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

The diode (G) functions as a rectifier, converting alternating current (AC) from the transformer into direct current (DC) suitable for charging the mobile phone.

The input voltage graph, $V_{IN}$, is a sine wave that fluctuates between positive and negative values as it alternates.

• The x-axis should represent time.
• The y-axis should represent voltage.
• The graph should feature a smooth oscillation characteristic of AC voltage.

The output voltage graph, $V_{XY}$, shows a pulsating pattern that indicates conversion from AC to DC.

• The graph should reflect the smoothing effect of the capacitor.
• The x-axis is time, and the y-axis is voltage, displaying a generally increasing voltage profile with minimal fluctuations.

Using the formula for energy stored in a capacitor, $E = \frac{1}{2} CV^2$, where C is the capacitance and V is the voltage.

Given:

• Capacitance, $C = 2200 \times 10^{-6} \, F$
• Voltage, $V = 16 \, V$

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

High voltage is used in electricity transmission to reduce power loss. By minimizing the current for a given power transmission, energy loss due to the resistance of the wires (which leads to heat generation) is significantly reduced. Higher voltage allows for efficient long-distance energy transportation.

The resistance of a wire can be calculated using the formula: $R = \frac{\rho L}{A}$, where:

• $\rho$ is the resistivity of aluminium ($2.8 \times 10^{-8} \, \Omega m$).
• $L$ is the length of the wire (3 km = 3000 m).
• $A$ is the cross-sectional area calculated as $A = \pi r^2$, with diameter 18 mm (radius = 0.009 m).

Calculating:

• Area, $A = \pi (0.009)^2 \approx 2.54 \times 10^{-4} \, m^2$.
• Now substituting: $R = \frac{(2.8 \times 10^{-8}) \times 3000}{2.54 \times 10^{-4}} \approx 0.33 \, \Omega$

Using the formula for power, $P = I^2R$, where $I$ is the current and $R$ is the resistance:

• With $I = 250 \, A$ and $R = 0.33 \, \Omega$: $P = (250)^2 \times 0.33 = 21,000 \, W$
• To find the total energy in ten minutes: $E = P \times t = 21,000 \times (10 \times 60) = 12,600,000 \, J$