DC Generator Working MCQ Quiz - Objective Question with Answer for DC Generator Working - Download Free PDF

Explanation:

To find the generated voltage of a long shunt compound generator delivering a load current of 50 A at 500 V, we need to calculate the voltage drops across the different components of the generator and then sum these drops to the terminal voltage.

Given Data:

• Load current, I_L = 50 A
• Terminal voltage, V_L = 500 V
• Armature resistance, R_a = 0.04 Ω
• Series field resistance, R_s = 0.025 Ω
• Shunt field resistance, R_sh = 125 Ω
• Brush voltage drop = 3 V

Step-by-Step Solution:

1. **Calculate the Shunt Field Current (I_sh):**

The shunt field current can be calculated using Ohm's Law:

I_sh = V_L / R_sh

Substituting the given values:

I_sh = 500 V / 125 Ω = 4 A

2. **Calculate the Armature Current (I_a):**

In a long shunt compound generator, the armature current is the sum of the load current and the shunt field current:

I_a = I_L + I_sh

Substituting the given values:

I_a = 50 A + 4 A = 54 A

3. **Calculate the Voltage Drop in the Armature (V_a):**

The voltage drop across the armature resistance can be calculated using Ohm's Law:

V_a = I_a × R_a

Substituting the given values:

V_a = 54 A × 0.04 Ω = 2.16 V

4. **Calculate the Voltage Drop in the Series Field (V_s):**

The voltage drop across the series field resistance can be calculated using Ohm's Law:

V_s = I_a × R_s

Substituting the given values:

V_s = 54 A × 0.025 Ω = 1.35 V

5. **Calculate the Total Voltage Drop (V_drop):**

The total voltage drop is the sum of the voltage drops in the armature, series field, and brushes:

V_drop = V_a + V_s + Brush voltage drop

Substituting the given values:

V_drop = 2.16 V + 1.35 V + 3 V = 6.51 V

6. **Calculate the Generated Voltage (E_g):**

The generated voltage is the sum of the terminal voltage and the total voltage drop:

E_g = V_L + V_drop

Substituting the given values:

E_g = 500 V + 6.51 V = 506.51 V

Conclusion:

The generated voltage of the long shunt compound generator is 506.51 V. Therefore, the correct option is:

Option 1: 506.51 V

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 516.48 V

This option is incorrect because it does not account for the correct calculation of the voltage drops across the armature, series field, and brushes. It likely overestimates the generated voltage.

Option 3: 520.66 V

This option is incorrect as it significantly overestimates the generated voltage. The total voltage drop calculation appears to be much higher than the actual value.

Option 4: 509.36 V

This option is incorrect because it slightly overestimates the generated voltage. The voltage drop calculation may have included additional factors that were not specified in the problem statement.

Conclusion:

Understanding the voltage drops and their contributions to the terminal voltage is crucial in accurately determining the generated voltage of a compound generator. The correct calculation involves summing the terminal voltage and the total voltage drop across the generator's components.

The correct answer is above option 3):(Field resistance is above critical resistance)

Concept:

Conditions for voltage build-up voltage of a self-excited or shunt Generator:

• There must be some residual magnetism in generator poles. Some amount of emf will be generated due to residual magnetism in the main poles. This small emf in turn produces additional flux to reinforce the original residual flux. This process is continued till the generator builds up the rated voltage.
• The connections of the field winding should be such that the field current strengthens the residual magnetism.
• The resistance of the filed circuit should be less than the critical resistance.
• Critical resistance is the total field resistance above which the generator fails to build up its voltage. In other words, the speed of the generator should be higher than the critical speed.
• Critical speed is the speed of the generator below which fails to build up its voltage. Under load conditions, the load resistance must be more than the critical load resistance.

​

According to the question circuit diagram of DC shunt generator can be drawn asNow, shunt field current (I_sh) is

\({I_{sh}} = \frac{V}{{{R_{sh}}}} = \frac{{250}}{{50}}\)

I_sh = 5 A

Now Armature current is

I_a = I_L + I_sh = 450 + 5

I_a = 455 A

Now generated emf in dc shunt generator on be calculated as

E = V + I_a R_a

E = 250 + 455 (0.01) = 250 + 4.55

E = 254.55 V

\(\begin{array}{l} No\;of\;turns = \frac{{No\;of\;conductors}}{2} = \frac{{720}}{2} = 360\\ No\;of\;coils = \frac{{360}}{3} = 120 \end{array}\)

No of coil sides = 2 × no of coils = 2 × 120 = 240

No of slots = 240/4 = 60 slots

For employing equaliser ring equipotential pitch value should be integer i.e. 2C/P should be integer.

C = Number of coils

P = Number of poles

\(\frac{{2C}}{P} = \frac{{2 \times 120}}{6} = 40\)

Hence equaliser rings can be employed.

