## Tuesday, 21 January 2020 ## DC Circuits Problems and Solutions

Problem 1:

Find the resistance of a copper wire of 0.75 km long and having a cross sectional area of 0.01 cm2. (Take ρ = 1.72 x 10-8 ohm-m).
Given data
Length of wire, l = 0.75 km = 0.75 x 103 m
Cross sectional area, a = 0.01 cm2 = 0.01 x 10-4 m2
Specific resistance, ρ = 1.72 x 10-8 ohm-m
Solution:
Resistance, R = ρl/a = 1.72 x 10-8 x 0.75 x 103/0.01 x 10-4
= 2.9 Ω

Problem 2:

Find the cross sectional area of a aluminum wire of 700m long and having a resistance of 0.24 ohm (Take ρ = 2.83 x 10-8 ohm-m)
Given data
Length of wire, l = 700 m
Resistance, R = 0.24 ohm
Specific resistance, ρ = 2.83 x 10-8 ohm-m
Solution:
Resistance, R = ρl/a
Cross sectional area, a = ρl/R = 2.83 x 10-8 x 700/0.24 = 1.81 x 10-5/0.24
a = 8.254 x 10-5 m2

Problem 3:

Find the resistance of a copper wire 1 km long and 0.5 cm diameter given the specific resistance of copper as 1.7 x 10-8 Ω-m

Given data
Length of wire in meter, l = 1 km = 1000 m
Specific resistance, ρ = 1.7 x 10-8 Ω-m

Solution:
Resistance, R = ρl/a = 1.7 x 10-8 x 1000/ 0.785 x 2.5 x 10-5
R = 0.866 ohm
Problem 4:
An electric iron is rated 1000 W and 240 V. Find the current drawn and resistance of the heating element.
Given data,
Power, P = V2/R
Resistance, R = V2/P = (240)2/1000 = 57.6 Ohms
Current, I = V/R = 240/57.6 = 4.17 amps

Problem 5:

Calculate the current and resistance of a 50 W, 100 V electric bulb.

Given Data
Power, P = 50 watts
Voltage, V = 100 volts

Solution:
Power, P = VI watts
Current, I = P/V = 50/100 = 0.5 amps
Resistance, R = V/I = 100/0.5 = 200 Ohms

Problem 6:

Calculate the power rating of a heater coil when used on 200V supply taking 4 amps

Solution:
Power, P = VI watts = 200 x 4 = 800 watts

Problem 7:

A filament lamp connected to a 230V DC supply draws 300 mA. Find the power absorbed by the lamp.

Solution:
Power P = VI watts = 230 x 300 x 10-3 = 69 watts

Problem 8:

A lamp having a resistance of 500 Ω works on 220 supply system, Determine the energy consumed by operating 30 days at the rate of 4 hours a day.

Solution:
Energy = Power x Energy watt-sec
Power, P = V2/R = 2202/500 = 96.8 watts
Total hours = 30 x 4 = 120
Energy = 96.8 x 120 = 11616 hr = 11.616 Kwhr

Problem 9:

A building has following loads : twenty 100 watt lamps operated 4 hours daily and thirty 60 watt lamps operated 3 hours daily, all connected to a 230 volt source. Calculate (a) the total power (b) the total current (c) the monthly consumption electrical energy (d) the monthly electrical energy charges at 35 paise per unit.

Solution:
(a) Total power, P = 20 x 100 + 30 x 60 = 3800 watts = 3.8 K
(b) Total current, I = P/V = 3800/230 = 16.5 amps
(c) Energy consumption daily = 20 x 100 x 4 + 30 x 60 x 3 = 13,400 watt-hours
Monthly consumption = 13400 x 30 Whr = 13.4 x 30 Kwh
(d) The monthly electrical energy charges at 35 paise per unit = 13.4 x 30 x 0.35 = Rs 140.70

Problem 10:

In the circuit shown in fig., Find the current , the voltage across each resistor and the power dissipated in each resistor.

