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A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal voltage of 220 V. The resistance of the armature shunt field and series field are 0.05 Ω, 10.0 Ω and 0.06 Ω respectively. The overall efficiency at the above load is 88%. The torqueexerted by the prime mover is ______
- a)315.1 N-m
- b)358.1 N-m
- c)277.3 N-m
- d)294.7 N-m
Correct answer is option 'B'. Can you explain this answer?
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A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal v...
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A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal v...
To find the torque exerted by the prime mover in a long shunt dynamo, we can use the formula:
\[T = \frac{P_{\text{output}}}{\omega}\]
Where:
T = Torque exerted by the prime mover (in N-m)
\(P_{\text{output}}\) = Output power of the dynamo (in watts)
\(\omega\) = Angular velocity (in radians/second)
Let's calculate the output power first:
\[P_{\text{output}} = P_{\text{terminal}} - P_{\text{losses}}\]
Where:
\(P_{\text{terminal}}\) = Terminal power (in watts)
\(P_{\text{losses}}\) = Power losses in the dynamo (in watts)
The terminal power can be calculated using the formula:
\[P_{\text{terminal}} = V_{\text{terminal}} \times I_{\text{terminal}}\]
Where:
\(V_{\text{terminal}}\) = Terminal voltage (in volts)
\(I_{\text{terminal}}\) = Terminal current (in amperes)
To find the terminal current, we can use Ohm's law:
\[I_{\text{terminal}} = \frac{V_{\text{terminal}}}{R_{\text{total}}}\]
Where:
\(R_{\text{total}}\) = Total resistance of the circuit (in ohms)
The total resistance can be calculated as:
\[R_{\text{total}} = R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field}}\]
Now, let's calculate the power losses:
\[P_{\text{losses}} = I_{\text{terminal}}^2 \times (R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field}})\]
Finally, we can calculate the torque exerted by the prime mover:
\[T = \frac{P_{\text{output}}}{\omega}\]
Given values:
\(P_{\text{output}} = 33 \, \text{kW} = 33000 \, \text{W}\)
\(\omega = 1000 \, \text{rpm} = \frac{1000}{60} \, \text{rad/s}\)
\(V_{\text{terminal}} = 220 \, \text{V}\)
\(R_{\text{armature}} = 0.05 \, \Omega\)
\(R_{\text{shunt field}} = 10.0 \, \Omega\)
\(R_{\text{series field}} = 0.06 \, \Omega\)
Let's substitute the values and calculate the torque:
\[I_{\text{terminal}} = \frac{V_{\text{terminal}}}{R_{\text{total}}} = \frac{220}{0.05 + 10.0 + 0.06} \, \text{A}\]
\[P_{\text{losses}} = I_{\text{terminal}}^2 \times (R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field
\[T = \frac{P_{\text{output}}}{\omega}\]
Where:
T = Torque exerted by the prime mover (in N-m)
\(P_{\text{output}}\) = Output power of the dynamo (in watts)
\(\omega\) = Angular velocity (in radians/second)
Let's calculate the output power first:
\[P_{\text{output}} = P_{\text{terminal}} - P_{\text{losses}}\]
Where:
\(P_{\text{terminal}}\) = Terminal power (in watts)
\(P_{\text{losses}}\) = Power losses in the dynamo (in watts)
The terminal power can be calculated using the formula:
\[P_{\text{terminal}} = V_{\text{terminal}} \times I_{\text{terminal}}\]
Where:
\(V_{\text{terminal}}\) = Terminal voltage (in volts)
\(I_{\text{terminal}}\) = Terminal current (in amperes)
To find the terminal current, we can use Ohm's law:
\[I_{\text{terminal}} = \frac{V_{\text{terminal}}}{R_{\text{total}}}\]
Where:
\(R_{\text{total}}\) = Total resistance of the circuit (in ohms)
The total resistance can be calculated as:
\[R_{\text{total}} = R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field}}\]
Now, let's calculate the power losses:
\[P_{\text{losses}} = I_{\text{terminal}}^2 \times (R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field}})\]
Finally, we can calculate the torque exerted by the prime mover:
\[T = \frac{P_{\text{output}}}{\omega}\]
Given values:
\(P_{\text{output}} = 33 \, \text{kW} = 33000 \, \text{W}\)
\(\omega = 1000 \, \text{rpm} = \frac{1000}{60} \, \text{rad/s}\)
\(V_{\text{terminal}} = 220 \, \text{V}\)
\(R_{\text{armature}} = 0.05 \, \Omega\)
\(R_{\text{shunt field}} = 10.0 \, \Omega\)
\(R_{\text{series field}} = 0.06 \, \Omega\)
Let's substitute the values and calculate the torque:
\[I_{\text{terminal}} = \frac{V_{\text{terminal}}}{R_{\text{total}}} = \frac{220}{0.05 + 10.0 + 0.06} \, \text{A}\]
\[P_{\text{losses}} = I_{\text{terminal}}^2 \times (R_{\text{armature}} + R_{\text{shunt field}} + R_{\text{series field
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Question Description
A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal voltage of 220 V. The resistance of the armature shunt field and series field are 0.05 Ω, 10.0 Ω and 0.06 Ω respectively. The overall efficiency at the above load is 88%. The torqueexerted by the prime mover is ______a)315.1 N-m b)358.1 N-mc)277.3 N-m d)294.7 N-mCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal voltage of 220 V. The resistance of the armature shunt field and series field are 0.05 Ω, 10.0 Ω and 0.06 Ω respectively. The overall efficiency at the above load is 88%. The torqueexerted by the prime mover is ______a)315.1 N-m b)358.1 N-mc)277.3 N-m d)294.7 N-mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long shunt dynamo running at 1000 rpm supplies 33 kW at a terminal voltage of 220 V. The resistance of the armature shunt field and series field are 0.05 Ω, 10.0 Ω and 0.06 Ω respectively. The overall efficiency at the above load is 88%. The torqueexerted by the prime mover is ______a)315.1 N-m b)358.1 N-mc)277.3 N-m d)294.7 N-mCorrect answer is option 'B'. Can you explain this answer?.
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