#### Page No 266:

#### 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:

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:

**(b) **The net
power consumed over a full cycle is given as:

*P* = *VI*

= 220 × 2.2 = 484 W

#### Question 7.2:

**(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, *V*_{0} = 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 7.3:

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, *ν* = 50 Hz

Angular frequency, *ω*=

Inductive reactance, *X*_{L} = *ω* *L*

Rms value of current is given as:

Hence, the rms value of current in the circuit is 15.92 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.

#### Answer:

Capacitance of
capacitor, *C* = 60 μF
= 60 × 10^{−6} F

Supply voltage, *V *=
110 V

Frequency, *ν* = 60 Hz

Angular frequency, *ω*=

Capacitive reactance

Rms value of current is given as:

Hence, the rms value of current is 2.49 A.

#### 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:

In the inductive circuit,

Rms value of current,* I* = 15.92 A

Rms value of voltage, *V* = 220 V

Hence, the net power absorbed can be obtained by the relation,

*P* = *VI* cos *Φ*

Where,

Φ = Phase
difference between *V* and *I*

For a pure inductive
circuit, the phase difference between alternating voltage and current
is 90° i.e., *Φ*= 90°.

Hence, *P* = 0
i.e., the net power is zero.

In the capacitive circuit,

Rms value of current, *I* = 2.49 A

Rms value of voltage, *V* = 110 V

Hence, the net power absorbed can ve obtained as:

*P* = *VI *Cos *Φ*

For a pure capacitive
circuit, the phase difference between alternating voltage and current
is 90° i.e., *Φ*= 90°.

Hence,* P *= 0
i.e., the net power is zero.

#### Question 7.6:

Obtain the resonant
frequency *ω**r *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:

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 7.7:

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 × 10^{3} rad/s.

#### 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?

#### Answer:

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:

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 the supply power equals the natural frequency of the
given *LCR* circuit.

Resistance, *R* =
20 Ω

Inductance, *L* =
1.5 H

Capacitance, *C* =
35 μF = 30 × 10^{−6 }F

AC supply voltage to
the *LCR* circuit, *V* = 200 V

Impedance of the circuit is given by the relation,

At resonance,

Current in the circuit can be calculated as:

Hence, the average
power transferred to the circuit in one complete cycle= *VI*

= 200 × 10 = 2000 W.

#### 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?

[*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:

The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency,
ν_{1} = 800 kHz =
800 × 10^{3} Hz

Upper tuning frequency,
ν_{2} = 1200 kHz =
1200× 10^{3} Hz

Effective inductance of
circuit *L* = 200 μH =
200 × 10^{−6} H

Capacitance of variable
capacitor for ν_{1 }is
given as_{: }

*C*_{1}

Where,

ω_{1} =
Angular frequency for capacitor *C*_{1}

Capacitance of variable
capacitor for ν_{2, }

*C*_{2}

Where,

ω_{2} =
Angular frequency for capacitor *C*_{2}

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

#### 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.

#### Answer:

Inductance of the inductor, *L *= 5.0 H

Capacitance of the capacitor, *C* = 80 μF = 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,

*V*_{0} = 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,

(*V*_{L})_{rms} = *I* × *ω*_{R}*L*

Where,

*I* = rms current

Potential drop across the capacitor,

Potential drop across the resistor,

(*V*_{R})_{rms} = *IR*

= × 40 = 230 V

Potential drop across the *LC *combination,

At resonance,

∴*V*_{LC}= 0

Hence, it is proved that the potential drop across the *LC *combination is zero at resonating frequency.

#### Page No 267:

#### 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?

#### Answer:

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 10^{3} 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)** *Q*^{1} = 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:

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:

Inductance of the
inductor,* L *= 0.50 H

Resistance of the
resistor, *R* = 100 Ω

Potential of the supply
voltage, *V* = 240 V

Frequency of the
supply, *ν* = 50 Hz

**(a) **Peak voltage
is given as:

Angular frequency of the supply,

ω
= 2 π*ν*

= 2π × 50 = 100 π rad/s

Maximum current in the circuit is given as:

**(b)** Equation for
voltage is given as:

*V* = *V*_{0} cos *ω**t*

Equation for current is given as:

*I* = *I*_{0} cos (*ω**t* − *Φ*)

Where,

Φ = Phase difference between voltage and current

At
time, *t *= 0.

