9.1 A simplified diagram of an electric generator is shown below - NSC Physical Sciences - Question 9 - 2019 - Paper 1

Electromagnetic induction occurs when a conductor, such as a coil of wire, moves through a magnetic field or when the magnetic field around a conductor changes. According to Faraday's law of electromagnetic induction, the induced emf in the coil is directly proportional to the rate of change of the magnetic flux through the coil. When the coil rotates in the magnetic field, the angle between the magnetic field lines and the normal to the coil changes, leading to a change in magnetic flux and thereby inducing an emf.

The output voltage graph for an AC generator is sinusoidal in shape. It typically varies between a positive peak and a negative peak over time. It can be represented as follows:

Voltage
  |
  |     /\         /\
  |    /  \       /  \
  |   /    \     /    \
  |  /      \   /      \
  |_/        \_/        \_
  ----------------------------------
                    Time

One complete cycle corresponds to one full rotation of the coil.

The rms (root mean square) voltage of an AC source is defined as the effective voltage which dissipates the same amount of energy in a resistive load as an equivalent DC voltage. Mathematically, it is expressed as: $V_{rms} = \frac{V_{max}}{\sqrt{2}}$ where $V_{max}$ is the peak voltage of the AC source.

To find the maximum peak voltage $V_{max}$ of the AC source, we first calculate the rms voltage:

Given that the energy dissipated in 10 s is 500 J in a 200 Ω resistor:

Using the formula for power: $P = \frac{W}{t}$ $P = \frac{500 \, J}{10 \, s} = 50 \, W$

Next, using the formula for power in terms of rms voltage: $P = \frac{V_{rms}^2}{R}$

Substituting $R = 200 \Omega$: $50 = \frac{V_{rms}^2}{200}$

This gives us: $V_{rms}^2 = 50 \times 200 = 10000 \, V^2$ $V_{rms} = 100 \, V$

Finally, we can determine the peak voltage: $V_{max} = V_{rms} \times \sqrt{2} = 100 \times \sqrt{2} \approx 141.42 \, V$