A DC shunt-wound generator running at 800 rpm has generated emf of 200 V. Assuming flux per pole is maintained constant, if the speed increases to 1500 rpm, the generated emf will be

EMF equation of DC motor

The EMF of the DC motor is given by:

\(E={NPϕ Z\over 60 A}\)

where, E = Generated EMF

N = Speed

P = No. of poles

ϕ = Flux per pole

Z = No. of conductors

A = No. of parallel paths

The no. of poles, conductors, and parallel paths are fixed for the DC motor. Hence, these always remain constant irrespective of the magnitude of the generated EMF.

Hence, the generated EMF depends upon the flux and the speed.

Calculation

\({E_1\over E_2}={N_1\over N_2}\times {\phi_1\over \phi_2}\)

For the DC shunt-wound generator, the flux remains constant.

Hence, \({E_1\over E_2}={N_1\over N_2}\)

Given, E₁ = 200 V at N₁ = 800 RPM

N₂ = 1500 RPM

\({200\over E_2}={800\over 1500}\)

E₂ = 375 V