Alternating Current
Class 12th Physics Part I CBSE Solution
Exercises
Question 1.A 100Ω resistor is connected to a 220 V, 50 Hz A.C. supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer:
Given:
Resistance, R = 100Ω
Voltage, Vrms = 220 V
Frequency of A.C. supply, F = 50 Hz
(a) The root mean square value of the current in the circuit can be calculated as follows:
Irms = Vrms/R
⇒ Irms = (220V)/(100Ω)
⇒ Irms = 2.20 A
(b) The net power consumed over a full cycle can be calculated as follows:
P = Voltage × Current
P = V × I
P = (220V) × (2.20A)
⇒ P = 484 W
Question 2.
The peak voltage of an A.C. supply is 300 V. What is the rms voltage?
Answer:
Given: Peak voltage of an A.C. supply, Vpeak = V0 = 300 V
The root mean square voltage can be calculated as follows:
Vrms = V0/√2
Vrms = (300V)/√2
⇒ Vrms = 212.1 V
Question 3.
The rms value of current in an A.C. circuit is 10 A. What is the peak current?
Answer:
Given: the root mean square value in an A.C. circuit, Irms = 10A A
The peak current can be calculated as follows:
Irms = I0 /√2
⇒ I0 = I√2 = (10 A)√2
⇒ I0 = 14.1 A
Question 4.
A 44 mH inductor is connected to 220 V, 50 Hz A.C. supply. Determine the rms value of the current in the circuit.
Answer:
Given: Inductance = 44 mH = 44 × 10-3H
Voltage V = 220 V
Frequency, v = 50 Hz
The diagram is:
Angular frequency, ω = 2πv
ω = 2 × 3.14 × 50Hz
Inductive reactance is given by the formula XL = ω × L
Substituting the value of ω from above
XL = 2 × 3.14 × 50Hz × 44 × 10-3H
The root mean square value of the current in the circuit is given as:
Irms = V/XL
Substituting the value of inductive reactance from equation (i), we get
⇒ 
⇒ Irms = 15.92 A
Question 5.
A 60 μF capacitor is connected to a 110 V, 60 Hz A.C. supply. Determine the rms value of the current in the circuit.
Answer:
Given: capacitance = 60 μF = 60 × 10-6 F
Voltage V = 110 V
Frequency f = 60 Hz
The diagram is:
Capacitive reactance is given by the following relation:
Xc = 1/ωC
Substituting ω = 2πv in the above formula, we get
Xc = 1/2πvC
Substituting the values, we get
⇒ 
Now, the root mean square value of the current can be calculated as follows:
Irms = V/Xc
Irms = 110V × 2 × 3.14 × (60Hz) × 60 × 10-6(F)
On calculating, we get
Irms = 2.49 A
Question 6.
In Exercises 3 and 4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer:
In inductive circuit
Root mean square value of current, I = 15.92 A
Root mean square value of voltage, V = 220 V
The net power absorbed by each circuit over a complete cycle is given by the solution:
Power = VI cosФ
Where Ф is the phase difference between the voltage and current
For a pure inductive circuit, the phase difference between voltage and current in 900.
Substituting the above values, we get
⇒ P = 0
In capacitor circuit
Root mean square value of current, I = 2.49 A
Root mean square value of voltage, V = 110 V
The net power absorbed by each circuit over a complete cycle is given by the solution:
Power = VI cos Ф
For a pure capacitor circuit, the phase difference between voltage and current in 900.
Substituting the above values, we get
⇒ P = 0
∴ Zero power is absorbed by each circuit.
Question 7.
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 μF and R = 10Ω. What is the Q-value of this circuit?
Answer:
Given: L = 2.0H
C = 32mF = 32 × 10-6 F
R = 10 Ω
The diagram of the question is:
Resonant frequency ωr is calculated as follows:
ωr = 125 s-1
Q- value of the LCR circuit can be calculated as follows:
On calculating, we get
Q = 25
Question 8.
A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
The capacitor C = 30 μF = 30 × 10 -6 F
Inductor L = 27 mH = 27 × 10-3 H
The diagram is given :
Angular frequency ωr is given by the following:
ωr = 1/√ LC
ωr = 1/{√(27 × 10-3H × 30 × 10-6F)}
on calculating, we get
ωr = 1.11 × 103 rad/s
Question 9.
