A 240 VDC series motor takes 40 A when giving its rated output at 1200 rpm. The combined value of armature and series field resistance is $0.3 \Omega .$ Find what resistance must be added to obtain the rated torque (a) at starting (b) at 800 rpm.

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Accepted Answer

[latex]\begin{array}{l} ext { Rated voltage, } V=240 V\ ext { Rated current, } I_{1}=40 A\ ext { Rated speed, } N_{1}=1200 rpm\end{array}[/latex]

[latex] ext { Back emf, } E_{b 1}=V-I_{a} R_{m}=240-40 imes 0.3=228 V[/latex]

(i) At start when speed is zero, the back emf [latex] E_{b 2}[/latex] will be zero and rated torque will be developed when current drawn by the motor is of rated value i.e.,

[latex]\begin{array}{l}\begin{array}{c} ext { Current, } I_{2}=40 A \ ext { Back emf, } E_{b 2}=V-I_{2}\left(R_{1}+R_{m} ight)\end{array}\ herefore ext { additional resistance, } R_{1}=\frac{V-E_{b 2}}{I_{2}}-R_{m}=\frac{240-0}{40}-0.3= 5 . 7 ext { ohm}\end{array}[/latex]

(ii) At 800 rpm, the rated torque will be developed only when current drawn by the motor is rated one i.e.,

[latex] I_{3}=40 A[/latex]

Since field current remains the same, back emf will be proportional to speed i.e.,

[latex] E_{b 2}=E_{b_{1}} imes \frac{N_{3}}{N_{1}}=228 imes \frac{800}{1200}=152 V[/latex]

[latex] herefore[/latex] required additional resistance, [latex] R_{2}=\frac{V-E_{b 3}}{I_{3}}-R_{m}=\frac{240-152}{40}-0.3=1.9 \Omega[/latex]

Question 5.36: A 240 VDC series motor takes 40 A when giving its rated outp...

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