A car battery is connected to an electric motor as shown - Scottish Highers Physics - Question 10 - 2015

To calculate the current (I) flowing in the circuit, we can use Ohm's law, which states:

$I = \frac{V}{R + r}$

Here, the total resistance in the circuit is the sum of the internal resistance (r) of the battery and the resistance (R) of the motor. Substituting the values:

$I = \frac{12.8}{0.050 + 6.0 \times 10^{-3}}$

Calculating:

$I = \frac{12.8}{0.050 + 0.006} = \frac{12.8}{0.056} \approx 228.57 A$

Thus, the current in the circuit when the motor is operating is approximately 230 A.

The wires have a large diameter to reduce their resistance, allowing for a larger current to flow without significant power loss. Larger wires also help prevent overheating and potential melting due to high current.

From the graph, the e.m.f. of the battery corresponds to the value of V when I = 0. Observing the graph, the e.m.f. of the battery is determined to be approximately 12.6 V.

To find the internal resistance of the battery, use the formula:

$r = \frac{V_{oc} - V}{I}$

Using the values from the graph and the calculated e.m.f. of 12.6 V, the internal resistance can be calculated as follows:

$r = \frac{12.6 - V}{I}$

Using data points from the graph, calculate r using either slope or specific points.

To find the initial charging current when switch S is closed, use the formula:

$I_{charge} = \frac{E_{charger} - E_{battery}}{R_{internal} + R_{battery}}$

Where E_{charger} = 15.0 V, E_{battery} = 11.5 V, R_{internal} = 0.45 Ω, and R_{battery} = 0.090 Ω.

Calculating:

$I_{charge} = \frac{15.0 - 11.5}{0.45 + 0.090} = \frac{3.5}{0.54} \approx 6.48 A$

As the battery charges, its e.m.f. increases closer to the charger’s e.m.f. This reduces the voltage difference available for charging, thereby decreasing the current. Additionally, as chemical reactions occur within the battery, internal resistance may also increase, further contributing to the reduction in charging current.