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Edexcel | Physics WPH0 | S 16 4 1 | 1 hour 30 minutes
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17
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A student is investigating capacitance. She sets up the circuit shown.

17
(a)
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3 Mark s

When the switch is closed there is a maximum current, which decreases to zero over

a period of time as the capacitor charges. Explain why.

Initially p.d. across C = 0V (producing maximum current)

Or initially p.d. across resistor is equal to e.m.f. of cell ( producing maximum

current )

As charge builds up on the capacitor the p.d. across resistor decreases

(reducing the current)

Or As charge builds up on the capacitor, this opposes the flow of

electrons/charge (reducing the current).

When capacitor is fully charged the p.d. across resistor is 0V (so current is

zero)

Or Eventually, the p.d. across C = e.m.f. of cell (so current is zero)

17
(b)
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2 Mark s

The student discharged the capacitor. She set the variable resistor to its maximum

resistance and closed the switch. As the capacitor charged, the student decreased

the resistance of the variable resistor so that the current remained constant until the

capacitor was fully charged.

A graph of current against time is shown.

17
(b) (I)
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3 Mark s

Determine the capacitance of the capacitor.

Capacitance =

17
(b) (II)
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17
(c)
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Capacitance can also be determined by measuring the current I at regular time

intervals, as a capacitor discharges through a resistor, and plotting a graph of ln I

against time.

17
(c) (I)
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3 Mark s

Explain how capacitance can be determined using this graph.

17
(c) (II)
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3 Mark s

A capacitor was discharged through a 390 Ω resistor. The capacitance of the

capacitor was calculated as 2200 μF.

Explain why the data for the graph for this circuit would be difficult to obtain

using an ammeter. Your answer should include a calculation.

Use of ? ? ??

t = 0.86 s

Statement recognising that the capacitor would discharge in a very short time

Or

t½ = 0.69RC

t½ = 0.59s

Statement recognising that the capacitor would discharge in a very short time

Or

time to fully discharge = 5RC

time to fully discharge = 4.29s

Statement recognising that the capacitor would discharge in a very short time

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