NCERT Solutions for Class 12 Physics Chapter 7 – Alternating Current
SimplyAcad is providing the NCERT Solutions Class 12 Alternating Current Chapter 7 in a detailed and precise manner to help students who are facing difficulty to understand the chapter. The solutions will allow the students to get a comprehensive knowledge of the methods and steps to approach the different questions present in the exercise.
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NCERT Solutions Class 12 Alternating Current Chapter 7
Question 7.1:
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 7.1:
Given :
The resistance R of the resistor is 100 Ω
The source voltage V is 220 V
The frequency of the supply is 50 Hz.
- a) To determine the value of the current in the connection, we can use the following relation:
𝐼=𝑉𝑅
Substituting values, we get
Therefore, the RMS value of the current in the connection is 2.20 A.
- b) The total power consumed over an entire cycle can be calculated using the following formula:
P = V x I
Substituting values in the above equation, we get
= 220 x 2.2 = 484 W
Therefore, the total power consumed is 484 W.
Question 7.2: (A) The peak voltage of an ac supply is 300V. What is the rms voltage?
(B) The rms value of current in an ac circuit is 10A. What is the peak current? Hint: “Relation between Irms and Im“
(A)Given,
Peak voltage (Vm)=300V
Vrms=Vm√2
Vrms=300√2=212.1V
Final Answer: 212.1V
(B)Given,
Rms current, lrms=10A
Irms=lm√2
Im=10√2=14.14A
Final Answer: 14.14A
Question 7.3: A 44 mH inductor is connected to a 220 V, 50 Hz AC supply. Determine the RMS value of the current in the circuit.
Solution- To determine the RMS value of the current, we first need to calculate the inductive reactance \( X_L \). The formula for the inductive reactance is: \[ X_L = \omega L = 2 \pi f L \] where: \begin{itemize} \item \( f \) is the frequency of the AC supply, \item \( L \) is the inductance of the inductor. \end{itemize} Given: \[ f = 50 \text{ Hz} \] \[ L = 44 \text{ mH} = 44 \times 10^{-3} \text{ H} \] Substitute these values into the formula: \[ X_L = 2 \pi \times 50 \times 44 \times 10^{-3} \] \[ X_L = 2 \times 3.14 \times 50 \times 44 \times 10^{-3} = 13.82 \text{ Ω} \] Next, we calculate the RMS value of the current \( I_{\text{rms}} \) using Ohm’s law for AC circuits: \[ I_{\text{rms}} = \frac{E}{X_L} \] where \( E \) is the RMS voltage of the supply. Given \( E = 220 \text{ V} \): \[ I_{\text{rms}} = \frac{220}{13.82} \approx 15.92 \text{ A}
Question 7.4: A 60 μF capacitor is connected to a 110 V, 60 Hz AC supply. Determine the RMS value of the current in the circuit. \section*
Solution
We know that XL=ωL=2πfL=2×3.14×50×44×10−3=13.82Ω
Question 7.5: In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.
Answer 7.5:
- a) We know that in the case of the inductive network,
The RMS current value is I = 15.92 A
The RMS voltage value is V = 220 V
Therefore, the total power taken can be derived by the following equation:
P = VI cos Φ
Here,
Φ is the phase difference between V and I
In the 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
Thus, 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 can be 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
Therefore, the net power absorbed by the circuit is zero.
Question 7.6:
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Solution:
Given: The inductance in the circuit is 2.0H.
The capacitance in the circuit is 32μF.
The resistance in the circuit is 10Ω.
To find: The resonant frequency and Q-value of the circuit.
The resonant frequency is given by
ωr=1√LC=1√2×32×10−6
⇒125rad/s
The Q value of the circuit is given by:
Q=1R√LC=110√232×10−6
⇒Q=25
Question 7.7:
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Question 7.8: 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 later time?
Solution
Capacitance of the capacitor, C = 30 μF = 30×10−6 F
Inductance of the inductor, L = 27 mH = 27 × 10−3 H
Charge on the capacitor, Q = 6 mC = 6 × 10−3 C
Total energy stored in the capacitor can be calculated by the relation,
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
Question 7.9: 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?