*V* = *V*_{0}(voltage is maximum)

For*ω**t* − *Φ* = 0 i.e., at time,

*I* = *I*_{0} (current is maximum)

Hence, the time lag between maximum voltage and maximum current is.

Now,
phase angle *Φ*is
given by the relation,

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

#### 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?

#### Answer:

Inductance of the
inductor, *L* = 0.5 Hz

Resistance of the
resistor, *R* = 100 Ω

Potential of the supply
voltages, *V* = 240 V

Frequency of the
supply,*ν*** = **10 kHz = 10^{4} Hz

Angular frequency, *ω* = 2π*ν*= 2π × 10^{4} rad/s

**(a) **Peak
voltage,

Maximum current,

**(b) **For phase
difference*Φ*, we have
the relation:

It
can be observed that *I*_{0} is very small in this case.
Hence, at high frequencies, the inductor amounts to an open circuit.

In
a dc circuit, after a steady state is achieved, *ω* = 0. Hence, inductor L behaves like a pure conducting object.

#### 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?

#### Answer:

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, *V*_{0} =

Maximum current is given as:

**(b) **In a capacitor circuit, the voltage lags behind the
current by a phase angle of*Φ*.
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.

#### 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 × 10^{3} Hz

Angular Frequency, *ω* = 2 π*ν*= 2 × π × 12
× 10^{3}03

= 24π ×
10^{3} 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.

#### 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:

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:

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:

Inductance, *L* =
80 mH = 80 × 10^{−3} H

Capacitance, *C *=
60 μF = 60 × 10^{−6 }F

Supply voltage, *V* = 230 V

Frequency, *ν* = 50 Hz

Angular frequency, *ω* = 2π*ν*= 100 π rad/s

Peak voltage, *V*_{0}^{ }=

**(a)** Maximum current is given as:

The negative sign appears because

Amplitude of maximum current,

Hence, rms value of current,

**(b)** Potential difference across the inductor,

*V*_{L}=* I *× ω*L*

= 8.22 × 100 π ×
80 × 10^{−3}

= 206.61 V

Potential difference across the capacitor,

**(c)** Average power consumed by the inductor is zero as actual
voltage leads the current by.

**(d)** Average power consumed by the capacitor is zero as voltage
lags current by.

**(e)** The total power absorbed (averaged over one cycle) is
zero.

#### 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:

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 *μ*F
= 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* = 2π*ν*= 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:

*P*_{R}= *I*^{2}*R*

= (7.25)^{2} × 15 = 788.44 W

Average power
transferred to capacitor, *P*_{C} = Average power
transferred to inductor, *P*_{L} = 0

Total power absorbed by the circuit:

= *P*_{R }*+
P*_{C}* + P*_{L}

= 788.44 + 0 + 0 = 788.44 W

Hence, the total power absorbed by the circuit is 788.44 W.

#### Page No 268:

#### Question 7.20:

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:

Inductance, *L* =
0.12 H

Capacitance, *C* =
480 nF = 480 × 10^{−9} F

Resistance, *R* =
23 Ω

Supply voltage, *V* = 230 V

Peak voltage is given as:

*V*_{0} = =
325.22 V

**(a)** Current flowing in the circuit is given by the relation,

Where,

*I*_{0} = maximum at resonance

At resonance, we have

Where,

ω_{R }*= *Resonance angular frequency

∴Resonant frequency,

And, maximum current

**(b)** Maximum
average power absorbed by the circuit is given as:

Hence, resonant frequency () is

**(c)** The power
transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, =

Where,

Hence, change in frequency,

∴

And,

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

**(d)** *Q*-factor of the given circuit can be obtained using
the relation,

Hence, the Q-factor of the given circuit is 21.74.

#### 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:

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:

*Q*-factor of the
series:

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 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) ** 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
statement is not true for rms voltage

It is true that in any ac circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for rms voltage because voltages across different elements may not be in phase.

**(b)** High induced
voltage is used to charge the capacitor.

A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.

**(c)** The dc signal will appear across capacitor *C* because for dc signals, the impedance of an inductor (*L*) is
negligible while the impedance of a capacitor (*C*) is very high
(almost infinite). Hence, a dc signal appears across *C*. For an
ac signal of high frequency, the impedance of *L* is high and
that of *C* is very low. Hence, an ac signal of high frequency
appears across *L*.

