Question

The current flowing through the armature of a d.c. shunt machine at maximum efficiency is given by :

• A. \(\sqrt{(P_i / R_a)}\)
• B. \(\sqrt{(R_a / P_i)}\)
• C. \(\sqrt{(P_i / R_a^2)}\)
• D. \(\sqrt{(R_a / P_i^2)}\)

Hint:

To easily remember maximum efficiency conditions for any electrical machine:
Set Copper Loss (\(I^2 R\)) = Constant Loss (\(P_i\)).
Solve for Current: \(I = \sqrt{\frac{P_i}{R}}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Efficiency (\(\eta\)) of any electrical machine is maximized when the variable losses equal the constant losses.
In a DC shunt machine (generator or motor), the variable losses are primarily the armature copper losses, and the constant losses include shunt field copper losses, iron losses, and mechanical losses.
Key Formula or Approach:
The condition for maximum efficiency is:
\[ \text{Variable Losses} = \text{Constant Losses} \]
Armature copper loss (variable loss) is:
\[ W_{cu} = I_a^2 R_a \]
Let constant losses be denoted by \(P_i\) (which includes iron loss and other constant losses).
At maximum efficiency:
\[ I_a^2 R_a = P_i \]

Step 2: Detailed Explanation:
Let us solve the equation for the armature current \(I_a\):
\[ I_a^2 = \frac{P_i}{R_a} \]
Taking the square root on both sides:
\[ I_a = \sqrt{\frac{P_i}{R_a}} \]
Where:
- \(I_a\) is the armature current.
- \(P_i\) is the constant loss (iron and other constant losses).
- \(R_a\) is the armature resistance.
Therefore, the armature current at maximum efficiency is \(\sqrt{P_i / R_a}\).
This matches Option (A).

Step 3: Final Answer:
The correct option is (A).