Suppose the initial charge on the capacitor in Exercise 7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
Given: Capacitance C = 30 uF = 30 × 10 -6F
Inductance L = 27 mH = 20 × 10 -6H
The charge on capacitor, Q = 6 mC = 6 × 10-3C
The diagram is given :
The total energy stored in the circuit can be calculated as followed:
E = 1/2 Q2C
E = 1/2 × (6 × 10-3C)2 × 30 × 10-6F
On calculating, we get
⇒ E = 0.6 J
At later time, total energy will be same i.e. E = 0.6 J as energy is shared between inductor and capacitor.
Question 10.
A series LCR circuit with R = 20Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V A.C. 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:
Given:
R = 20 Ω
L = 1.5 H
C = 35 μF = 35 × 10-6F
Voltage supplied V = 200 V
The diagram given is:
Impedance of the circuit can be calculated as follows:
At resonance condition i.e. when ωL = 1/ωC
Z = R = 20 Ω
Current is given by the relation
I = V/R
I = 200V /20Ω
⇒ I = 10A
Therefore, the average power transferred to the circuit in one complete cycle can be calculated as follows:
Power = VI
Power = 200V × 10A
Power = 2000 W.
Question 11.
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?
Answer:
Given: frequency range of radio waves = 800 kHz to 1200 kHz
Therefore, low tuning frequency of radio wave v1 = 800 kHz, or 800 × 10-3 Hz
Upper tuning frequency v2 = 1200 kHz or 1200 × 10-3 Hz
Inductance L = 200 μH or 200 × 10-6H
Capacitance of the variable capacitor for low tuning frequency v1 is given by the following relation:
C1 = 1/ ω12L……………..(i)
Where ω is equal to 2πv1
Substituting in equation (i), we get
C1 = 1/{(2π × 800 × 10-3Hz)2 × 200 × 10-6H}
On calculating we get
C1 = 1.9809 × 10-10F
Or C1 = 198.1 pF
Capacitance of the variable capacitor for upper tuning frequency v2 is given by the following relation:
C2 = 1/ ω22L……………..(ii)
Where ω is equal to 2πv2
Substituting in equation (i), we get
C2 = 1/{(2π × 1200 × 10-3Hz)2 × 200 × 10-6H}
On calculating we get
⇒ C2 = 88.04 pF
Therefore, the range of its variable capacitor is from 88.04 pF to 198.1 pF.
Question 12.
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40Ω.
(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) Given: L = 5.0 H
C = 80 Μf = 80 × 10-6F
R = 40Ω
Voltage source V = 230 V
The diagram is given as:
Resonance angular frequency can be calculated as follows
ωR = 1/√LC
ωR = 1/ √(5H × 80 × 10-6F)
On calculating, we get
ωR = 50 rad/s
Therefore, source frequency which drives the circuit in resonance frequency ωR = 50 rad/s.
(b) Impedance can be calculated by using the formula:
At resonance condition i.e. when 
Z = R = 40 Ω
At resonating frequency, the amplitude of the current can be calculated using the formula:
Where V0 is the peak voltage
Z is the impedance
V0 = √2Vrms
I0 = (√2 × 230V)/40Ω
⇒I0 = 8.13 A
The impedance of the circuit is 40 Ω and the amplitude of current at the resonating frequency is 8.13 A.
(c) The root mean square potential drop across the inductor is given by the following formula:
(VL) rms = I × 1/ωR L
Where I = I0/√ 2
I = √ 2V/√2Z
Substituting values, we get
I = 230V/40Ω = 230/40 A
Now,
(VL) rms = I × 1/ωR L
(VL)r.m.s.
(VL)r.m.s = 1437.5 V
Potential drop across capacitor can be calculated as follows:
(VC) rms = I × 1/ω R C
(VC)r.m.s.
(VC)r.m.s = 1437.5 V
Potential drop across resistor can be calculated as follows:
(VR) rms = I × R
(VR) r. m. s
(VR) rms = 230 V
Potential drop across LC combination can be calculated as follows:
VLC
At resonance condition i.e. when ωL = 1/ωC
Therefore, VLC = 0
Additional Exercises
Question 1.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:
Inductor L = 20 mH
In henry L = 20 × 10-3H
capacitor C = 50 μF = 50 × 10-6F
Charge Q = 10 mC = 10 × 10-3C
(a) Total energy stored in the capacitor can be calculated as follows:
E = 1/2 Q2C
Substituting values in above formula, we get
E = 
On calculating, we get
E = 1 Joule
Yes, the total energy will be conserved during LC oscillations because there is no resistor connected in the circuit.
