A shunt generator running at 1000 r.p.m. has generated e.m.f. as 200 V. If the speed increases to 1200 r.p.m., what will be the generated e.m.f.?

Steps to Solve

1. Identify the relationship between speed and e.m.f.

We know that the e.m.f. generated by a shunt generator is directly proportional to the speed of the generator. This can be expressed as: $$ E \propto N $$ where ( E ) is the e.m.f. and ( N ) is the speed in revolutions per minute (r.p.m.).

2. Establish the original e.m.f. and speed

Assume the original e.m.f. generated at 1000 r.p.m is ( E_1 ). Thus, we can write: $$ E_1 = k \cdot 1000 $$ where ( k ) is the constant of proportionality.

3. Calculate the new e.m.f. at increased speed

At the new speed of 1200 r.p.m, we denote the new e.m.f. as ( E_2 ). From the proportional relationship: $$ E_2 = k \cdot 1200 $$

4. Expressing the new e.m.f. in terms of the old e.m.f.

To find the new e.m.f. in terms of the original e.m.f. ( E_1 ), we can divide the equations: $$ \frac{E_2}{E_1} = \frac{k \cdot 1200}{k \cdot 1000} $$ This simplifies to: $$ \frac{E_2}{E_1} = \frac{1200}{1000} = 1.2 $$

5. Deriving the final calculation for e.m.f.

We can now express ( E_2 ): $$ E_2 = 1.2 E_1 $$

This means that the new e.m.f. is 1.2 times the original e.m.f. generated at 1000 r.p.m.