12. (a) A student sets up the circuit shown - Scottish Highers Physics - Question 12 - 2019

To calculate the internal resistance r, we use the formula for EMF in terms of circuit potential differences:

$E = V + Ir$

Substituting the known values:

$1.5 = 1.3 + (0.8)r$

Rearranging gives:

$r = \frac{(1.5 - 1.3)}{0.8} = \frac{0.2}{0.8} = 0.25 \, \Omega$

When the switch is closed, a current flows through the circuit. This current creates a voltage drop across the internal resistance of the cell, which reduces the voltage available to the external circuit. Consequently, the reading on the voltmeter decreases.

Using the total resistance in the circuit, we have:

Total resistance, R = (1.2 + 2.4) Ω = 3.6 Ω.

The total voltage is 9.0 V. Using Ohm's Law:

$I = \frac{E}{R} = \frac{9.0}{3.6} = 2.5 \, A$

Power dissipated in the lamp can be calculated using:

$P = I^2 R$

Substituting in the current (I = 2.5 A) and the resistance of the lamp (R = 2.4 Ω):

$P = (2.5)^2 \cdot 2.4 = 15 \, W$