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Date May 2016 Marks available 2 Reference code 16M.2.HL.TZ0.7
Level Higher level Paper Paper 2 Time zone Time zone 0
Command term Calculate and State Question number 7 Adapted from N/A

Question

An uncharged capacitor in a vacuum is connected to a cell of emf 12V and negligible internal resistance. A resistor of resistance R is also connected.

At t=0 the switch is placed at position A. The graph shows the variation with time t of the voltage V across the capacitor. The capacitor has capacitance 4.5μF in a vacuum.

On the axes, draw a graph to show the variation with time of the voltage across the resistor.

[2]
a.

(i) The time constant of this circuit is 22s. State what is meant by the time constant.

(ii) Calculate the resistance R.

[2]
b.

A dielectric material is now inserted between the plates of the fully charged capacitor. State the effect, if any, on

(i) the potential difference across the capacitor.

(ii) the charge on one of the capacitor plates.

[2]
c.

(i) The permittivity of the dielectric material in (c) is twice that of a vacuum. Calculate the energy stored in the capacitor when it is fully charged.

(ii) The switch in the circuit is now moved to position B and the fully charged capacitor discharges. Describe what happens to the energy in (d)(i).

[3]
d.

Markscheme

general shape starting at 12 V
crosses at 6 V

Line must not touch time axis for MP2.

Allow tolerance of one square in 12 V (start) and 6 V (crossing).

a.

(i)
the time for the voltage/charge/current «in circuit» to drop to \(\frac{1}{e}\) or 37% of its initial value

«as the capacitor discharges»

OR

time for voltage/charge/current «in circuit» to increase to \(\left( {1 - \frac{1}{e}} \right)\) or 63% of its final value

«as the capacitor charges» 

(ii)
\(R =  \ll \frac{{22}}{{4.5 \times {{10}^{ - 6}}}} =  \gg 4.9 \times {10^6}\Omega \)

b.

(i)
no change
OR
«remains at» 12 V

(ii)
increases
OR
doubles

Allow “doubles” in the light of (d).

 

c.

(i)
recognises that new capacitance is 9.0 μF 
\(E =  \ll \frac{1}{2}C{V^2} = \frac{1}{2} \times 9.0 \times {10^{ - 6}} \times {12^2} \gg  = 0.65{\rm{mJ}}\) or 6.5×10–4J 

Allow 11.8 V (value on graph at t=100s).

(ii)
energy goes into the resistor/surroundings
OR
«energy transferred» into thermal/internal energy form 

Do not accept “dissipated” without location or form.
Do not allow “heat”.

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Additional higher level (AHL) » Topic 11: Electromagnetic induction » 11.3 – Capacitance

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