A graph of emf against time is provided, showing a maximum voltage of 94,3 V - NSC Physical Sciences - Question 8 - 2017 - Paper 1

To find the average power (P_average), we can use the root mean square voltage (V_rms). Since the root mean square current (I_rms) is 3% of the root mean square voltage:

1. The formula for average power is:

   $P_{average} = V_{rms} imes I_{rms}$

2. First, calculate V_rms:

   $V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{94.3}{\sqrt{2}} \approx 66.7 \text{ V}$

3. Now calculate I_rms:

   $I_{rms} = 0.03 \times V_{rms} \approx 0.03 \times 66.7 \approx 2.0 \text{ A}$

4. Calculate average power:

   $P_{average} = 66.7 \times 2.0 \approx 133.9 \text{ W}$

To convert the AC generator to a DC generator, one must replace the slip rings with a split-ring commutator. This will allow the generator to rectify the alternating current (AC) into direct current (DC).

To calculate the resistance of the light bulb (R), we can use the formula based on power (P) and voltage (V):

1. Use the value from one of the ratings. Let's use the 90 W light bulb and assume it operates at 230 V (if not stated, this is a common assumption):

   $P = \frac{V^2}{R}$

2. Rearranging gives:

   $R = \frac{V^2}{P}$

3. Substituting the known values:

   $R = \frac{230^2}{90} \approx 593.33 \Omega$

The brightness of the light bulb will be TOO BRIGHT because the power of the generator exceeds the power rating of the light bulb. This means the bulb receives more energy than it is designed to handle, resulting in increased brightness and potentially damaging the bulb.