8.1 What does 94,3 V on the graph represent? 8.1.2 Calculate the average power for this generator - NSC Physical Sciences - Question 8 - 2017 - Paper 1

To calculate the average power ( (P_{average}) ) for the generator, we use the formula:

$P_{average} = V_{rms} \times I_{rms}$

The root mean square (RMS) voltage is given by:

$V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{94,3}{\sqrt{2}} = 66,7 \, V$

Given that the root mean square current is 3% of the root mean square voltage, we find:

$I_{rms} = 0,03 \times V_{rms} = 0,03 \times 66,7 = 2,00 \, A$

Now substituting these values to find average power:

$P_{average} = 66,7 V \times 2,00 A = 133,9 W$

To convert the AC generator to a DC generator, the following alterations must be made:

1. Replace the slip rings with a split-ring commutator. This will allow for the conversion of the alternating current to direct current by rectifying the output.

To find the resistance (R) of the light bulb, we use the formula:

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

Here we can rearrange it to find R:

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

Given the ratings of 90 W and 56 W, we can find the resistance at different power ratings. Using the higher value of 90 W:

Let’s assume a voltage of 90 V:

$R = \frac{90^2}{90} = 90 \Omega$

Now using the lower power rating of 56 W:

$R = \frac{90^2}{56} = 144,64 \Omega$

The brightness of the light bulb will change based on the power supplied by the generator. Since the power of the generator is greater than the power of the light bulb (if we assume it operates at 90 W), it will be 'TOO BRIGHT'.

The generator can provide more energy than the light bulb can handle, leading to increased brightness which may burn out the bulb. Therefore, the correct choice is 'TOO BRIGHT.'