Question

A dc generator running at 30 rev/s generates an e.m.f. of 200 V. Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.

• A. 87.5%
• B. 85.5%
• C. 75.5%
• D. 70.5%

Hint:

In DC machines, emf varies directly with both speed and magnetic flux. Any reduction in speed must be compensated by an increase in flux.

Answer 1

The Correct Option is A

Step 1: If flux stayed unchanged, running at \(20\) rev/s instead of \(30\) rev/s would scale emf down by the speed ratio: \( 200 \times \dfrac{20}{30} = 133.33 \) V.
Step 2: To reach \(250\) V from this \(133.33\) V baseline, flux alone must make up the rest: required scale-up \( = \dfrac{250}{133.33} = 1.875 \).
Step 3: This \(1.875\times\) scale-up in flux is a \(0.875\times\), i.e. \(87.5\%\), increase over the original flux.
\[ \boxed{87.5\%} \]

Answer 2

A third approach checks each option by assuming it is the correct percentage increase, computing the resulting emf at \(20\) rev/s, and seeing which one actually reproduces the required \(250\) V.

Baseline relationship: for a fixed constant of proportionality, \( E = k \Phi N \), and originally \(200 = k \Phi_1 (30)\), so \(k\Phi_1 = 6.667\).

1. 87.5% increase: New flux \( = 1.875\, \Phi_1 \). Resulting emf at \(20\) rev/s: \( 6.667 \times 1.875 \times 20 = 250.0 \) V, exactly matching the required generated emf.
2. 85.5% increase: New flux \(= 1.855\,\Phi_1\). Resulting emf: \(6.667 \times 1.855 \times 20 \approx 247.3\) V, short of \(250\) V.
3. 75.5% increase: New flux \(=1.755\,\Phi_1\). Resulting emf: \(6.667 \times 1.755 \times 20 \approx 234.0\) V, well short of \(250\) V.
4. 70.5% increase: New flux \(=1.705\,\Phi_1\). Resulting emf: \(6.667\times1.705\times20\approx227.3\) V, the largest shortfall among the options.

Only the \(87.5\%\) increase reproduces the required \(250\) V exactly when run at \(20\) rev/s.

Therefore, the correct answer is 87.5%.