Question:

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 lamps 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 true for the rms voltage.

(b) The capacitor is used in the primary circuit of the induction coil to store energy and enable the coil to produce a higher voltage.

(c) The dc signal will appear across the capacitor and the ac signal will appear across the inductor due to the capacitors ability to block dc signals and the inductors ability to pass ac signals.

(d) If the connection is to an ac line, the insertion of the iron core in the choke will cause the lamp to dim because the iron core will decrease the inductance of the choke coil, reducing the current flowing through the lamp.

(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it provides a low-impedance path for the high-frequency current, allowing more current to flow through the tube and providing a more consistent light output. An ordinary resistor cannot be used instead of the choke coil because it would not provide a low-impedance path for the high-frequency current.

Question:

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:

a) The source frequency for which current amplitude is maximum is the resonant frequency of the circuit, which is given by:

f_res = 1/(2π√(LC))

f_res = 1/(2π√(0.12 x 480 x 10^-9)) = 1.04 x 10^4 Hz

The maximum current amplitude is given by:

I_max = V/R = 230/23 = 10 A

b) The source frequency for which the average power absorbed by the circuit is maximum is also the resonant frequency of the circuit, which is 1.04 x 10^4 Hz. The maximum power absorbed by the circuit is given by:

P_max = (V^2/R) x (Q^2/R) = (230^2/23) x (Q^2/23)

Where Q is the quality factor of the circuit.

c) The frequencies of the source for which the power transferred to the circuit is half the power at the resonant frequency are known as the half-power frequencies, which are given by:

f_hp = f_res / √2

f_hp = 1.04 x 10^4 / √2 = 7.35 x 10^3 Hz

The current amplitude at these frequencies is given by:

I_hp = V/R = 230/23 = 10 A

d) The quality factor (Q-factor) of the given circuit is given by:

Q = ω_res x L/R

Q = (2π x 1.04 x 10^4 x 0.12)/23 = 13.2

Question:

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? Exercise: [ 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 0.5 Ωper km.The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.]

Answer:

1.The power transmitted through the two wire line is given by the equation: P = (V^2)/R = (440^2)/(0.5*15) = 73600 kW

2.The power received by the town is given by: P = (V^2)/R = (220^2)/(0.5*15) = 16800 kW

3.The power lost in the transmission line is given by: P_loss = P_transmitted - P_received = 73600 - 16800 = 56800 kW

4.The power received by the town after the 4000-220 V step-down transformer is given by: P = (V^2)/R = (220^2)/(0.5*15) = 16800 kW

5.The power lost in the transformer is given by: P_loss = P_transmitted - P_received = 73600 - 16800 = 56800 kW

6.The power supplied to the town is given by: P_supplied = P_received - P_loss = 16800 - 56800 = 800 kW

7.High voltage transmission is preferred because it reduces the power losses in the transmission line. This is because the power loss is proportional to the square of the voltage. Therefore, by increasing the voltage, the power loss is reduced.

Question:

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 0.5 Ωper 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) The line power loss in the form of heat can be estimated by using the formula P = I^2R, where P is the power loss, I is the current and R is the resistance. Therefore, the line power loss is equal to (800/220)^20.5*15, which is equal to 8.2 kW.

b) The plant must supply 800 kW + 8.2 kW = 808.2 kW.

c) The step up transformer at the plant is a device that increases the voltage from 220 V to 440 V.

Question:

A 60μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

Step 1: Calculate the capacitive reactance of the capacitor. Xc = 1/(2πfC) Xc = 1/(2π60 Hz60 μF) Xc = 683.2 Ω

Step 2: Calculate the total impedance of the circuit. Z = √(R2 + X2) Z = √(0 + 683.22) Z = 683.2 Ω

Step 3: Calculate the rms current. I = V/Z I = 110 V/683.2 Ω I = 0.161 A (rms)

Question:

A coil of inductance 0.50 H and resistance 100Ω 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:

a) Maximum current in the coil = (240V)/(100Ω + j(2π500.50H)) = 2.4A

b) Time lag between voltage maximum and current maximum = (2π0.50H)/(100Ω + j(2π50*0.50H)) = 0.0314 radians = 1.8 degrees

Question:

A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Answer:

1. Determine the capacitive reactance of the capacitor: XC = 1/(2πfC) = 1/(2π(30x10-6)) = 526 Ω

2. Determine the inductive reactance of the inductor: XL = 2πfL = 2π(27x10-3) = 168.5 Ω

3. Determine the total reactance of the circuit: XT = XC + XL = 526 + 168.5 = 694.5 Ω

4. Determine the angular frequency of free oscillations of the circuit: ω = 1/√(LC) = 1/√((30x10-6)(27x10-3)) = 694.5 rad/s

Question:

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:

1. Calculate the natural frequency of the circuit: f = 1/(2π√(LC)) = 1/(2π√(1.5 x 10^-3 x 35 x 10^-6)) = 7.67 Hz

2. Calculate the average power transferred to the circuit in one complete cycle: P = (V^2)/R = (200^2)/20 = 2000 W

Question:

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:

a) The total energy stored initially is equal to the energy stored in the capacitor, which is given by 0.5CV^2, where C is the capacitance and V is the voltage. The energy stored in the capacitor is 0.55010^-610^-3^2 = 2.510^-7 J. Yes, this energy is conserved during LC oscillations.

b) The natural frequency of the circuit is given by 1/2π√(LC), where L is the inductance and C is the capacitance. The natural frequency of the circuit is 1/2π√(2010^-35010^-6) = 159.15 Hz.

c) (i) The energy stored completely electrical is when the voltage across the capacitor is maximum. This occurs at a time of t=π/2, which is 0.78 seconds.

(ii) The energy stored completely magnetic is when the current through the inductor is maximum. This occurs at a time of t=3π/2, which is 4.71 seconds.

d) The total energy is shared equally between the inductor and the capacitor at times t=π and t=2π, which are 1.57 seconds and 3.14 seconds respectively.

e) If a resistor is inserted in the circuit, the energy eventually dissipated as heat is equal to the energy stored in the inductor, which is given by 0.5LI^2, where L is the inductance and I is the current. The energy dissipated as heat is 0.52010^-3*I^2, where I is the current.

Question:

(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) The RMS voltage of an AC supply is equal to the peak voltage divided by the square root of two, which is equal to 212.13 V.

(b) The peak current of an AC circuit is equal to the RMS current multiplied by the square root of two, which is equal to 14.14 A.

Question:

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:

a) The rms value of current in the circuit is 2.2 A.

b) The net power consumed over a full cycle is 48.4 W.

Question:

Do you agree that it is difficult not to go along with someone who is very strong and powerful? Express your views frankly and clearly

Answer:

1. Yes, I agree that it can be difficult not to go along with someone who is very strong and powerful.

2. This is because powerful people often have the ability to influence or manipulate those around them, which makes it difficult to resist their wishes.

3. However, I believe it is important to stand up for oneself and to make decisions based on one’s own values and beliefs, rather than simply following the dictates of someone else.

4. This can be difficult, especially when faced with someone who is strong and powerful, but in the long run it is important to be true to oneself and to make decisions that are in line with one’s own values and beliefs.

Question:

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?

Answer:

1. Resonant frequency, ωr = 1/√(LC)

2. ωr = 1/√(2.0 H * 32 μF)

3. ωr = 1/√(0.064)

4. ωr = 25.12 rad/s

5. Q-value = ωr * √(L/R)

6. Q-value = 25.12 * √(2.0/10)

7. Q-value = 25.12 * √(0.2)

8. Q-value = 25.12 * 0.447

9. Q-value = 11.26

Question:

In question (a) and (b) what is the net power absorbed by each circuit over a complete cycle. Explain your answer. (a). A 44mH inductor is connected to 220V,50Hz ac supply. Determine the rms value of the current in the circuit. (b). A 60F capacitor is connected to a 110V,60Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

(a) The net power absorbed by the circuit over a complete cycle is zero. This is because the inductor stores energy in the form of a magnetic field during the positive half cycle and releases it during the negative half cycle, thus balancing the power absorbed and released.

(b) The net power absorbed by the circuit over a complete cycle is the product of the rms value of the current and the voltage. The rms value of the current can be calculated using Ohm’s law and the capacitive reactance formula: I_rms = V/(X_C) = 110/(1/(2πfC)) = 110/(1/(2π6060F)) = 0.86A. Therefore, the net power absorbed by the circuit over a complete cycle is 110V * 0.86A = 94.6W.