Solution:
Total resistance, R = R1 + R2 + R3 = 6 + 9 + 10 = 25 Ω
I = V/R = 50/25 = 2 amps
Voltage drop across 6 Ω resistor, V1 = IR1 = 2 x 6 = 12 volts
Voltage drop across 9 Ω resistor, V2 = IR2 = 2 x 9 = 18 volts
Voltage drop across 10 Ω resistor, V3 = IR3 = 2 x 10 = 20 volts
Power dissipated in 6 Ω resistor, P1 = I2R1 = 4 x 6 = 24 watts
Power dissipated in 9 Ω resistor, P2 = I2R2 = 4 x 9 = 36 watts
Power dissipated in 10 Ω resistor, P3 = I2R3 = 4 x 10 = 40 watts

Problem 11:

Two resistors resistance 6 Ω and 12 Ω are connected in series. Find the equivalent Also find the equivalent resistance when they are connected in parallel.

Given data

Case 1: Series combination of 6 Ω and 12 Ω resistors.
R1 = 6 Ω, R2 = 12 Ω
Solution:
Equivalent resistance, R = R1 + R2 = 18 Ω

Case 2: Parallel combination of 6 Ω and 12 Ω resistors.
R1 = 6 Ω, R2 = 12 Ω

Equivalent resistance, 1/R = 1/ R1 + 1/R2
R = R1R2/(R1 + R2) = 6 x 12/(6 + 12) = 72/18 = 4 Ω

Problem 12:

In the circuit shown in figure , find the total resistance and current flowing through each branch.

Given data, R1 = 5 Ω, R2 = 10 Ω, R3 = 20 Ω

Solution:
Equivalent resistance, 1/R = 1/ R1 + 1/R2 + 1/R3 = 0.2 + 0.1 + 0.05 = 0.35
Total resistance, R = 2.86 Ω
The current flowing through 5 Ω resistor, I1 = V/R1 = 25/5 = 5 amps.
The current flowing through 10 Ω resistor, I2 = V/R2 = 25/10 = 2.5 amps
The current flowing through 20 Ω resistor, I3 = V/R3 = 25/20 = 1.25 amps

Problem 13:

Resistors of values 2, 3, 4, and 5 ohm are connected in parallel. If the total power absorbed by all the resistors is 200w, find the voltage applied to the circuit.
Power, P = 200 watts
R1 = 2 Ω, R2 = 3 Ω, R3 = 4 Ω, R4 = 5 Ω

Solution:
Total Resistance, 1/R = 1/ R1 + 1/R2 + 1/R3 + 1/R4 = 0.5 + 0.33 +0.25 + 0.2
1/R = 1.28,  R = 1/1.28 = 0.78 ohm
Power, P = V2/R
V2 = PX R
V = (P X R) = (200 X 0.78) = 12.5 volts

Problem 14:

Two resistors are connected in parallel across a 200V mains take a total current of 10amps. The power dissipated in one of the resistors is 1200. watts. Find the value of each resistor.

Given data:
Voltage, V = 200 volts
Current, I = 10 amps
Power dissipated in R1 of the resistors P1 = 1200 watts

Solution:
I1 = P1/V = 1200/ 200 = 6 amps.
Resistors, R1 = V/ I1 = 200/6 = 33.33 Ω
Total current, I = I1 + I2
10 = 6 + I2
I2 = 4 amps
Resistors, R2 = V/ I2 = 200/4 = 50 Ω ## Introduction to DC Circuits

Introduction to DC Circuits

In the study of electric circuits we are interesting in the flow of electricity from one device to another. A simplest of electric devices will have a pair of terminals. Electricity enters at one terminals and leaves from another.

The energy associated with an electric system is measured by quantities like electric charge, potential and current. Historically electric charges were first observed by rubbing certain dry substances together. Atomic physicists now picture the electric charges as one of the building blocks of the universe. The negatively charged electron is one of the constituents of the atom Its mass is 9.107 ×10-28 gm. As the electron is very small unit, the practical unit of charge is the coulomb.

1 coulomb = 9.107 ×10-28 × Qe

Where Qe is the electronic charge

BASIC CONCEPT

All matter consists of minute particles called molecules. Molecules are made up of more minute particles called atoms. Inside the atom there a central nucleus and a number of electrons around the nucleus. The nucleus contains protons and neutrons. Protons are positively charged particles and neutrons do not carry any charge. The electron carries a negative charge.

In a normal atom, the positive charge on the nucleus is exactly equal to the total negative charge due to all the electrons surrounding the nucleus.

Every atom in which the positive and negative charges are equal has no net positive and negative charge.

