The field and armature resistance of a 500 V D C series motor is 0.2 ohm and 0.3 ohm respectively. The motor runs at 500 rpm when drawing a current of 49 A. If the load torque varies as the square of the speed, determine the value of the external resistance to be added in series with the armature

Question

The field and armature resistance of a 500 V D C series motor is 0.2 ohm and 0.3 ohm respectively. The motor runs at 500 rpm when drawing a current of 49 A. If the load torque varies as the square of the speed, determine the value of the external resistance to be added in series with the armature for the motor to run at 450 rpm Assume linear magnetisation.

Accepted Answer

[latex]\begin{array}{l} ext { Here, } N_{1}=500 rpm ; I_{a 1}=40 A ; V=600 V ; N_{2}=450 rpm \T \propto N^{2} ; R_{a}=0.3 \Omega ; R_{s e}=0.2 \Omega ; \phi \propto I_{a} \E_{b 1}=V-I_{a 1}\left(R_{a}+R_{s e} ight)=600-40(0.3+0.2)=580 V \T \propto \phi I_{a} ext { or } T \propto I_{a}^{2}\end{array}[/latex]

and [latex] T \propto N^{2}[/latex]

[latex] herefore \quad I_{a}^{2}=N^{2}[/latex]

and [latex] \left(\frac{I_{a 2}}{I_{a 1}} ight)^{2}=\left(\frac{N_{2}}{N_{1}} ight)^{2}[/latex]

[latex]\begin{array}{l} ext { or }\I_{a 2}=\frac{N_{2}}{N_{1}} imes I_{a 1}=\frac{450}{500} imes 40=36 A\end{array}[/latex]

[latex] ext { Let } R ext { be the resistance added in series with the armature }[/latex]

[latex]E_{b 2}=V-I_{a 2}\left(R_{a}+R_{s c}+R ight)=600-36\left(0.3^{0}+0.2+R ight)=582-36 R[/latex]

[latex] ext { Now } \quad \frac{N_{2}}{N_{1}}=\frac{E_{b 2}}{\phi_{2}} imes \frac{\phi_{1}}{E_{b 1}} ext { or } \frac{N_{2}}{N_{1}}=\frac{E_{b 2}}{I_{a 2}} imes \frac{I_{a 1}}{E_{b 1}}[/latex]

or [latex] \frac{450}{500}=\frac{(582-36 R)}{36} imes \frac{40}{580}[/latex]

or [latex] 582-36 R=\frac{450 imes 36 imes 580}{500 imes 40}[/latex]

or [latex] \quad 582-36 R =469.8[/latex] or [latex] R= 3 . 1 1 7 \Omega[/latex]

Question 5.35: The field and armature resistance of a 500 V D C series moto...

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