Concept:

The EMF equation of a DC Machine is

\({E_b} = \frac{{NPϕ Z}}{{60A}}\)

From the above equation,

\(N \propto \frac{{{E_b}}}{ϕ }\)

In a dc generator, induced emf is

Eg = V + IaRa

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Given-

V = 200, R_a = 0.2 Ω, I_a = 20 A, ϕ₂ = 1.1 ϕ₁

Let ϕ₁ = Generator flux, ϕ₂ = Motor flux, N₁ = Generator speed, N₂ = Motor speed 

∴ \({E_g} = 200 + 20 \times 0.2 = 204\ V\)

\(\begin{array}{l} {E_b} = 200 - 20 \times 0.2 = 196\ V\\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_b}}}{{{E_g}}} \times \frac{{{ϕ _1}}}{{{ϕ _2}}} = \frac{{196}}{{204}} \times \frac{{{ϕ _1}}}{{1.1{ϕ _1}}} = 0.87 \end{array}\)

Concept:

The EMF equation of a DC Machine is

\({E_b} = \frac{{NPϕ Z}}{{60A}}\)

From the above equation,

\(N \propto \frac{{{E_b}}}{ϕ }\)

In a dc generator, induced emf is

Eg = V + IaRa

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Given-

V = 200, R_a = 0.2 Ω, I_a = 20 A, ϕ₂ = 1.1 ϕ₁

Let ϕ₁ = Generator flux, ϕ₂ = Motor flux, N₁ = Generator speed, N₂ = Motor speed 

∴ \({E_g} = 200 + 20 \times 0.2 = 204\ V\)

\(\begin{array}{l} {E_b} = 200 - 20 \times 0.2 = 196\ V\\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_b}}}{{{E_g}}} \times \frac{{{ϕ _1}}}{{{ϕ _2}}} = \frac{{196}}{{204}} \times \frac{{{ϕ _1}}}{{1.1{ϕ _1}}} = 0.87 \end{array}\)

According to the question circuit diagram of DC shunt generator can be drawn asNow, shunt field current (I_sh) is

\({I_{sh}} = \frac{V}{{{R_{sh}}}} = \frac{{250}}{{50}}\)

I_sh = 5 A

Now Armature current is

I_a = I_L + I_sh = 450 + 5

I_a = 455 A

Now generated emf in dc shunt generator on be calculated as

E = V + I_a R_a

E = 250 + 455 (0.01) = 250 + 4.55

E = 254.55 V

\(\begin{array}{l} No\;of\;turns = \frac{{No\;of\;conductors}}{2} = \frac{{720}}{2} = 360\\ No\;of\;coils = \frac{{360}}{3} = 120 \end{array}\)

No of coil sides = 2 × no of coils = 2 × 120 = 240

No of slots = 240/4 = 60 slots

For employing equaliser ring equipotential pitch value should be integer i.e. 2C/P should be integer.

C = Number of coils

P = Number of poles

\(\frac{{2C}}{P} = \frac{{2 \times 120}}{6} = 40\)

Hence equaliser rings can be employed.

Concept:

The EMF equation of a DC Machine is

\({E_b} = \frac{{NPϕ Z}}{{60A}}\)

From the above equation,

\(N \propto \frac{{{E_b}}}{ϕ }\)

In a dc generator, induced emf is

Eg = V + IaRa

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Given-

V = 200, R_a = 0.2 Ω, I_a = 20 A, ϕ₂ = 1.1 ϕ₁

Let ϕ₁ = Generator flux, ϕ₂ = Motor flux, N₁ = Generator speed, N₂ = Motor speed 

∴ \({E_g} = 200 + 20 \times 0.2 = 204\ V\)

\(\begin{array}{l} {E_b} = 200 - 20 \times 0.2 = 196\ V\\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_b}}}{{{E_g}}} \times \frac{{{ϕ _1}}}{{{ϕ _2}}} = \frac{{196}}{{204}} \times \frac{{{ϕ _1}}}{{1.1{ϕ _1}}} = 0.87 \end{array}\)

The correct answer is above option 3):(Field resistance is above critical resistance)

Concept:

Conditions for voltage build-up voltage of a self-excited or shunt Generator:

• There must be some residual magnetism in generator poles. Some amount of emf will be generated due to residual magnetism in the main poles. This small emf in turn produces additional flux to reinforce the original residual flux. This process is continued till the generator builds up the rated voltage.
• The connections of the field winding should be such that the field current strengthens the residual magnetism.
• The resistance of the filed circuit should be less than the critical resistance.
• Critical resistance is the total field resistance above which the generator fails to build up its voltage. In other words, the speed of the generator should be higher than the critical speed.
• Critical speed is the speed of the generator below which fails to build up its voltage. Under load conditions, the load resistance must be more than the critical load resistance.