Electrons revolve around the nucleus in elliptical orbits. The first orbits has a maximum of 2 electrons, second 8 electrons , third 18 electrons and so on. Number of electrons present in the orbit is 2n2 , where ‘n’ is the orbit number.

IMPORTANT DEFINITIONS

1. Current

Flow of electrons in any conductor is called current. It is represented by the letter ‘I’
The unit of the current is ampere
Current , I = Charge/ Time
I = Q/T
Where ‘Q’ is in coulomb and ‘t’ is in second.

2. One Ampere

One ampere can be defined as one coulomb of charge is transferred in one second.

3. Voltage

The difference of potentials between two points is called voltage or potential difference.
The voltage represented by the letter ‘V’
The unit of the voltage is volt.

4. One volt

One volt is defined as the potential difference across a resistance of one ohm carrying a current of one ampere

5. Resistance

The opposition offered by a substance to the flow of current is called resistance.
The resistance represented by the letter ‘R’
The unit of the resistance is Ohm (Ω)

Laws of Resistance

The resistance (R) of a conductor
is directly proportional to its length (l)
is inversely proportional to its area of cross section (a)
depends up on the nature of material
depends on it temperature
R α l/a
R = ρl/a
Where ρ (Rho) is a constant for the material called resistivity (or) specific resistance

Conductance , G

Conductance is the reciprocal of resistance.
The conductance represented by the letter ‘G’
The unit of the conductance is mho
G = 1/R = a/ ρl = σa/l
Where σ = 1/ρ
where σ (sigma) is called the conductivity.
Conductivity
The reciprocal of resistivity is called conductivity.

6. Ohm’s Law

The current flowing in an electric circuit depends on two quantities, the applied emf and the resistance in the circuit. The connecting the three factors was established first by G.S. Ohm in 1826 and is popularly known as the Ohm’s law.
Statement of Ohm’s Law
At constant temperature, the current flowing through the conductor is directly proportional to the potential difference (voltage) between the two ends of the conductor

i.e., V α I
V = RI
V = IR
Where ‘R’ is the resistance of the conductor in ohms

Explanation:

Consider a simple closed circuit as shown in figure.
Let ‘V’ be the supply voltage (dc)
‘I’ be the current flowing through the circuit (amp)
‘R’ be the resistance of the circuit (ohm)
According to Ohm’s law, V α I
V = RI
V/I = R (constant)
The unit of resistance is Ohm

7. Power

Power is the rate of doing work.
Power = Work done/Time
The power represented by the letter ‘P’
The unit of the power is watt or joules / second
Power is the product of voltage and current
Power, P = voltage x current
P = VI watts (or)
P = (IR) I
P = I2R watts (or)
P = V (V/ R)
P = V2/R watts
The unit of power can also be expressed in kilowatts (Kw) , 1 Kw = 1000 watts

8. Energy:

Energy is the capacity to do work. Energy is the product of the power ant time
Energy = Power x time
= P x t
= VI x t (or)
Energy = I2R x t (or)
Energy = V2 x t/ R
The unit of energy is watt-sec. Watt-sec. is the smallest unit of energy. So kilo watt hour (Khr) is generally used. ## Characteristics of DC Series Motor

1. Electrical Characteristics :

In a d.c. series motor, the armature current and series field current are same and therefore Φ Ia. The armature torque is equal to

Ta Φ Ia or
Ta Ia2

This is true till the point of magnetic saturation. When Ia is zero, torque is also zero and when I is small torque is also small. Since is proportional to square of the armature current the curve is parabolic. After saturation Φ is almost constant and for any increase in armature current torque increases linearly i.e., Ta Ia. The shaft torque is less than armature torque by rotational losses, The curve is shown in Fig.

2. Speed and Armature Current Characteristics :

For a d.c. series motor

Eb = V - Ia (Ra +Rse) and Eb = ΦN x ZP/60A
or Eb = KaΦN
Where Ka = ZP/60A

Substituting the values,

KaΦN = V - Ia (Ra +Rse)
or N = V - Ia (Ra +Rse)/ KaΦ
or N = V/ KaΦ - Ia (Ra +Rse)/ KaΦ

But for a dc series motor, Φ Ia, therefore

N = V/ KbIa(Ra +Rse)/ Kb ; where Kb = Ia/Ka

From the above equation, it is seen that neglecting armature reaction and with saturation the speed-current characteristics of d.c. series motor is hyperbolic as shown in Fig. When la increases Φ also increase but due to demagnetisation effect of armature reaction and saturation, the air gap flux tends to remain constant and for constant flux Φ the term V/KaΦ remains constant and the term la [Ra + Rse]/ KaΦ increases with armature current linearly.
Thus for larger values of armature current, the curve takes a straight line path. At no load the armature current is small and so the armature drop Ia (Ra + Rse) and can be considered negligible as compared to terminal voltage.

