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A 220V series motor that has a total resistance (field winding and armature winding) of 0.2 ohms and runs at 1000rpm and takes full load current of 200A. The speed of the motor at which it develops half of the full load torque is?

I was doing this problem but got stuck as I could not find the value of the back EMF developed at half of the full load torque.

Here is my go at it:

$$ V = 220\mathrm{V} \\ I_{full-load} = 200\mathrm{A}\\ R = 0.2\Omega\\ N_{full-load}=1000\mathrm{rpm}$$ $$$$ $$E_{b_{full-load}} = V - I_{full-load}\times R\\ E_{b_{full-load}} = 220-(200)(0.2)\\ E_{b_{full-load}} = 180\mathrm{V}$$ $$$$ In a DC series motor, $$\phi \propto I $$ $$$$ $$E_b = \frac{Zp}{60A}N\phi\\ E_b = k\cdot N\phi\\ E_b = k^{'}\cdot NI\\ T=E_b I\cdot (\frac{60}{2\pi N})\\ T=c\cdot \frac{E_b I}{N}$$ $$$$ $$\frac{T^{'}}{T}=\frac12=\frac{\frac{E_b^{'}I^{'}}{N^{'}}}{\frac{E_bI}{N}}$$ from, $$E_b = k^{'}\cdot NI$$ $$N^{'}=N\cdot\frac{E_b^{'}}{E_b}\cdot\sqrt{\frac{T^{'}}{T}}$$ Only value for back EMF at half of full load torque is required to obtain the speed.
The answer is 1518 rpm.

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