A capacitor C is charged using a supply of e.m.f. 8.0 V. It is then discharged through a resistor R.

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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}

State two uses of capacitors in electrical circuits, other than for the smoothing of direct current.

Question 4

(a) State two uses of capacitors in electrical circuits, other than for the smoothing of direct current. [2]

(b) The combined capacitance between terminals A and B of the arrangement shown in Fig. 7.1 is 4.0 μF.

Fig. 7.1

Two capacitors each have capacitance C and the remaining capacitors each have

capacitance 3.0 μF.

The potential difference (p.d.) between terminals A and B is 12 V.

(i) Determine the capacitance C. [2]

(ii) Calculate the magnitude of the total positive charge transferred to the arrangement. [2]

(iii) Use your answer in (ii) to state the magnitude of the charge on one plate of

1. a capacitor of capacitance C,

2. a capacitor of capacitance 3.0 μF.

[2]

[Total: 8]

Reference: Past Exam Paper – June 2016 Paper 42 Q7

Solution:

(a) e.g.

storing energy

blocking d.c.

in oscillator circuits

in tuning circuits

in timing circuits

(b)

(i)

{For the capacitors in parallel, combined capacitance = 3 + 3 = 6.0 μF

The 2 capacitors C are connected in series to the parallel combination.

Combined capacitance between A and B = 4.0 μF

1 / 4 = 1/6 + 1/C + 1/C }

1/6 + 1/C + 1/C = 1/4

C = 24 μF                 

(ii)

{The total charge is determined using the total capacitance and the applied potential difference.

Combined capacitance between A and B = 4.0 μF

p.d. between A and B = 12 V}

Q = CV

Q = 4.0×10^-6 × 12

Q = 48 μC

(iii)

1.

{Current is the flow of charge. So, we need to identify the flow of current on the capacitors.

Current flows into the two capacitors C. So, the total charge appears on C.

The total charge is received by the capacitors in series.}

48 μC

2.

{At the junction of the parallel combination, the current splits equally. So, the charge is shared by the capacitors in parallel.}

24 μC