A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is ——-
Given that V = 300 R_{a}\,=\,0.5\,\,\Omega N_{0}=900\;\mathrm{rpm},\;I_{a0}=2\,\mathrm{A},\,I_{a1}=15\,\mathrm{A}
Under no load condition:
E_{\mathrm{b0}}=V-I_{\mathrm{a0}}\,R_{\mathrm{a}}=300-(2)(0.5)=299\,\,V
Under load condition:
E_{\mathrm{b1}}=300-I_{\mathrm{al}}R_{\mathrm{a}}=300-(15)(0.5)=292.5~\mathrm{V}
We know that
\frac{N_{1}}{N_{0}}=\frac{E_{\mathrm{bl}}}{E_{\mathrm{b0}}}\Rightarrow N_{1}=\frac{292.5}{299}\times900=880\;\mathrm{rpm}