Photo AI
A student carries out a series of experiments to investigate alternating current - Scottish Highers Physics - Question 12 - 2018
Question 12
A student carries out a series of experiments to investigate alternating current. (a) A signal generator is connected to an oscilloscope and a circuit as shown. Th... show full transcript
Worked Solution & Example Answer:A student carries out a series of experiments to investigate alternating current - Scottish Highers Physics - Question 12 - 2018
Step 1
(i) Determine the peak voltage of the output of the signal generator.
Answer
To find the peak voltage, we can refer to the Y-gain setting on the oscilloscope, which is 1.0 V/div. The output waveform shows 3 divisions vertically indicating the peak voltage:
Peak Voltage = Number of Divisions × Y-gain
Thus, the peak voltage of the output is 3.0 V.
Step 2
(ii) Determine the frequency of the output of the signal generator.
Answer
The timebase setting on the oscilloscope is 0.5 s/div. The waveform shows that it completes one full cycle in 5 divisions:
Frequency can be calculated using the formula: f = rac{1}{T} Where T is the time period of one cycle.
Time for one cycle = Number of divisions × Timebase setting
Thus, the frequency is: f = rac{1}{2.5 ext{ s}} = 0.4 ext{ Hz}
Step 3
(iii) Explain these observations.
Answer
The red LED lights up only when the ammeter gives a positive reading, indicating that current flows through it in the forward direction. Conversely, the green LED lights up when the ammeter indicates a negative reading, meaning the current is now flowing in the opposite direction, thereby forward biasing the green LED instead. This behavior is due to the two LEDs being positioned in opposite directions, allowing current to conduct only when they are forward biased. The changing polarity of the alternating current leads to these observations as each LED responds to the direction of current.
Step 4
(b) Determine the r.m.s. voltage across the 82 Ω resistor.
Answer
In this case, we need to calculate the r.m.s. voltage using:
oot{2}{}}$$ We have already determined the peak voltage to be 3.0 V: $$V_{rms} = 3.0 ext{ V} imes rac{1}{ oot{2}{}} \\ = 3.0 ext{ V} imes 0.707 \approx 2.12 ext{ V}$$ Thus, the r.m.s. voltage across the 82 Ω resistor is approximately 2.12 V.