Question

1. determine the electric energy transferred by the generator to the motor if a current of 35a passes through it during 3s.
2. calculate the efficiency of the generator.

the drive motor of a winch is powered by a generator of electromotive force e = 144v and internal resistance r = 1ω.

Explanation:

Step1: Recall the formula for power and energy transfer

The electromotive force (emf) of the generator is \( E = 144\,\text{V} \), internal resistance \( r=1\,\Omega \), and current \( I = 35\,\text{A} \). The total power generated by the generator is \( P_{\text{total}}=EI \), and the power dissipated in the internal resistance is \( P_{\text{loss}} = I^{2}r \). The power transferred to the motor (useful power) is \( P_{\text{useful}}=P_{\text{total}} - P_{\text{loss}} \).

Step2: Calculate total power

\( P_{\text{total}}=EI = 144\,\text{V}\times35\,\text{A}=5040\,\text{W} \)

Step3: Calculate power loss

\( P_{\text{loss}}=I^{2}r=(35\,\text{A})^{2}\times1\,\Omega = 1225\,\text{W} \)

Step4: Calculate useful power

\( P_{\text{useful}}=P_{\text{total}}-P_{\text{loss}}=5040 - 1225=3815\,\text{W} \)

Step5: Calculate efficiency of energy transfer (for part 1)

Efficiency \( \eta_{1}=\frac{P_{\text{useful}}}{P_{\text{total}}}\times100\%=\frac{3815}{5040}\times100\%\approx75.7\% \)

Step6: Calculate efficiency of the generator (for part 2)

The generator's efficiency is also given by the ratio of useful power to total power, so it is the same as the energy transfer efficiency to the motor (since the loss is in the generator's internal resistance). So \( \eta_{2}=\eta_{1}\approx75.7\% \)

Answer:

1. The electric energy transferred to the motor (efficiency of energy transfer) is approximately \( 75.7\% \).
2. The efficiency of the generator is approximately \( 75.7\% \).