0.2 Ω
The question involves the application of Ohm's law to a DC generator. We are given the following information:
- Load current (I) = 25 A
- Terminal voltage (V) = 200 V
- Generated EMF (E) = 205 V
The difference between the generated EMF (E) and the terminal voltage (V) represents the voltage drop across the armature resistance (\(R_a\)). This voltage drop is caused by the current flowing through the armature's internal resistance.
Therefore, we can write:
Voltage drop across armature = Generated EMF - Terminal voltage
\(E - V = I \times R_a\)
Solving for \(R_a\), we get:
\(R_a = \frac{E - V}{I}\)
Substituting the given values:
\(R_a = \frac{205 V - 200 V}{25 A}\)
\(R_a = \frac{5 V}{25 A}\)
\(R_a = 0.2 \, Ω\)
Therefore, the armature resistance is 0.2 Ω.
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