-up transformer is 440 V – 40 V=300 V. It is clear that percentage power loss is greatly reduced by high voltage transmission. In Question 7.25, this power loss is (600/1400) × 100 = 43%. Here it is only (6/806) × 100 = 0.74%.