Then N = V/KaΦ, But Φ Ia,therefore N = V/Kbla

On no load when the armature current tends to zero, the speed tends to infinity. Therefore, the no load speed of d.c. series motor is dangerously high and due to this reason the d.c. series motor must not be started without a load. The curve is shown in Fig.

3. Speed-Torque Characteristics :

The torque of d.c. series motor is proportional to the square of armature current i.e.

Ia2 Ta or
Ia √Ta

Neglecting the armature reaction and saturation

N = V/KbIa  or  N = V/Kb√Ta

Squaring both sides

N2 =  V2/KcTa
or Ta  = V2/N2Kc
or Ta ∝ 1/N2

The speed torque characteristics is also hyperbolic in nature. But with saturation and armature reaction large torque requires larger currents and these large currents tends to make the air gap flux constant and the effect is Ta Ia instead of Ta Ia2. The curves approaches straight line as shown in Fig.

## Sunday, 19 January 2020 ## Characteristics of DC Shunt Motor

CHARACTERISTICS OF DC MOTORS:

The characteristics curves of d.c. motor are curves drawn to show the relation between armature current, speed and torque. The following are the general characteristics of interest in a d.c. motor.

a. Electrical characteristics : This shows a relation between torque developed and armature current in a d.c. motor.
b. Speed and armature current characteristics : This shows the relation as to how the speed varies with armature current.
c. Speed-Torque characteristics : This is also mechanical characteristics. It gives the relation between speed and torque in a d.c. motor.

Characteristics of D.C. Shunt Motor

1. Electrical Characteristics :

In a d.c. shunt motor, shunt field flux is dependent on terminal voltage V and the shunt field resistance. Assuming that the terminal voltage is constant, the shunt field flux is also constant i.e., Φ is constant, then

Torque developed is given by = EbIa/ω
Torque developed in armature = 0.159 [ΦZla] P/A = IaK1

Where K1 = 0.159 x ZΦP/A is a constant.

The torque developed in armature (Ta) Ia.
It is a linear relation and practically a straight line front the origin. Since shaft torque is equal to the armature torque minus the rotational losses, the shaft torque line also follows the armature torque line as shown in dotted in Fig.

2. Speed and Armature Current Characteristics:

For a dc shunt motor assuming that flux remaining constant

Eb = V – IaRa = ZΦNP/60A
Or V – IaRa = N x ZΦP/60A = K2N
Where, K2 = ZΦP/60A
Or V – IaRa = K2N

The terminal voltage V and armature resistance are constant. If Ia = 0, then N is maximum and as Ia increases N decreases. As la increases, load on the shaft also increases. Invariably the speed of the shunt motor decreases with the increase of load as shown in Fig.

3. Speed-Torque Characteristics :

In a d.c. shunt motor Ta Ia and V - laRa = K2N

From these two equations, it is evident that as torque increases the armature current also increases proportionately and the net value of the back e.m.f. decreases. Decrease in back will result in decrease in speed. Hence, when speed is zero, back e.m.f. is zero, armature current is also zero and torque is maximum. The curve is shown in Fig. As speed increases armature current decreases to minimum but does not become zero and torque decrease. Therefore the characteristics line does not follow the speed axis.

## Wednesday, 8 January 2020 ## Voltage and Speed Relation in DC Motor

VOLTAGE EQUATION OF DC MOTORS

1. DC Shunt Motor:

In a dc shunt motor, the field winding is connected across the armature and the supply is given to the armature. As such the current for the field winding is taken from the supply. The armature current is therefore equal to the difference of line current and the shunt field current. Therefore,

Ia – IL - Ish
Ish = V/Rsh
Eb = V – IaRa

Where,
Ish is the shunt field current
Rsh is the shunt field resistance.

