**Q7.26 **Do the same exercise as above with the replacement of the earlier transformer by a 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?

Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS current in the wireline
Now,
a) power loss in the line
b)
Power supplied by plant = 800 kW + 6 kW = 806kW.
c)
Voltage drop in the power line =
Total voltage transmitted from the plant = 300+40000=40300
as power is generated at 440V, The rating of the power plant is 440V-40300V.
We prefer high...

**Q7.25 (c) **A small town with a demand of of electric power at is situated away from an electric plant generating power at . The resistance of the two wire line carrying power is per km. The town gets power from the line through a step-down transformer at a sub-station in the town. Characterise the step up transformer at the plant.

Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS Current in the wireline
Now,
Voltage drop in the power line =
Total voltage transmitted from the plant = 3000+4000=7000
as power is generated at 440V, The rating of the power plant is 440V-7000V.

**Q7.25 (b) **A small town with a demand of of electric power at is situated away from an electric plant generating power at . The resistance of the two wire line carrying power is per km.The town gets power from the line through a step-down transformer at a sub-station in the town.

How much power must the plant supply, assuming there is negligible power loss due to leakage?

Power required
The total resistance of the two-wire line
Input voltage
Output voltage:
RMS current in the wireline
Now,
Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

**Q7.25 (a) **A small town with a demand of of electric power at is situated away from an electric plant generating power at . The resistance of the two wire line carrying power is per km. The town gets power from the line through a step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.

Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS Current in the wireline
Now,
Power loss in the line
Hence, power loss in line is 600kW.

**Q7.23 ** A power transmission line feeds input power at to a stepdown transformer with its primary windings having turns. What should be the number of turns in the secondary in order to get output power at ?

Given,
Input voltage:
Number of turns in the primary coil
Output voltage:
Now,
Let number of turns in secondary be
Now as we know, in a transformer,
Hence the number of turns in secondary winding id 400.

**Q7.22 (e)** Answer the following questions:

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?

We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.

**Q 7.22 (d) **Answer the following questions:

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.

For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect in the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

**Q 7.22 (c) **Answer the following questions:

An applied voltage signal consists of a superposition of a 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 and the ac signal across .

For a high frequency, the inductive reactance and capacitive reactance:
Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L.
Similarly
For DC, the inductive reactance and capacitive reactance:
Hence DC signal appears across Capacitor only.

**7.22 (b) ** Answer the following questions: A capacitor is used in the primary circuit of an induction coil.

Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

**Q7.22 (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?

Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

**Q7.21 **Obtain the resonant frequency and -factor of a series circuit with , , and . 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.

The inductance of the inductor
The capacitance of the capacitor
The resistance of the resistor
Now,
Resonant frequency
Q-Factor of the circuit
Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing,
we have to change the resistance of the resistor to half of its value, that is

**Q7.20 (d) **A series LCR circuit with , , is connected to a variable frequency supply. 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?

As
Power
Power will be half when the current is times the maximum current.
As,
At half power point :
here,
On putting values, we get, two values of for which
And they are:
Also,
The current amplitude at these frequencies

**Q7.20 (c) **A series LCR circuit with , , is connected to a variable frequency supply. What is the factor of the given circuit?

The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67
Q-factor of any circuit is given by
Hence Q-factor for the circuit is 21.74.

**Q7.20 (b) **A series LCR circuit with , , is connected to a variable frequency supply.

What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,
Hence when source frequency will be equal to the natural frequency, the power absorbed will be...

**Q7.20 (a) **A series LCR circuit with , , is connected to a variable frequency supply.

What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

The inductance of the inductor
The capacitance of the capacitor
The resistance of the resistor
Voltage supply
Frequency of voltage supply
As we know,
the current amplitude is maximum at the natural frequency of oscillation, which is
Also, at this frequency,
SO,
The maximum current in the circuit :
Hence maximum current is 14.14A.

**Q7.19 **Suppose the circuit in Exercise 7.18 has a resistance of . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

The inductance of the inductor
The capacitance of the capacitor
The resistance of a resistor
Voltage supply
Frequency of voltage supply
As we know,
Impedance
Current flowing in the circuit :
Now,
Average power transferred to the resistor:
Average power transferred to the inductor = 0
Average power transferred to the capacitor = 0:
Total power absorbed by circuit :
Hence circuit...

**Q7.18 (e) **A circuit containing a inductor and a capacitor in series is connected to a , supply. The resistance of the circuit is negligible. What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it...

**Q7.18 (d) **A circuit containing a inductor and a capacitor in series is connected to a , supply. The resistance of the circuit is negligible. What is the average power transferred to the capacitor?

As we know,
Average power where is the phase difference between voltage and current.
Since in the circuit, phase difference is , the average power is zero.

**Q7.18 (c) ** A circuit containing a inductor and a capacitor in series is connected to a , supply. The resistance of the circuit is negligible

(c) What is the average power transferred to the inductor?

Since
Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.

**Q7.18 (b) **A circuit containing a inductor and a capacitor in series is connected to a , supply. The resistance of the circuit is negligible. Obtain the rms values of potential drops across each element.

As we know,
RMS potential drop across an element with impedance Z:
SO,
RMS potential difference across inductor:
RMS potential drop across capacitor

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