A technician sets up a circuit as shown - Scottish Highers Physics - Question 13 - 2016

The maximum energy (E) stored in the capacitor can be calculated using the formula:

$E = \frac{1}{2} C V^2$

Where:

• C is the capacitance (150 mF = 150 \times 10^{-3} F)
• V is the voltage (12 V)

Substituting the values:

$E = \frac{1}{2} \times (150 \times 10^{-3}) \times (12)^2$

Calculating this gives:

$E = \frac{1}{2} \times (150 \times 10^{-3}) \times 144$ $E = 10.8 J$

To find the maximum discharge current (I), we first need to determine the total resistance (R) in the circuit.

The resistances in series are:

• 56 Ω and 19 Ω.

Thus,

$R_{total} = 56 + 19 = 75 \: \Omega$

Using Ohm's law, where the initial potential difference (V) across the capacitor is 12 V:

$I = \frac{V}{R} = \frac{12}{75} = 0.16 A$

Replacing the 150 mF capacitor with a 47 mF capacitor decreases the capacitance. According to the formula for energy storage, which is given by:

$E = \frac{1}{2} C V^2$

A smaller capacitance results in less energy stored. Consequently, the lamp will stay lit for a shorter time compared to when using the larger capacitor. This is because less stored energy means that it will discharge more quickly once the switch is moved to position Q.