AQA Exam -- oscilloscope question concept

AI Thread Summary
The discussion centers on understanding voltage readings in an AC circuit involving a resistor and capacitor. It highlights that the voltage across the resistor can be negative while the capacitor's voltage remains positive due to the direction of current flow during charging and discharging phases. The oscilloscope's reference point affects how these voltages are measured, leading to potential confusion about their signs. Additionally, the circuit's specifics, including the waveform and reference points, are crucial for accurate interpretation. Overall, the relationship between the resistor and capacitor voltages is influenced by the current direction and the oscilloscope's measurement setup.
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Homework Statement
aqa paper question images below
https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74083A-QP-JUN17.PDF
3.2 question
Relevant Equations
no relevant equation i think
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can anyone explain why the voltage reading is negative for the resistor over a time period , yet the capacitor reading always stays positive ?
In AC generation the voltages goes from negative to positive, how exactly does this affect the circuit components voltages? I am not sure how AC affects the components of the circuit When charging and discharging? i understand that the capacitor is charging and discharging and the general shape of the curve.
thank you in advance
 
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Think about which way the current is flowing as the charged capacitor now discharges through the resistor (during the time when generator is at zero volts)
 
scottdave said:
Think about which way the current is flowing as the charged capacitor now discharges through the resistor (during the time when generator is at zero volts)
upwards? Is the voltage direction of the resistor always opposite to the capacitor?
essentially does the resistor current flow in the opposite direction to the capacitors?
 
Please don't think of voltage as having a "direction". Given a certain reference point, another point may have higher or lower voltage than the reference. This would read as a positive or negative voltage. You need to note that the oscilloscope has moved it's reference point. The voltage at the "top" of the resistor is equal to the output voltage of the signal generator in this circuit (when the switch is closed). So think about how the oscope is measuring across the resistor.
 
Are you sure the circuit in Figure 9 (question 02.3) applies to question 02.6? It says the waveform is shown in Figure 10 (not Figure 12). Question 02.6 is not about AC (a sinusoidal waveform), it is a voltage alternating between a constant positive value and zero.
 
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