2. Back e.m.f. in DC Series Motor:
In a D.C. series motor, the applied voltage V has to supply the ohmic drop in armature (IaRa) and also series field voltage drop (IseRse) ie.,

V = Eb + IaRa + IseRse

Since the field winding is in series with the armature, the current in the armature and series field are same. Therefore,
IL = Ia = Ise

Voltage equation can be rewritten as under,
V = Eb + Ia(Ra + Rse)

Where,
V is the applied voltage in volts
Ia(Ra + Rse) is the voltage drop in the armature and series field winding.

Multiplying both sides by Ia

VIa = IaEb + Ia2Ra + Ia2Rse
is the power equation

Where

VIa is the power input in watts
IaEb is the electrical equivalent of mechanical power developed
Ia2Ra is the copper loss in the armature
Ia2Rse is the copper loss in the series field winding

3. Back emf in DC Compound motor:

In DC Compound motor there are two possible connections. One when the shunt winding is connected parallel to the armature, it is called short shunt compound motor and when the shunt winding is connected parallel to the supply, it is called long shunt compound motor. The two types of connections are shown in figure.

In the long shunt compound motor, the voltage equation of the armature circuit is given by,

V = Eb + IaRa + IseRse

But, Ia = Ise

Therefore, V = Eb + Ia(Ra + Rse)
And the current equation is IL = Ish + Ia

Multiplying the voltage equation both sides by Ia
IaV = IaEb + Ia2Ra + Ia2Rse

Where IaV is the power input to the armature circuit. The total power taken by the motor is equal to the sum of the power input to armature circuit and shunt field circuit. That is,

ILV = IaV + IshV

The electrical power equivalent to mechanical is IaEb.

In the short shunt compound generator, the voltage equation is given by,

V = Eb + IaRa + IseRse
And, IL = Ise = Ia + Ish
Where, IL = Line current in amps

Ia = Armature current in amps
Ish = Shunt field current in amps
= Vsh/Rsh = Eb – IaRa/Rsh

The electrical power develop equivalent to mechanical is IaEb. In all the cases the back emf is also equal to,

Eb = ZΦNP/60A

SPEED EQUATION OF DC MOTOR:

The speed equation of dc motors can be deduced from the voltage equation itself. Thus we can find the voltage and speed relation in dc motor. The back emf equation for a motor is given by,

Eb = V – IaRa

But, Eb = ZΦNP/60A
or Eb = K1ΦN where, K1 = ZP/60A
or Eb ΦN or N ∝ Eb/Φ

That is, N (V – IaRa)/Φ

From the above it is seen that for the speed to change either

a. the field flux has to be changed or
b. the armature current has to be changed. Change of armature current means change of load conditions.

For a dc series motor,
N (V – Ia(Ra + Rse))/ Φ

It should be noted that for the field flux to change, a resistance called the diverter is added in parallel to the series field and for the armature current to change a resistance is connected in series to the series field.

## Saturday, 4 January 2020 ## Torque in DC Motor Formula

Torque in DC Motor with Formula

Turning or twisting moment of a force about the axis is called torque in DC Motor. The unit of torque is Newton-metre. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius R metres acted upon by a force on the circumference with F Newtons to cause the pulley to rotate at a speed of N rev. per second as shown in Figure.

Then as per the definition torque = F x r Newton-metre
Work done by this force in one revolution (Force x distance) = F x 2πr Joules
If N is the speed in r.p.s. then Work done/second = F x 2πr x N = (F.r) x 2πN
2πN is called the angular velocity and is represented by ‘ω'. It is measured in radians/second.

Power developed (Pm) = Two Joules/second or watts

The torque developed at the shaft in DC Motor is always lesser than the torque developed at the armature because of losses.

The formula of torque developed by the armature = Ta.2πN watts
The mechanical power developed in the armature =, Eb la watts

But the mechanical power developed in the armature is equal to the armature torque i.e.,

Ebla = Ta.2πN/60 where N is in r.p.m. and

Nrpm/60 = Nrps

But Eb = ZΦNP/60A

Substituting for Eb

ZΦNP/60A x la = Ta.2πN/60 or
Ta = ZΦ la/2π x P/A
= 1/2π Φ Zla x P/A = 0.159 Φ Zla [P/A] N-m
= 0.159/9.81 x Φ Zla x P/A kg-m
= 0.0162 Φ Zla x [P/A] kg-m

This is also known as the gross torque. The torque developed at the shaft is the useful torque and is known as shaft torque (Tsh).

