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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
(a)(i) Determine the peak voltage of the output of the signal generator.
Answer
To find the peak voltage, we use the oscilloscope settings. Given that the Y-gain setting is 1.0 V/div and the oscillation shows 3 divisions from the baseline to the peak, the peak voltage can be calculated as:
Peak Voltage = Number of divisions × Y-gain = 3 × 1.0 V = 3.0 V.
Step 2
(a)(ii) Determine the frequency of the output of the signal generator.
Answer
To find the frequency, we first need to determine the time period (T) from the timebase setting. Since the timebase is set at 0.5 s/div and there are 2 full cycles displayed across 10 divisions:
Time period for one cycle (2 cycles = 10 divisions):
T = (10 divisions) × (0.5 s/div) = 5.0 s
Frequency (f) is the reciprocal of the time period:
f = 1/T = 1/5.0 s = 0.20 Hz.
Step 3
(a)(iii) Explain these observations.
Answer
The observation relates to the behavior of LEDs in relation to current direction. LEDs will only conduct electricity when they are forward biased, meaning that the positive terminal of the power source is connected to the anode of the LED and the negative terminal to the cathode.
When the ammeter gives a positive reading, it indicates that the current is flowing in one direction, thus forward biasing the red LED. Conversely, when the ammeter shows a negative reading, the current is flowing in the opposite direction, which forward biases the green LED. This is why each LED lights only under its respective current condition.
Step 4
(b) Determine the r.m.s. voltage across the 82.Ω resistor.
Answer
To find the r.m.s. voltage (V_r.m.s) across the resistor, we first calculate the total resistance (R_total) in the circuit, which includes the resistor of 68 Ω and the 82 Ω resistor:
R_total = 68 + 82 = 150 Ω.
Using Ohm's Law and remembering that the peak voltage from part (a)(i) is 3.0 V, the r.m.s. voltage can be calculated by:
V_r.m.s = V_peak / √2 = 3.0 V / √2 ≈ 2.12 V.
The r.m.s. voltage across the 82 Ω resistor is the same since the signal generator's internal resistance is negligible.