Question:

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:

1. Calculate the turn ratio: Turn ratio = 4000/N Where N is the number of turns in the secondary.

2. Calculate the output voltage: Output voltage = Turn ratio × Input voltage Output voltage = (4000/N) × 2300

3. Solve for N: N = 4000/(2300/230) N = 870

Question:

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) Maximum current in the circuit = 110 V / (40 Ω + (1/100 μF)) = 2.75 A

b) Time lag between the current maximum and voltage maximum = (1/2π)(1/√(1/100 μF x 40 Ω)) = 0.0198 s

Question:

A coil of inductance 0.50 H and resistance 100Ω 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? 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) Maximum current in the coil = 240V/(100Ω + (2π * 50Hz * 0.50H)) = 4.8 A

(b) Time lag between voltage maximum and current maximum = (2π * 0.50H)/(100Ω + (2π * 50Hz * 0.50H)) = 0.063 seconds

At very high frequency, the inductor in the circuit nearly amounts to an open circuit because the inductance is inversely proportional to the frequency. This means that as the frequency increases, the inductance decreases and the current in the circuit decreases. As a result, the inductor acts as an open circuit.

In a DC circuit after the steady state, an inductor behaves as a short circuit because the inductance is constant and the current is allowed to flow freely. This means that the inductor acts as a short circuit, allowing the current to flow freely.

Question:

A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

1. Calculate the impedance (Z) of the inductor: Z = 2πfL = 2π(50 Hz)(44 mH) = 28.2 Ω

2. Calculate the rms value of the current (I): I = V/Z = 220 V/28.2 Ω = 7.80 A

Question:

At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m^3s^−1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8ms^−2 ).

Answer:

1. Calculate the potential energy of the water flow: PE = mgh = (100 m^3s^−1)(300 m)(9.8 ms^−2) = 2.94 x 10^7 J/s

2. Calculate the available electric power: P = PE x efficiency = (2.94 x 10^7 J/s)(0.6) = 1.76 x 10^7 W

Question:

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: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Answer:

1. Calculate the natural frequency of the LC circuit (f): f = 1 / (2π√(LC))

2. Substitute the values of L (200 μH) and C (variable) into the equation to calculate the range of the variable capacitor: C = 1 / (2πf√(LC))

3. Substitute the frequency range of the MW broadcast band (800 kHz to 1200 kHz) into the equation to calculate the range of the variable capacitor: C = 1 / (2π(800 kHz to 1200 kHz)√(200 μH))

4. The range of the variable capacitor is therefore: C = 0.0000000125 to 0.00000001875 μF

Question:

A circuit containing a 80 mH inductor and a 60μF 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 (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [Average implies average value over one cycle.]

Answer:

(a) The current amplitude and rms values can be obtained by using the formula:

I = V/√(L/C) = 230/√(80mH/60μF) = 4.74 A

The rms current value = 4.74 A

(b) The rms value of potential drops across each element can be obtained by using the formula:

V_Inductor = IL = 4.74A80mH = 378.4 V

V_Capacitor = IC = 4.74A60μF = 284.4 V

(c) The average power transferred to the inductor can be calculated as:

P_Inductor = V_InductorI = 378.4V4.74A = 1799.2 W

(d) The average power transferred to the capacitor can be calculated as:

P_Capacitor = V_CapacitorI = 284.4V4.74A = 1350.4 W

(e) The total average power absorbed by the circuit can be calculated as:

P_Total = P_Inductor + P_Capacitor = 1799.2 W + 1350.4 W = 3149.6 W

Question:

Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C=27μF, 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:

1. Calculate the resonant frequency of the series LCR circuit using the formula: f_res = 1/(2π√(LC))

f_res = 1/(2π√(3.02710^-6)) f_res = 1/(2π√(0.00081)) f_res = 1/(2π*0.02883) f_res = 218.2 Hz

2. Calculate the Q-factor of the series LCR circuit using the formula: Q = 1/(R/2L)

Q = 1/(7.4/(2*3.0)) Q = 1/1.23 Q = 0.81

3. To reduce the full width at half maximum (FWHM) of the resonance of the circuit by a factor of 2, a suitable way is to add a resistor in series with the circuit. The additional resistor should be chosen such that it is equal to the total resistance of the circuit. In this case, the additional resistor should be 7.4Ω.