A circuit is set up as shown - Scottish Highers Physics - Question 18 - 2018

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

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

where:

• E is the energy in joules (J)
• C is the capacitance in farads (F)
• V is the voltage in volts (V)

Given:

• Capacitance, C = 220 μF = 220 \times 10^{-6} F
• Voltage, V = 12 V

Substituting the values, we have:

$E = \frac{1}{2} (220 \times 10^{-6}) (12)^2$

Calculating this:

$E = \frac{1}{2} (220 \times 10^{-6}) (144)$

$E = 0.01584 J = 1.584 \times 10^{-2} J$

Rounding this to one significant figure gives us approximately:

1.6 × 10^1 J