Define capacitance - Leaving Cert Physics - Question 8 - 2015

When the switch is closed, the light bulb lights up indicating that current flows through the circuit and the capacitor begins charging. If the capacitor is connected properly, the light bulb will illuminate briefly as the capacitor charges.

If a 12 V d.c. supply is used instead, the bulb would not light up continuously. Instead, it might flash briefly as the capacitor charges up to the supply voltage, but will remain off while the capacitor is fully charged since there will be no current flow through a fully charged capacitor.

These observations indicate that:

1. Capacitors store electrical charge.
2. They can hold a charge and block direct current (d.c.) when fully charged, while allowing alternating current (a.c.) to pass through by continuously charging and discharging.

To calculate the capacitance (C), we can use the formula: $C = \frac{Q}{V}$ Substituting the values:

• Charge (Q) = 0.8 C
• Voltage (V) = 12 V

So, $C = \frac{0.8 \, \text{C}}{12 \, \text{V}} = 0.0667 \, \text{F} = 66.7 \, \mu F$

To demonstrate that energy is stored in a charged capacitor, perform the following experiment:

1. Apparatus: Connect a charged capacitor to a light bulb or a resistor.
2. Procedure:
   • Charge the capacitor using a battery or power supply until it reaches a specified voltage.
   • Disconnect the power supply and quickly connect the charged capacitor to the light bulb.
3. Observation/Conclusion: The light bulb will illuminate momentarily, showing that energy stored in the capacitor is released, thus proving that a charged capacitor can store energy.

For the radio, the property of capacitors used is 'filtering capabilities' which allows for the selection of specific frequencies. For the camera flash, the property of capacitors used is 'storing energy' that is released quickly to produce a bright flash of light.