Question

If an alternator has a flux per pole of \(0.05\ \text{Wb}\), a frequency of \(50\ \text{Hz}\), and \(100\) turns per phase (assume pitch factor and distribution factor is \(1\)), the RMS phase emf is approximately

• A. \(222\ \text{V}\)
• B. \(1110\ \text{V}\)
• C. \(555\ \text{V}\)
• D. \(444\ \text{V}\)

Hint:

For alternators, remember the RMS phase emf formula \(E=4.44f\phi TK_pK_d\).

Answer 1

The Correct Option is B

Concept: The RMS emf induced per phase in an alternator is: \[ E=4.44 f\phi T K_pK_d \] where \(f\) is frequency, \(\phi\) is flux per pole, \(T\) is turns per phase, \(K_p\) is pitch factor, and \(K_d\) is distribution factor.

Step 1: Given: \[ f=50\ \text{Hz} \] \[ \phi=0.05\ \text{Wb} \] \[ T=100 \] \[ K_p=1,\qquad K_d=1 \]

Step 2: Apply the alternator emf equation. \[ E=4.44 f\phi T K_pK_d \] \[ E=4.44\times 50\times 0.05\times 100\times 1\times 1 \]

Step 3: Calculate step by step. \[ 50\times 0.05=2.5 \] \[ 2.5\times 100=250 \] \[ E=4.44\times 250 \] \[ E=1110\ \text{V} \] Therefore: \[ \boxed{1110\ \text{V}} \]