(b) Natural frequency of the circuit can be calculated using the following expression:
V = 1/(2π√LC)
V = 1/[2 × 3.14 × √ (20 × 10-3H × 50 × 10-6F)]
On calculating, we get
V = 159.24 Hz
Natural angular frequency is given by the following expression:
ωr = 1/√ LC
or ωr = 1/√ (20 × 10-3H × 50 × 10-6F)
ωr = 103 rad/s
(c) (i) Time period T = 1/v
T = 1/159.24 Hz
Or T = 6.29 ms
Therefore, for time period T = 6.29 ms, the total charge on the capacitor can be calculated by the following relation:
Q’ = Q cos 2π/T
Since energy stored is electrical, therefore Q’ = Q
Therefore, energy stored is electrical at time t = 0, T/2, T, 3T/2
(ii) Since when electrical energy is zero, the magnetic energy will be maximum
Therefore, energy stored is electrical at time t = T/4, 3T/4, 5T/4
(d) Q’ is equal to the charge on the capacitor when the total energy is shared between capacitor and inductor
When the energy is shared between the capacitor and inductor then the total energy stored in the capacitor is given as half of the maximum energy
⇒ 
⇒ Q’ = Q/√ 2………………(1)
But as we know, Q’ = Q cos 2π/T
Substituting eqn (1) in above
Thus, the total energy is shared between capacitor and inductor at time t = T/8, 3T/8, 5T/8
(e) If a resistor is inserted in the circuit, then total energy is eventually dissipated as heat.
Question 2.
A coil of inductance 0.50 H and resistance 100 W is connected to a 240 V, 50 Hz A.C. 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: Inductance = 0.50 H
Resistance R = 100 Ω
Voltage V = 240 V
Frequency = 50 Hz
The diagram is:
(a) Peak voltage is given by the following formula:
V0 = √2 V
V0 = √2 × 240V
On calculating, we get
V0 = 339.41 V
Angular frequency is given by the formula ω = 2πv
Substituting value of v in above formula
ω = 2 × π × 50(Hz) = 100 π rad/s.
The maximum current in the coil can be calculated as follows:
On calculating, we get
I0 = 1.82 A
(b) The equation for voltage and current are as follows:
V = V0 cos ωt
I = I0 cos (ωt-Ф)
At time t = 0, V = V0
At t = Ф/ ω, I = I0
Therefore, the time lag between maximum current and maximum voltage is Ф/ ω. Phase angle is given by the following relation:
tan Ф = ωL/R
tan Ф = (2 π × 50(Hz) × 0.5(H))/100
On calculating, we get
tan Ф = 1.57
or Ф = 57.50
in rad Ф = 57.5π/180 rad
ωt = 57.5π/180
or 
On calculating, we get
t = 3.2 ms
Question 3.
Obtain the answers (a) to (b) in Exercise 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:
Given: Inductor = 0.5 Hz
Resistor R = 100 Ω
Voltage V = 240 V
Frequency v = 10 kHz
In Hz, the frequency v = 104 Hz
(a) Angular frequency is given by the following relation
ω = 2πv
ω = 2π × 104 rad/s
Peak voltage is given by the relation: V = √2 × V
On substituting the values
V = 240√2V
The maximum current is given by the relation:
Substituting the values we get
⇒ 
On calculating, we get
⇒ I0 = 1.1 × 10-2A
(b) tan Ф = ωL/R
tan Ф = 
on calculating, we get
or Ф = = 89.820
in rad, Ф = 89.82 π/180 rad
ωt = 89.82π/180
or 
On calculating, we get
t = 25 μs
Since I0 is very small, therefore at higher frequency the inductor is at open circuit.
In case of a dc circuit, when steady state is obtained i.e. ω = 0. Therefore, the inductor behaves like a conducting object.
Question 4.