Solution
Capacitance of the capacitor, C = 30 μF = 30×10−6 F
Inductance of the inductor, L = 27 mH = 27 × 10−3 H
Charge on the capacitor, Q = 6 mC = 6 × 10−3 C
Total energy stored in the capacitor can be calculated by the relation,
Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.
Question 7.10:
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?
Solution
Lower tuning, f1=800×103 Hz
Upper tuning, f2=1200×103 Hz
Question 7.11:
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, 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.
Inductance of the inductor, L = 5.0 H
Capacitance of the capacitor, C = 80 μH = 80 × 10−6 F
Resistance of the resistor, R = 40 Ω
Potential of the variable voltage source, V = 230 V
(a) Resonance angular frequency is given as:
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
(b) Impedance of the circuit is given by the relation,
At resonance,
Amplitude of the current at the resonating frequency is given as: 
Where,
V0 = Peak voltage
Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) Rms potential drop across the inductor,
(VL)rms = I × ωRL
Where,
I = rms current
Potential drop across the capacitor,
Potential drop across the resistor,
(VR)rms = IR
= 
× 40 = 230 V
Potential drop across the LC combination,
At resonance,
∴VLC= 0
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.
Question 7.12:
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?
Inductance of the inductor, L = 20 mH = 20 × 10−3 H
Capacitance of the capacitor, C = 50 μF = 50 × 10−6 F
Initial charge on the capacitor, Q = 10 mC = 10 × 10−3 C
(a) Total energy stored initially in the circuit is given as:
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
(b)Natural frequency of the circuit is given by the relation,
Natural angular frequency,
Hence, the natural frequency of the circuit is 103 rad/s.
(c) (i) For time period (T
), total charge on the capacitor at time t, 
For energy stored is electrical, we can write Q’ = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =
(ii) Magnetic energy is the maximum when electrical energy, Q′ is equal to 0.
Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time, 
(d) Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = 
(maximum energy).
Hence, total energy is equally shared between the inductor and the capacity at time, 
(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.
Question 7.13:
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?
Inductance of the inductor, L = 20 mH = 20 × 10−3 H
Capacitance of the capacitor, C = 50 μF = 50 × 10−6 F
Initial charge on the capacitor, Q = 10 mC = 10 × 10−3 C
(a) Total energy stored initially in the circuit is given as:
Hence, the total energy stored in the LC circuit will be conserved because there is no resistor connected in the circuit.
(b)Natural frequency of the circuit is given by the relation,
Natural angular frequency,
Hence, the natural frequency of the circuit is 103 rad/s.
(c) (i) For time period (T), total charge on the capacitor at time t,
For energy stored is electrical, we can write Q’ = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical at time, t =
(ii) Magnetic energy is the maximum when electrical energy, Q′ is equal to 0.
Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time,
(d) Q1 = Charge on the capacitor when total energy is equally shared between the capacitor and the inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor = (maximum energy).
Hence, total energy is equally shared between the inductor and the capacity at time,
(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the circuit. The resistance damps out the LC oscillation.
Question 7.14: 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?
(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 7.15:
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?
Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R = 40 Ω
Supply voltage, V = 110 V
(a) Frequency of oscillations, ν= 60 Hz
Angular frequency, 
For a RC circuit, we have the relation for impedance as:
Peak voltage, V0 = 
Maximum current is given as:
(b) In a capacitor circuit, the voltage lags behind the current by a phase angle ofΦ. This angle is given by the relation:
Hence, the time lag between maximum current and maximum voltage is 1.55 ms.
Question 7.16: 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.
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 × 10303
= 24π × 103 rad/s
Peak voltage, 
Maximum current, 
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 7.17: 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.
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:
Where,
ω = Angular frequency
At resonance,
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:
Rms current flowing through capacitor C is given as:
Rms current flowing through resistor R is given as:
Question 7.18 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.
