Correct Answer - Option 1 : 400 rpm
Given that \(N = 600\;rpm,\;{V_t} = 240\;V,\;{I_a} = 30\;A,\;{R_a} = 0.5\;{\rm{\Omega }}\)
\(\begin{array}{l}
{E_b} = {V_t} - {I_a}{R_a}\\
{E_b} = 240 - 30 \times 0.5 = 225\;V
\end{array}\)
Now 2.5 ohm external resistance added to armature in series.
\(\begin{array}{l}
\Rightarrow {I_{a\left( {new} \right)}} = 0.5 + 2.5 = 3\;{\rm{\Omega }}\\
\Rightarrow {E_{b\left( {new} \right)}} = 240 - 30 \times 3 = 150\;V
\end{array}\)
We know that, \({E_b} \propto N\phi \)
In shunt motor the flux is constant.
\(\begin{array}{l}
\Rightarrow \frac{{{E_{b\left( {new} \right)}}}}{{{E_b}}} = \frac{{{N_{new}}}}{N}\\
{N_{new}} = \frac{{150}}{{225}} \times 600 = 400\;rpm
\end{array}\)