A 100 μF capacitor in series with a 40Ω 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:
Given: capacitor C = 100 μf = 100 × 10-6F
Resistance R = 40Ω
Voltage V = 110 V
Frequency v = 60 Hz
(a) Angular frequency is given by the relation: ω = 2πv
ω = 2π × 60 rad/s
The impedance is given by the relation
Or Z2 = 
Peak voltage is given by the relation V0 = V √2
Or V0 = 110 √ 2 V
The maximum current is given by the following relation:
I0 = V0/Z
I0 = 
I0 = 
On calculating, we get
I0 = 3.24 A
(b) Since in capacitor circuit, the voltage lags behind the current by a phase angle Ф. Therefore, the phase angle is given by the expression
tanФ = 
tanФ = 1/ωCR
substituting the values, we get
tanФ = 1/(120π(rad/s) × 10-4(F) × 40 (Ω))
On calculating, we get
tanФ = 0.6635
or Ф = 33.56°
in radians, Ф = (33.56 π) /180
Time lag is given by the relation
t = Ф/ω
t = 
on calculating, we get
t = 1.55 × 10-3s
on converting in ms, we get time lag t = 1.55 ms.
Question 5.
Obtain the answers to (a) and (b) in Exercise 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:
Given: Capacitance C = 100 μF
In Farad, Capacitance C = 100 × 10-6F
Resistance = 40Ω
Voltage V = 110 V
Frequency = 12 kHz
In Hz, frequency v = 12 × 103Hz
Angular frequency ω = 2πv = 2 × π × 12 × 103(Hz)
ω = 24 × 103 rad/s
Peak voltage is calculated as follows:
V0 = V/ √2
V0 = 110√2V
The peak current is calculated as follows:
I0 =
I0 = 
On calculating, we get
I0 = 3.9 A
Since for RC circuit, the voltage lags behind the current by a phase angle of Ф
tanФ =
tanФ = 1/ωCR
substituting the values, we get
tanФ = 1/(24π × 103(Hz) × 100 × 10-6(F) × 40 (Ω))
On calculating, we get
tanФ = 1/96π
or Ф = 0.20
in radians, Ф = (0.2 π) /180
At high frequency, Ф approaches to zero and capacitor acts as a conductor.
In d.c. circuit, ω is zero, and the steady state is attained and therefore, capacitor C tends to be an open circuit.
Question 6.
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 11 for this frequency.
Answer:
Given: L = 5.0 H
capacitor C = 80 × 10-6F
Resistance = 40 Ω
Voltage = 230 V
Impedance is expressed as:
At resonance condition, 
ω = 1/ √LC
substituting the values, we get
ω = 1/ √(5(H) × 80 × 10-6(F))
On calculating, we get
ω = 50 rad/s
Since, the magnitude of impedance is maximum at 50 rad/s due to which the total current is minimum.
The root mean square current flowing through the inductor is given by the relation:
IL = V/ωL
Substituting the values, we get
IL = 230(V)/(50 (rad/s) × 5 (H))
On calculating, we get
IL = 0.92 A
Root mean square value of the current through capacitor is given by the following relation:
Ic = ωCV
Ic = 50rad/s × 80 × 10-6F × 230V = 0.92 A
Root mean square value of the current through resistor is given by the following relation:
IR = V/R
IR = 230V/40Ω
IR = 5.75A
Question 7.
A circuit containing a 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:
Given: L = 80 mH = 80 × 10-3 H
capacitor C = 60 Μf = 60 × 10-6 F
Voltage V = 230 V
Frequency v = 50Hz
(a) Peak voltage is given by the formula:
V0 = V/√ 2
V0 = 230√2 V
Substituting the values we get
On calculating, we get
I0 = -11.63 A.
The amplitude of maximum current is given by the following relation:
I0 = 11.63 A.
The root mean square value of the current I = I0/√ 2
⇒ I = -11.63(A)/√ 2
⇒ I = - 8.22 A
(b) The potential difference across the inductor can be calculated as follows:
VL = I × ω × L = 8.22(A) × 100 π(Hz) × 80 × 10-3 (F) = 206.61 V
Potential difference across the capacitor is given by the formula as follows:
Vc = I × 1/ωC
Substituting values we get
Vc = 8.22(A) × 1/ 100 π(Hz) × 60 × 10-6(F)
On calculating, we get
Vc = 436.3 V
(c) Since the actual voltage leads the current by π/2, so the average power consumed by the inductor is zero.