BHP (metric) = Tshaft x 2πN/735.5 where N is in r.p.s.

Or Tsh = (BHP)m x 735.5/2πN

The motor output is given by Tsh x 2πN watt

Tsh = output in watts/2πN  , where N is in r.p.s.
= 60/2π x Output/N = 9.55 x Output/N  N-m
Where N is in r.p.m

The difference between armature torque and shaft torque is the torque lost.

Torque lost = 0.159 x iron and frictional loss/N  N-m

Case 1 Shunt Motor:

The flux is constant in shunt motor and is dependent on the terminal voltage and shunt field resistance. Therefore,
Ta Ia

Case 2 Series Motor:

In the case of a series motor the armature and series field currents are same i.e., Ia = Ise = IL, therefore
Φ Ia and Ta Ia2
and also the torque in dc motor formula,

Ta = EbIa/2πN = 0.159 EbIa/N  N-m
= 0.0162 EbIa/N kg-m
= 0.0162/N x Power developed in armature Kg-m

Difference between Developed Torque, Useful Torque and Lost Torque

The whole of the armature torque is not available for doing any useful work because a certain percentage of it is required for supplying iron and frictional losses in the motor. Therefore, the net torque available is the shaft torque. The difference between the armature torque and shaft torque is the torque lost. The torque developed in the armature is called gross torque and the torque which is available at the shaft which is much less than the torque developed in the armature is called shaft torque or useful torque.

## Wednesday, 1 January 2020 ## Significance of Back EMF in DC Motor

Significance of Back EMF in a DC Motor

As soon as the armature starts rotating inside a magnetic field due to the current in the conductor, the following conditions are set in

a. The conductor is in motion
b. The flux of the main pole exists
c. While the conductor is in motion, it also cuts the main pole flux

This condition creates a situation for inducing e.m.f. in the conductor. The direction of induced e.m.f. as found by the Fleming's Right hand rule, is in direct opposition to the applied voltage across the armature conductor. It is therefore termed as back e.m.f. (Eb). The magnitude of back e.m.f. (back emf in dc motor formula) is given by :

Eb = ZΦNP / 60A

Where,
Z is the number of conductors in the armature
Φ is the flux of the main pole in Webers
N is the speed of rotation of the shaft in r.p.m
P is the number of poles of the machine
A is the number of parallel path of the winding
A = P for Lap winding
A = 2 for Wave winding

This back e.m.f. acts in opposite direction to the applied voltage in the armature as shown in Figure. Hence, the net voltage across the armature would be the difference between the applied voltage and back e.m.f.

i.e., Net voltage across armature = (V — Eb) Volts

Where,
V is the applied voltage across the armature
Eb is the back e.m.f.
If the resistance of the armature is Ra, then the armature current is given by
Armature current = Net voltage across armature /Armature resistance

Ia =(V – Eb)/ Ra amperes

It should be remembered that the magnitude of back e.m.f. is dependent upon the speed of rotation of the shaft (N). If the speed is zero, the back e.m.f. is also zero and as the speed increases the magnitude of back e.m.f, also increases. At stand still or starting position, the shaft is stationary and therefore the back e.m.f. is zero. Rewriting the equation for applied voltage

laRa = V—Eb or

V= Eb +IaRa

The voltage V applied across the armature has to

a. overcome the back e.m.f. Eb and
b. supply the armature ohmic drop laRa

This is known as the voltage equation of d.c. motor. Multiplying the equation on both sides by laRa, power equation is obtained.

Power = VIa = EbIa + la2Ra

Where,
VIa is the electrical input to the armature
Ebla is the electrical equivalent of mechanical power developed in the armature
Ia2Ra is the copper loss in the armature

Hence, it would be seen that some power is wasted in the armature out of the input and the rest only is converted into mechanical power within the armature. The gross mechanical power developed by the motor is given by

Pm = VIa— Ia2Ra

Differentiating both sides with respect to la and equating the result to zero,

dPm/dIa = V — 2IaRa =0

Therefore

IaRa = V/2

Thus, the gross mechanical power developed by the d.c. motor will be maximum when the back e.m.f. is equal to the half of the applied voltage. If this condition is taken into account, then half the input would be wasted in the form of heat and taking into other losses like frictional and magnetic losses, the motor efficiency would fall to 50%. As such this condition is not realised in practice.