The diagram shows a battery pack powering a model rescue boat motor. Table 1 gives the terminal voltage of the battery pack at different motor currents. The...
Question 1 Report
The diagram shows a battery pack powering a model rescue boat motor. Table 1 gives the terminal voltage of the battery pack at different motor currents. The student uses a data logger so that each reading is taken after the motor speed has become steady. The battery has internal resistance, so some energy is transferred thermally inside the battery when current flows.
| motor current / A | terminal voltage / V |
|---|---|
| 0.50 | 6.0 |
| 1.00 | 5.8 |
| 1.50 | 5.6 |
(a) Calculate the resistance of the motor when its current is 1.00 A. [2]
(b) Calculate the electrical energy transferred to the motor in 120 s at 1.00 A. [3]
(c) Explain why the terminal voltage decreases as the motor current increases. [3]
(d) Calculate the internal resistance of the battery pack using the first and last rows of Table 1. [2]
Answer Details
(a) At \(1.00\ \mathrm{A}\), the terminal voltage is \(5.8\ \mathrm{V}\). Therefore:
\[R=\frac{V}{I}=\frac{5.8\ \mathrm{V}}{1.00\ \mathrm{A}}=5.8\ \Omega\]
Motor resistance = \(5.8\ \Omega\) [2].
(b) The power delivered to the motor is:
\[P=VI=5.8\ \mathrm{V}\times1.00\ \mathrm{A}=5.8\ \mathrm{W}\]
\[E=Pt=5.8\ \mathrm{W}\times120\ \mathrm{s}=696\ \mathrm{J}\]
Energy transferred to the motor = \(696\ \mathrm{J}\) [3].
(c) The battery has internal resistance [1]. A larger current produces a larger voltage drop across this internal resistance [1]. More energy is transferred thermally inside the battery, leaving a lower terminal voltage for the motor [1].
(d) From the first and last readings:
\[\Delta V=6.0-5.6=0.4\ \mathrm{V}\]
\[\Delta I=1.50-0.50=1.00\ \mathrm{A}\]
\[r=\frac{\Delta V}{\Delta I}=\frac{0.4}{1.00}=0.40\ \Omega\]
Internal resistance = \(0.40\ \Omega\) [2].
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