(d) Since voltage lags the current by π/2, so the average power consumed by the capacitor is zero.
(e) The total power over a complete cycle is zero.
Question 8.
Suppose the circuit in Exercise 18 has a resistance of 15Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Answer:
Given: Average power transferred to the resistor = 788.44 W
Average power transferred to the capacitor = 0 W
Power absorbed by the circuit = 788.44 W
Inductance L = 80 mH
In henry, Inductance L = 80 × 10-3H
Capacitance C = 60 μF = 60 × 10-6F
Resistance R = 15 Ω
Voltage V = 230 V
Frequency v = 50 Hz
Angular frequency is calculated as follows:
ω = 2πv
ω = 2 × 3.14 × 50(Hz) = 100π rad/s
The impedance can be calculated as follows:
On calculating, we get
Z = 31.728 Ω
Current can be calculated as follows: I = V/Z
I = 230 (A)/31.728 (Ω) = 7.25 A
The average power transferred to resistor can be calculated as follows:
Pr = I2R
Substituting values, we get
Pr = (7.25)2(A) × 15(Ω)
Pr = 788.44 W
The average power transferred to capacitor is equal to the average power transferred to resistor = 0
Total power absorbed = PR + Pc + PL = 788.4W
Question 9.
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23Ω 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:
(a) Given: L = 0.12 H
C = 480 nF
Converting into Farad, we get 480 × 10-9F
R = 23Ω
Voltage = 230 V
Peak voltage can be calculated as follows:
V0 = √2V
Substituting values we get
V0 = 230√2
V0 = 325.22 V
Current flowing through the circuit is given by the following:
I0 = 
At resonating frequency, ωR L - 1/ωRC = 0
ωR = 1/√LC
= 1/ √0.12 (H) × 480 × 10-9 (F) = 4166.67 rad/s
Resonant frequency can be calculated using the formula:
vR = ωR/2π
vR = 4166.67(Hz)/2 × 3.14
vR = 663.48 Hz
Maximum current is calculated as follows:
(I0)max = V0/R = 325.22(Hz)/23 = 14.14 A
(b) The maximum power absorbed by the circuit can be calculated as follows:
Pav = 1/2(I0)2maxR
Substituting the values, we get
Pav = 1/2 × (14.14)2(A) × 23 (Ω)
Pav = 2299.33 W
Therefore, the resonating frequency is 663.48 Hz.
(c) The power transferred is equal to the half of the power at the resonating frequency.
Frequency at which power is half = ωR �∆ω
Or = 2π (vR �∆v)
Where ∆ω = R/2L
Substituting the values we get
∆ω = 23/2(Ω) × 0.12(H)
∆ω = 95.83 rad/s
Therefore, change in frequency is written as follows:
∆v = 1/2π × ∆ω
Substituting the values, we get
∆v = 95.83(rad/s)/2π = 15.26 Hz
vR + ∆v is calculated as follows:
663.48(Hz) + 15.26(Hz) = 678.74 Hz
vR-∆v is calculated as follows:
663.48(Hz) – 15.26(Hz) = 648.22 Hz
The current amplitude is calculated as follows:
I’ = 1/√2 × (I0)max
I’ = 14.14(A)/ √2
I’ = 10 A
(d) Q factor can be calculated as follows:
Q = ωRL/R
Q = (4166.67 rad/s) × 0.12H/23Ω
⇒Q = 21.74
Question 10.
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4Ω. 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:
Given: L = 3.0 H
C = 27 μF,
In Farad, capacitance C = 27 × 10-6F
R = 7.4Ω
Angular frequency is given by the relation:
ωR = 1/√LC
ωR = 1(√27 × 10-6F × 3 H)
On calculating, we get
ωR = 111.11 rad/s
Q factor can be calculated as follows:
Q = ωRL/R
Q = (111.11(rad/s) × 3H)/7.4(Ω)
⇒Q = 45.0446
The sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2 by reducing R to half.
i.e. R = R/2 = 7.4(Ω)/2 = 3.7Ω
Question 11.
Answer the following question:
In any A.C. 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?
Answer:
The statement is not true for root mean square voltage. In any A.C. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit but is not true for root mean square voltage because the voltages are not in the same phase across different elements.
Question 12.
Answer the following question:
A capacitor is used in the primary circuit of an induction coil.
