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 × Q_e

Where Q_e 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 ∝ I_a. The armature torque is equal to

T_a ∝ Φ I_a or

T_a ∝ I_a²

This is true till the point of magnetic saturation. When I_a 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., T_a ∝ I_a. 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

E_b = V - I_a (R_a +R_se) and E_b = ΦN x ZP/60A

or E_b = K_aΦN

Where K_a = ZP/60A

Substituting the values,

K_aΦN = V - I_a (R_a +R_se)

or N = V - I_a (R_a +R_se)/ K_aΦ

or N = V/ K_aΦ - I_a (R_a +R_se)/ K_aΦ

But for a dc series motor, Φ ∝ I_a, therefore

N = V/ K_bI_a – (R_a +R_se)/ K_b ; where K_b = I_a/K_a

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 l_a 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/K_aΦ remains constant and the term l_a [R_a + R_se]/ K_aΦ 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 I_a (R_a + R_se) and can be considered negligible as compared to terminal voltage.

Then N = V/K_aΦ, But Φ ∝ I_a,therefore N = V/K_bl_a

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.

I_a² ∝ T_a or

I_a ∝ √T_a

Neglecting the armature reaction and saturation

N = V/K_bI_a  or  N = V/K_b√T_a

Squaring both sides

N² =  V²/K_cT_a 

or T_a  = V²/N²K_c 

or T_a ∝ 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 T_a ∝ I_a instead of T_a ∝ I_a². The curves approaches straight line as shown in Fig.

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 = E_bI_a/ω

Torque developed in armature = 0.159 [ΦZl_a] P/A = I_aK₁

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

The torque developed in armature (T_a) ∝ I_a.

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

E_b = V – I_aR_a = ZΦNP/60A

Or V – I_aR_a = N x ZΦP/60A = K₂N

Where, K₂ = ZΦP/60A

Or V – I_aR_a = K₂N

The terminal voltage V and armature resistance are constant. If I_a = 0, then N is maximum and as I_a 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 T_a ∝ I_a and V - l_aR_a = 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.

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,

I_a – I_L - I_sh

I_sh = V/R_sh

E_b = V – I_aR_a

Where,

I_sh is the shunt field current

R_sh 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 (I_aR_a) and also series field voltage drop (I_seR_se) ie.,

V = E_b + I_aR_a + I_seR_se

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

I_L = I_a = I_se

Voltage equation can be rewritten as under,

V = E_b + I_a(R_a + R_se)

Where,

V is the applied voltage in volts

I_a(R_a + R_se) is the voltage drop in the armature and series field winding.

Multiplying both sides by I_a

VI_a = I_aE_b + I_a²R_a + I_a²R_se

is the power equation

Where

VI_a is the power input in watts

I_aE_b is the electrical equivalent of mechanical power developed

I_a²R_a is the copper loss in the armature

I_a²R_se 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 = E_b + I_aR_a + I_seR_se

But, I_a = I_se

Therefore, V = E_b + I_a(R_a + R_se)

And the current equation is I_L = I_sh + I_a

Multiplying the voltage equation both sides by I_a

I_aV = I_aE_b + I_a²R_a + I_a²R_se

Where I_aV 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,

I_LV = I_aV + I_shV

The electrical power equivalent to mechanical is I_aE_b.

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

V = E_b + I_aR_a + I_seR_se

And, I_L = I_se = I_a + I_sh

Where, I_L = Line current in amps

I_a = Armature current in amps

I_sh = Shunt field current in amps

= V_sh/R_sh = E_b – I_aR_a/R_sh

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

E_b = 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.

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 (P_m) = 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 = T_a.2πN watts

The mechanical power developed in the armature =, E_b l_a watts

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

E_bl_a = T_a.2πN/60 where N is in r.p.m. and

N_rpm/60 = N_rps

But E_b = ZΦNP/60A

Substituting for E_b

ZΦNP/60A x l_a = T_a.2πN/60 or

T_a = ZΦ l_a/2π x P/A

= 1/2π Φ Zl_a x P/A = 0.159 Φ Zl_a [P/A] N-m

= 0.159/9.81 x Φ Zl_a x P/A kg-m

= 0.0162 Φ Zl_a 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 (T_sh).

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

Or T_sh = (BHP)_m x 735.5/2πN

The motor output is given by T_sh x 2πN watt

T_sh = 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,

T_a ∝ I_a

Case 2 Series Motor:

In the case of a series motor the armature and series field currents are same i.e., I_a = I_se = I_L, therefore

Φ ∝ I_a and T_a ∝ I_a²

and also the torque in dc motor formula,

T_a = E_bI_a/2πN = 0.159 E_bI_a/N  N-m

= 0.0162 E_bI_a/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.

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 :

E_b = 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 — E_b) Volts

Where,

V is the applied voltage across the armature

E_b is the back e.m.f.

If the resistance of the armature is R_a, then the armature current is given by

Armature current = Net voltage across armature /Armature resistance

I_a =(V – E_b)/ R_a 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

l_aR_a = V—E_b or

V= E_b +I_aR_a

The voltage V applied across the armature has to

a. overcome the back e.m.f. E_b and

b. supply the armature ohmic drop l_aR_a

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

Power = VI_a = E_bI_a + l_a²R_a

Where,

VI_a is the electrical input to the armature

E_bl_a is the electrical equivalent of mechanical power developed in the armature

I_a²R_a 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

P_m = VI_a— I_a²R_a

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

dP_m/dI_a = V — 2I_aR_a =0

Therefore

I_aR_a = 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.