Given,

i. Separately excited DC motor

ii. Speed, $\mathrm{N}_{2}=1.5 \mathrm{~N}_{1}$ $=1.5 \mathrm{~N}_{\mathrm {rated }}$

iii. Torque, $\mathrm{T}_{2}=0.25 \mathrm{~T}_{1}$ = $0.25 \mathrm{~T}_{\mathrm{rated }}$

iv. Negligible armature resistance drop, $\mathrm{I}_{\mathrm{a}} \mathrm{R}_{\mathrm{a}}=0$

Since armature resistance drop is negligible, the back EMF will be constant.

We know that, $\mathrm{E}_{\mathrm{b}}=\mathrm{k} \phi \mathrm{N}$

$\mathrm{E}_{\mathrm{b}} \propto \phi \mathrm{N}$

$\frac{E_{ b 2}}{E_{ b 1}}=\frac{\phi_ 2 * N _2}{\phi_ 1 * N _1}$

$\frac{E_{ b 2}}{E _{b 1}}=\frac{\phi_ 2 * 1.5 N _1}{\phi _1 * N_ 1}$ $\left[\mathrm{E}_{\mathrm{b} 1}=\mathrm{E}_{\mathrm{b} 2}\right]$

$\phi_{2}=0.67 \phi_{1}$

Again, $\mathrm{T} \propto \phi \mathrm{l}_{\mathrm{a}}$

$\frac{T_{2}}{T_{1}}=\frac{\phi_{2} * I_{2}}{\phi_{1} * I_{1}}$

$\frac{0.25 T_{1}}{T_{1}}=\frac{0.67 \phi_{1} * I_{2}}{\phi_{1} * I_{1}}$

$\frac{I_{2}}{I_{1}}=0.373$

$\frac{I_{2}}{I_{1}}-1=0.373-1$

$\frac{I_{2}-I_{1}}{I_{1}}=-0.627$ i.e $-62.7 \%$ (decrement)

Hence, the required percentage change in excitation would be -62.7%

Transformers