Solution
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:
Where,
ω = Angular frequency
At resonance,
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:
Rms current flowing through capacitor C is given as:
Rms current flowing through resistor R is given as:
Question 7.19:
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 7.19:
Inductance, L = 80mH = 80 x 10-3 H
Capacitance, C = 60μF = 60 x 10-6 F
Supply voltage, V = 230 V
Frequency, ν = 50 Hz
Resistance of the resistor, R = 15 Ω
Angular frequency of the signal, ω = 2 πv = 2π x 50 = 100 π rad/s
The impedance of the circuit,
The average power transferred to resistance is given as
PR = I2 R
= (7.25)2 x 15 = 788.44 W
Average power transferred to the capacitor, PC = 0
Average power transferred to the capacitor, PL= 0
Therefore, the total power absorbed by the circuit = PR + PC + PL = 788.44 + 0 + 0 = 788. 44 W
Question 7.20:
A series LCR circuit with L = 0.12 H, C = 480 nF, and R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which the current amplitude is maximum? Obtain this maximum value.
(b) What is the source frequency for which the average power absorbed by the circuit is maximum? Obtain the value of this maximum power.
(c) For which frequencies of the source are 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 7.20:
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 x 10-9 F
Resistance, R = 23 Ω
Supply voltage, V = 230 V
Peak voltage is given as
Vo = V√2 = 230√2 = 325.22 V
(a) Current flowing the circuit is given by the relation
ωR is the resonance angular frequency
ωR = 1/√LC
= 1/√0.12 x 480 x 10-9 = 4166. 67 rad/s
Resonant frequency, νR = ωR/2π = 4166. 67/2 x 3.14 = 663.48 Hz
Maximum current, (I0)max = V0/R = 325.22/23 = 14.14 A
(b) Maximum average power absorbed by the circuit is
PMax = (1/2) (I0)2max R
= (1/2) x (14.14)2 x 23 = 2300 W
(c) The power transferred to the circuit is half the power at the resonant frequency
P = Vrms Irms cos Φ
= [(Vrms)2/Z ] x (R/Z)
= (Vrms2 x R)/Z2
Pmax = (Vrms)2/Z
Given P = (1/2) Pmax
Therefore, we get
(Vrms2 x R)/Z2 =(Vrms)2/2Z
2R2 = Z2
2R2 = R2 + (XC– XL)2
XC– XL = ± R
Let us take XC– XL = + R
(1/Cω) – Lω = R
1 – LCω2 = RCω
LCω2 + RCω -1 = 0
ω = 4263.63 rad/s
Let us consider
XC– XL = – R
(1/Cω) – Lω = – R
1 – LCω2 = RCω
LCω2 – RCω -1 = 0
= (CR)2 – 4 (LC)
Substituting the values, we get a negative value. Therefore, it will not be considered.
So, ω = 4263.63 rad/s
Therefore, 2πf = ω
⇒ f = 4263.63/(2 x 3.14) =4263.63/6.28 = 678.57 Hz
At this frequency, the current frequency is given as,
I’ = (1/√2) x (I0)max
= 14.14/√2 = 10 A
(d) Q factor,
Question 7.21:
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 7.21:
Inductance, L = 3.0 H
Capacitance, C = 27μF = 27 x 10-6 F
Resistance, R = 7.4 Ω
For the LCR series circuit, the resonant frequency of the source is
Q-factor of the series
Q = ωL/R
= (111.11 x 3)/7.4 = 45.04
To improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2, we should reduce the resistance to half.
= R/2 = 7.4/2 = 3. 7 Ω
Question 7.22:
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) Why is a capacitor 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 coils needed in the use of fluorescent tubes with AC mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer 7.22:
(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 a 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 fluorescent tube without wasting power. A resistor would waste power as heat.
Question 7.23:
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 7.23:
The 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 7.24:
At a hydroelectric power plant, the water pressure head is at the 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 7.24:
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
The 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 7.25:
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 substation 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 7.25:
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 7.26:
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 7.26:
The 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%.
These are provided solutions of Alternating Current Chapter 7 compiled together at one place according to the latest syllabus 2024-25.
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