**(d)** If an iron core is inserted in the choke coil (which is in
series with a lamp connected to the ac line), then the lamp will glow
dimly. This is because the choke coil and the iron core increase the
impedance of the circuit.

**(e)** A choke coil is needed in the use of fluorescent tubes
with ac mains because it reduces the voltage across the tube without
wasting much power. An ordinary resistor cannot be used instead of a
choke coil for this purpose because it wastes power in the form of
heat.

#### Question 7.23:

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:

Input voltage, *V*_{1} = 2300

Number of turns in
primary coil, *n*_{1} = 4000

Output voltage, *V*_{2} = 230 V

Number of turns in
secondary coil = *n*_{2}

Voltage is related to the number of turns as:

Hence, there are 400 turns in the second winding.

#### Question 7.24:

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

#### Answer:

Height of water
pressure head, *h* = 300 m

Volume of water flow
per second, *V* = 100 m^{3}/s

Efficiency of turbine
generator, *n* = 60% = 0.6

Acceleration due to
gravity, g = 9.8 m/s^{2}

Density of water, *ρ* = 10^{3} kg/m^{3}

Electric power
available from the plant = *η* × *hρ*g*V*

= 0.6 × 300 × 10^{3} × 9.8 × 100

= 176.4 × 10^{6} W

= 176.4 MW

#### 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 wire line 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:

Total electric power
required, *P* = 800 kW = 800 × 10^{3} W

Supply voltage, *V* = 220 V

Voltage at which
electric plant is generating power, *V*‘ = 440 V

Distance between the
town and power generating station, *d* = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the
wires, *R* = (15 + 15)0.5 = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station.

Input voltage, *V*_{1} = 4000 V

Output voltage, *V*_{2} = 220 V

Rms current in the wire lines is given as:

**(a)** Line power loss = *I*^{2}*R*

= (200)^{2} × 15

= 600 × 10^{3} W

= 600 kW

**(b)** Assuming that the power loss is negligible due to the
leakage of the current:

Total power supplied by the plant = 800 kW + 600 kW

= 1400 kW

**(c)** Voltage drop in the power line = *IR* = 200 ×
15 = 3000 V

Hence, total voltage transmitted from the plant = 3000 + 4000

= 7000 V

Also, the power generated is 440 V.

Hence, the rating of the step-up transformer situated at the power 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:

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, *V*_{1} = 40000 V

Output voltage, *V*_{2} = 220 V

Total electric power
required, *P* = 800 kW = 800 × 10^{3} W

Source potential, *V* = 220 V

Voltage at which the
electric plant generates power, *V*‘ = 440 V

Distance between the
town and power generating station, *d* = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the
wire lines, *R* = (15 + 15)0.5 = 15 Ω

*P* = V_{1}*I*

Rms current in the wire line is given as:

**(a)** Line power loss = *I*^{2}*R*

= (20)^{2} × 15

= 6 kW

**(b)** Assuming that the power loss is negligible due to the
leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

**(c)** Voltage drop in the power line = *IR *= 20 × 15
= 300 V

Hence, voltage that is transmitted by the power plant

= 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is

440 V − 40300 V.

Hence, power loss during transmission =

In the previous exercise, the power loss due to the same reason is. Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

**NCERT Solutions for Class 12 Physics Chapters**

**Physics Part I**

- Chapter 1 – Electric Charges And Fields
- Chapter 2 – Electrostatic Potential And Capacitance
- Chapter 3 – Current Electricity
- Chapter 4 – Moving Charges And Magnetism
- Chapter 5 – Magnetism And Matter
- Chapter 6 – Electromagnetic Induction
- Chapter 7 – Alternating Current
- Chapter 8 – Electromagnetic Waves

**Physics Part II**

- Chapter 1 – Ray Optics And Optical Instruments
- Chapter 2 – Wave Optics
- Chapter 3 – Dual Nature Of Radiation And Matter
- Chapter 4 – Atoms
- Chapter 5 – Nuclei
- Chapter 6 – Semiconductor Electronics: Materials Devices and Simple Circuits
- Chapter 7 – Communication Systems

**NCERT Solutions for Class 12-science:**