Power in Electric Circuits — Question 9

Question 9

A lifting motor separates useful power from resistive and other losses

A motor draws 8.00 A at 48.0 V while lifting mass 20.0 kg at steady speed 1.20 m/s. Its winding resistance is .500 $\Omega$. Use g=9.80 m/s$^2$.

1. Find electrical input, useful output and efficiency.

2. Find winding heating and all remaining losses, then find the lifting speed if all non-winding losses vanished at the same electrical input and current.

Question 9 - Solution

A lifting motor separates useful power from resistive and other losses

(a) Method and calculation

$$P_{\mathrm{in}}=VI=384\,\mathrm W,\quad P_{\mathrm{useful}}=mgv=235.2\,\mathrm W.$$$$\eta=235.2/384=.6125=61.25\%.$$

(b) Interpretation and further calculation

$$P_{\mathrm{winding}}=I^2R=32.0\,\mathrm W,\quad P_{\mathrm{other}}=384-235.2-32=116.8\,\mathrm W.$$$$v_{\mathrm{idealized}}=\frac{384-32}{20(9.80)}=1.79592\,\mathrm{m/s}.$$

Physical check

Electrical input is not all converted to heat: mechanical output must be included in the energy balance.