The diagram shows a circuit used in a charger for a mobile phone - Leaving Cert Physics - Question 8 - 2013

The function of the diode (G) in this circuit is to allow current to flow in one direction only. This prevents the backflow of current, ensuring that the capacitor (H) is charged correctly and that the circuit operates safely by converting AC into DC.

(i) the input voltage The input voltage will be depicted as a sinusoidal wave that varies periodically over time, showing positive and negative cycles for the AC input.

(ii) the output voltage, Vx-y. The output voltage graph will show a smoother waveform, indicating the smoothing effect of the capacitor, with periods of time where it remains constant between peaks of the input voltage, reflecting a DC output.

The maximum energy (E) that can be stored in a capacitor is calculated using the formula: $E = \frac{1}{2}CV^2$ Where: C = capacitance = $2200 \times 10^{-6}$ F V = voltage = 16 V

Calculating this, we have: $E = \frac{1}{2} \times 2200 \times 10^{-6} \times (16)^2 = 0.2816 J$

High voltage is used in electricity transmission over long distances to minimize power loss. The higher the voltage, the lower the current for a given power, according to the formula: $P = IV$ This reduces the resistive losses in the transmission lines, since power loss due to resistance can be expressed as: $P_{loss} = I^2R$ Thus, minimizing current minimizes energy lost as heat.

To calculate the resistance (R) of the wire, use the formula: $R = \frac{\rho L}{A}$ Where: ρ = resistivity of aluminium = $2.8 \times 10^{-8} \Omega m$ L = length of the wire = 3000 m A = cross-sectional area of the wire = \pi (r^2), where r is the radius. For a diameter of 18 mm, the radius is 9 mm = 0.009 m.

Thus, the area A is: $A = \pi (0.009)^2 = 2.54 \times 10^{-4} m^2$ Now substituting values into the resistance formula: $R = \frac{2.8 \times 10^{-8} \times 3000}{2.54 \times 10^{-4}} \approx 0.33 \Omega$

Using the power formula, first compute the total electrical energy (E) converted to heat in ten minutes using: $E = Pt$ Where: P = I^2 R and I = 250 A, R calculated previously is 0.33 Ω. Thus: $P = (250)^2 \times 0.33 = 20625 W$ For t = 600 seconds (10 minutes): $E = 20625 \times 600 = 12375000 J$