9.1 State Ohm's law in words - NSC Physical Sciences - Question 9 - 2017 - Paper 1

To draw the line of best fit, extend a straight line that closely follows the linear trend of the plotted points. Make sure the line intersects both the voltage and current axes.

The emf (ε) of the battery is 5.5 V, which corresponds to the y-intercept of the graph.

The internal resistance (r) of the battery can be calculated using coordinates from the line drawn on the graph. Using Ohm's law, we find r = 1.2 Ω.

The current (I) flowing through the 8 Ω resistor can be calculated using the formula:

$I = \frac{V}{R} = \frac{21.84 \text{ V}}{8 \text{ Ω}} = 2.73 \text{ A}$

The equivalent resistance (R_eq) of the parallel resistors can be found using:

$\frac{1}{R_{eq}} = \frac{1}{30 \text{ Ω}} + \frac{1}{20 \text{ Ω}} + \frac{1}{8 \text{ Ω}}$

Calculating the above gives:

$R_{eq} = 12 \text{ Ω}$

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

$V_{terminal} = V_{emf} - V_{lost}$

Using the values:

$r = \frac{V_{lost}}{I} = 1.98 \text{ Ω}$

The heat (H) dissipated can be calculated using:

$H = I^2 R t$

Substituting the values gives:

$H = (2.73 ext{ A})^2 \times 8 \text{ Ω} \times 0.2 ext{ s} = 29.81 ext{ J}$