Capacitor Questions

Q10. A student sets up the circuit shown in the diagram.

(a)

(i) She moves the switch from X to Y. Explain what happens to the capacitor.

The capacitor discharges (loses charge) exponentially. It will lose charge very rapidly at first and then the rate of charge movement will decrease with time.

(2 marks max)

(ii) Sketch a graph to show how current varies with time from the moment the switch touches Y. Indicate typical values of current and time on the axes of your graph.

V = IR

Imax = Vmax/R = 10/5000 = 0.0020A = 2.0 mA

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See the notes on capacitor discharge

Time constant for the circuit = RC = 5000 x 10 x 10-6 = 0.050s

and 'half life' is roughly 0.7RC = 0.035s = 35 ms

Therefore when plotting your sketch graph you need to indicate this data!

 

(3 marks)

(iii) Describe how the graph would appear when the switch is moved back to X.

It would be the same graph shape but the line would be on the negative side of current axis because the current would flow in the opposite direction

(2 marks)

(b) Calculate the maximum energy stored on the capacitor in this circuit.

Energy stored = ½CV2

Energy stored = ½ (10 ×10-6) 102

Rnergy stored = 5 x10-4 J

(2 marks)

(c) The student wants to produce a time delay equal to the time it takes for the potential difference across the capacitor to fall to 0.07 of its maximum value. Calculate this time delay.

V = V0e-t/RC

0.7 = 10 e-t/RC

0.07 = e-t/RC

ln 0.07 = -t/RC

t = -ln 0. 7 x 0.05

t = 0.13 s

(2 marks)

(Total 11 marks)