Answer:
This statement is true that A capacitor is used in the primary circuit of an induction coil. This is due to the reason, that when a circuit is broken, capacitor is charged through high induced voltage so as to avoid the sparkling.
Question 13.
Answer the following question:
An applied voltage signal consists of a superposition of a dc voltage and an A.C. 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 A.C. signal across L.
Answer:
Dc signal will appear across capacitor. This is because impedance of an inductor is negligible whereas the impedance of the capacitor is high and almost infinite. An A.C. singal of high frequency appears across inductor because for an A.C. signal of high frequency, the impedance of inductor is high while impedance of capacitor is low.
Question 14.
Answer the following question:
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 A.C. line.
Answer:
When an iron core is connected in series with the lamp connected to the A.C. line then the choke coil and iron will increase the impedance and the lamp will glow dim.
Question 15.
Answer the following question:
Why is choke coil needed in the use of fluorescent tubes with A.C. mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer:
Choke coil needed in the use of fluorescent tubes with A.C. mains because choke coil reduces the voltage across the tube and doesn’t waste the power whereas while using ordinary resistor, it waste the power in form of heat dissipation. Therefore, we can’t use an ordinary resistor instead of choke coil.
Question 16.
A power transmission line feeds input power at 2300 V to a stepdown 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:
Given: Input voltage = 2300 V
No. of turns = 4000
Output voltage = 230 V
The relation between voltage and no. of turns is written as follows:
V1/V2 = n1/n2
Substituting the values
2300(V)/230(V) = 4000/ n2
On calculating, we get
n2 = 400
Question 17.
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:
Given: height of water pressure h = 300 m
Height of water flow = 100 m3s–1
Efficiency of turbine n = 60%
Acceleration due to gravity g = 9.8 ms–2
Density of water = 103Kg/m3
The electric power available from the plant is given by the relation:
h × η × р × g × V
substituting the values, we get
300m × 0.6 × 103 × 9.8ms–2 × 100 m3s–1 = 176.4 MW
Question 18.
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 wire line carrying power is 0.5Ω 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:
(a) Given: Power P = 800 kW
Voltage = 220 V
Voltage at which power is generated P’ = 440 V
Distance between town and power generating station = 15 km
The resistance of the two-wire line carrying power is 0.5 Ω per km.
Total resistance of wire = (15(km) + 15(km))0.5 (Ω per km)
∴ Rtotal = 15 Ω
Input voltage v1 = 4000 V
Output voltage V2 = 220 V
The root mean square current in the lines can be calculated as follows:
I = P/V1
= (800(kW) × 103)/4000(V) = 200 A
Line power loss can be calculated as follows:
P = I2Rtotal
= (200)2A × 15Ω
On calculating we get P = 600 Kw
(b) Let the power loss due to current leakage is negligible
Total power supplied by the power plant = 800kW + 600kW = 1400 kW
(c) Voltage drop V = IRTotal
Substituting the values, we get
200V × 15Ω = 3000V
Therefore, the total voltage supplied = 3000V + 4000V = 7000V
And since the power generated = 440 V
Therefore, the rating should be 440V - 7000 V
Question 19.
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:
Given: Rating of step down transformed = 40,000-220 V
Input voltage V1 = 40000 V
Output voltage V2 = 220 V
Total electric power P = 800 KW
In watt, P = 800 × 103 W
Potential V = 220 V
Voltage of power generated by the plant V’ = 440 V
Distance between town and power generating station = 15 km
The resistance of the two wire line carrying power is 0.5Ω per km.
Total resistance of wire = (15(km) + 15(km))0.5 = 15 Ω
Input voltage v1 = 4000 V
Output voltage V2 = 220 V
The root mean square current in the lines can be calculated as follows:
I = P/V1
= (800 × 103W)/40000V = 20 A
Line power loss can be calculated as follows:
P = I2R
= (20A)2 × 15Ω = 6 kW
B. Let the power loss due to current leakage is negligible
Total power supplied by the power plant = 800kW + 6kW = 806 Kw
C. Voltage drop V = IR
Substituting the values, we get
20A × 15Ω = 3000V
Therefore, the total voltage supplied = 300V + 40000V = 40300V
And since the power generated = 440 V
Therefore, the rating should be 440 V- 40300 V