Question 5
A capacitor C is charged using a supply of e.m.f. 8.0 V.
It is then discharged through a resistor R.
The circuit is shown in Fig. 5.1.
Fig. 5.1
The variation with time t of the potential
difference V across the resistor R during the
discharge of the capacitor is shown in Fig. 5.2.
Fig. 5.2
(a) During the first 1.0 s of the discharge of the capacitor,
0.13 J of energy is transferred to the resistor R.
Show that the capacitance of the capacitor C is 4500 μF.
[3]
(b) Some capacitors, each of capacitance 4500 μF with a
maximum working voltage of 6 V, are available.
Draw an arrangement of these capacitors that could
provide a total capacitance of 4500 μF for use in the circuit of Fig. 5.1. [2]
Reference: Past Exam Paper – June 2008 Paper 4 Q5
Solution:
(a)
{From the
graph,}
at t = 1.0 s, V = 2.5 V
Energy of
charged capacitor = ½ CV2
{Note that
this formula is for the energy stored in a charged capacitor (not the
energy transferred to the resistor).
Initially
(at t = 0), energy of charged capacitor = ½ × C × 82 J
At time t = 1 s, energy of charged capacitor = ½ × C × 2.52 J
During this
time, the energy is lost by the capacitor is transferred to the resistor.
Energy
transferred = 0.13 J = energy of capacitor at t=0 – energy of capacitor at t=1}
0.13 = ½ × C × (8.02 – 2.52)
C = 4500 μF
(b)
{When capacitors are connected in series, the
equivalent capacitance decreases and when connected in parallel, the equivalent
capacitance decreases.
Arranging 2 such capacitors reduces the
capacitance to half. So, by connecting another branch of 2 capacitors in
parallel, the capacitance doubles, resulting in a capacitance of 400 μF again.
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