​

According to the question circuit diagram of DC shunt generator can be drawn asNow, shunt field current (I_sh) is

\({I_{sh}} = \frac{V}{{{R_{sh}}}} = \frac{{250}}{{50}}\)

I_sh = 5 A

Now Armature current is

I_a = I_L + I_sh = 450 + 5

I_a = 455 A

Now generated emf in dc shunt generator on be calculated as

E = V + I_a R_a

E = 250 + 455 (0.01) = 250 + 4.55

E = 254.55 V

\(\begin{array}{l} No\;of\;turns = \frac{{No\;of\;conductors}}{2} = \frac{{720}}{2} = 360\\ No\;of\;coils = \frac{{360}}{3} = 120 \end{array}\)

No of coil sides = 2 × no of coils = 2 × 120 = 240

No of slots = 240/4 = 60 slots

For employing equaliser ring equipotential pitch value should be integer i.e. 2C/P should be integer.

C = Number of coils

P = Number of poles

\(\frac{{2C}}{P} = \frac{{2 \times 120}}{6} = 40\)

Hence equaliser rings can be employed.

Explanation:

To find the generated voltage of a long shunt compound generator delivering a load current of 50 A at 500 V, we need to calculate the voltage drops across the different components of the generator and then sum these drops to the terminal voltage.

Given Data:

• Load current, I_L = 50 A
• Terminal voltage, V_L = 500 V
• Armature resistance, R_a = 0.04 Ω
• Series field resistance, R_s = 0.025 Ω
• Shunt field resistance, R_sh = 125 Ω
• Brush voltage drop = 3 V

Step-by-Step Solution:

1. **Calculate the Shunt Field Current (I_sh):**

The shunt field current can be calculated using Ohm's Law:

I_sh = V_L / R_sh

Substituting the given values:

I_sh = 500 V / 125 Ω = 4 A

2. **Calculate the Armature Current (I_a):**

In a long shunt compound generator, the armature current is the sum of the load current and the shunt field current:

I_a = I_L + I_sh

Substituting the given values:

I_a = 50 A + 4 A = 54 A

3. **Calculate the Voltage Drop in the Armature (V_a):**

The voltage drop across the armature resistance can be calculated using Ohm's Law:

V_a = I_a × R_a

Substituting the given values:

V_a = 54 A × 0.04 Ω = 2.16 V

4. **Calculate the Voltage Drop in the Series Field (V_s):**

The voltage drop across the series field resistance can be calculated using Ohm's Law:

V_s = I_a × R_s

Substituting the given values:

V_s = 54 A × 0.025 Ω = 1.35 V

5. **Calculate the Total Voltage Drop (V_drop):**

The total voltage drop is the sum of the voltage drops in the armature, series field, and brushes:

V_drop = V_a + V_s + Brush voltage drop

Substituting the given values:

V_drop = 2.16 V + 1.35 V + 3 V = 6.51 V

6. **Calculate the Generated Voltage (E_g):**

The generated voltage is the sum of the terminal voltage and the total voltage drop:

E_g = V_L + V_drop

Substituting the given values:

E_g = 500 V + 6.51 V = 506.51 V

Conclusion:

The generated voltage of the long shunt compound generator is 506.51 V. Therefore, the correct option is:

Option 1: 506.51 V

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 2: 516.48 V

This option is incorrect because it does not account for the correct calculation of the voltage drops across the armature, series field, and brushes. It likely overestimates the generated voltage.

Option 3: 520.66 V

This option is incorrect as it significantly overestimates the generated voltage. The total voltage drop calculation appears to be much higher than the actual value.

Option 4: 509.36 V

This option is incorrect because it slightly overestimates the generated voltage. The voltage drop calculation may have included additional factors that were not specified in the problem statement.

Conclusion:

Understanding the voltage drops and their contributions to the terminal voltage is crucial in accurately determining the generated voltage of a compound generator. The correct calculation involves summing the terminal voltage and the total voltage drop across the generator's components.

When working as a generator

E₁ = V + I_aR_a

= 240 + (24 × 0.25)

= 246 V

When working as motor

E₂ = V - I_aR_a

= 240 – (24 × 0.25)

= 234 V

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{N_1}{\phi _1}}}{{{N_2}{\phi _2}}}\)

\(\frac{{246}}{{234}} = \frac{{{N_1}}}{{{N_2}}}\frac{1}{{1.2}}\)

\(\frac{{{N_2}}}{{{N_1}}} = \frac{1}{{1.2}} \times \frac{{234}}{{246}}\)

Load resistance \(= \frac{{150}}{{200}} = 0.75{\rm{\Omega }}\)

E₁ = 150 + (200 × 0.25) + 2

= 202 V

N₁ = 1500 rpm, N₂ = 1000 rpm

At 1000 rpm, \({E_2} = 202 \times \frac{{1000}}{{1500}}\)

= 134.67 V

V = 134.67 – 0.25 I – 2

I = Load current

V = 132.67 – 0.25 I

\(I = \frac{V}{R} = \frac{{132.67 - 0.25I}}{{0.75}}\)