CAPACITORS - 2 Saturday, 18 July 2020CAPACITORS -2

CAPACITORS IN PARALLEL

The capacitors in the circuit shown below are in parallel and therefore there is the same potential difference ( ) across each. For a charged capacitor and therefore

Adding gives

If the total charge ( ) is written as , this becomes

A single capacitor which has the same effect as these three must store charge, , when the potential difference across its plates is , and therefore has a capacitance, , given by

It follows therefore that three capacitors whose capacitances are and , and which are in parallel, have a total (effective) capacitance , given by

CAPACITORS IN SERIES

The capacitors shown are in series.

On connection , the battery draws electrons from plate A and leaves it with a positive charge . This induces a charge, , on B. B and M together with their connection wire constitute an isolated conductor , and the charge which appears on B results from electrons moving from M to B. M is left with charge and this induces charge on

N and so on , so the situation is as shown on the diagram on the left.

Since , , we have :

V Q = CV

Q1 = C1V

Q2 = C2V

Q3 = C3V

Q1 + Q2 + Q3 = (C1 + C2 + C3)V

Q1 + Q2 + Q3 Q

Q = (C1 + C2 + C3)V

QV

C

Q = CV

C1, C2 C3C,

C = C1 + C2 + C3

+Q−Q1

, + Q, , − Q,

C = QV

1 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020

Adding gives

i.e.

A single capacitor which has the same effect as these three would have a capacitance , given by

It follows from equations that three capacitors whose capacitances are and and which are connected in series , have a total (effective ) capacitance ,

given by

COMPARISON OF CAPACITOR AND RESISTOR NETWORK

Note, in particular , that the capacitance of a series combination of capacitors is less than the smallest individual capacitance , and the resistance of a parallel combination of resistors is less than the smallest individual resistance.

VAB =QC1

VMN =QC2

VXY =QC3

VAB + VMN + VXY = Q(1C1

+1C2

+1C3

)

V = Q(1C1

+1C3

+1C3

)

C,

V =QV

C1, C2,C3 C,

1C

=1C1

+1C2

+1C3

Capacitor network Resistor network

Series Connection

Parallel Connection Same PD

1R

=1R1

+1R2

+1R3

V = IR

Same currentR = R1 + R2 + R3

V = IR

Same PDC = C1 + C2 + C3

Q = CV

Same charge

1C

=1C1

+1C2

+1C3

Q = CV

2 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020ENERGY STORED IN A CAPACITOR

Once the charging of a capacitor has begun , the addition of electrons to the negative plate involves doing work against the repulsive forces of the electrons which are already there. Equally, the removal of electrons from the positive plate requires that work is done against the attractive forces of the positive charges on that plate. The work which is done is stored in the form of electrical potential energy.

Consider a capacitor whose capacitance is Suppose that it is partially charged , so that the charge on its plates is and the potential difference between the plates is . Suppose now, that the charge increases to ( ). This involves moving a charge from one plate to the other. If is small , can be considered unchanged by this process, in which case the work done , , is give as

The total work done , , in increasing the charge from to is therefore given by

i.e.

Writing for and making use of leads to

where

= the energy stored in the charged capacitor (J)

the charge on the plates (C)

the PD across the plates (V)

the capacitance (F)

C .Q

V Q + δQδQ δQ V

δW

δW = VδQ

∴ δW =QC

δQ

W 0 QO

W = ∫Qo

0

QC

dQ

W =Q2

O

2CQ Qo Q = CV

W =Q2

2C=

12

CV 2 =12

QV

W

Q =

V =

C =

3 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020JOINING TWO CAPACITORS

The figure shows two charged capacitors whose capacitances are and .

On closing S, charge flows until the potential difference across each capacitor is the same . The final charge, potential difference and energy of each capacitor can be calculated by making use of :

(i) there is no change in the amount of charge

(ii) the two capacitors acquire the same potential difference

(iii) the capacitors are in parallel therefore the capacitance of the

combination is given by

It turns out that unless the initial potential differences of the capacitors are the same , the total energy stored by the capacitors decrease when they are joined together.

(We will illustrate the above while solving some problems related to the circuit.)

Energy is dissipated as heat in the connecting wires when charge flows from one capacitor to the other, and this accounts for the decrease in stored energy. (If two capacitors are placed in contact without using connecting wires , the ‘lost’ potential energy goes into producing a spark as the terminals of the capacitors approach each other.)

Exercises

Question 1

Calculate the charges on the capacitors and the potential die fear never across each.

Solution :

The capacitance of the parallel combination of Z and Y is given by as

C1 C2

C = C1 + C2

CZYC = C1 + C2

CZY = 2μF + 4μF = 6μF

4 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020The circuit is therefore equivalent to a capacitor in series with a capacitor

and the total capacitance is given by

i.e.

For the circuit as a whole the total charge , is given by

i.e.

coulombs

Therefore

Charge on X = coulombs

and

Charge on Z+Y = coulombs

(note that [X} and {Z+Y] are in series and have the same charge )

For X:

If the PD across X is , then from

It follows therefore that the PD across each of Y and Z is

For Y:

If the charge on Y is , then by

coulombs

For Z:

coulombs

3μF 6μF

C1C

=1C1

+1C2

1C

=1

3 × 10−6+

16 × 10−6

=3

6 × 10−6

C = 2 × 10−6F

Q,

Q = CV

Q = (2 × 10−6) × (120)

Q = 240 × 10−6

240 × 10−6

240 × 10−6

VX Q = CV

120 × 10−6 = (3 × 10−6) × (VX)

VX = 80V

120 − 80 = 40V

QY Q = CV

QY = (40) × (2 × 10−6)

QY = 80 × 10−6

QZ = (40) × (4 × 10−6)

QZ = 160 × 10−6

5 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020

Question 2

A capacitor (X) is charged by a 40V supply and is then connected across an uncharged capacitor (Y). Calculate :

(i) the final PD across each

(ii) the final charge on each

(iii) the initial and final energies stored by the capacitor

Solution :

If the initial charge on X is , then

i.e. coulombs

The situation after the capacitors have been connected are shown below , where and are the final charges on and respectively and is the final PD.

The capacitors are in parallel , and therefore the total capacitance is given as

Total capacitance =

The total charge is unchanged , and therefore , and therefore

Total charge = coulombs

Applying to the combination gives

i.e.

5μF20μF

Qo

Qo = (5 × 10−6) × (40)

Qo = 200 × 10−6

QX QY X Y V

5 + 20 = 25μF

200 × 10−6

Q = CV

200 × 10−6 = (V ) × (25 × 10−6)

V = 8V

6 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020For X:

coulombs

For Y:

coulombs

The energy of a charged capacitor is given by , and therefore

Initial energy =

Final energy =

Question 3

Calculate effect of doubling the separation of the plates of a parallel-plate capacitor on the energy stored by the capacitor:

(i) when the capacitor is isolated ,

(ii) when the capacitor is connected to a battery

In each case account for the energy changes which occur.

Solution

The capacitance of a parallel-plate capacitor is given by , and therefore

doubling the separation of the plates halves the capacitance.

Isolated Capacitor

When the capacitor is isolated the charge on the plates is constant. Since , halving causes to double. The energy stored in the capacitor is given by

and therefore , since is halved and is constant , the energy stored by

the capacitor doubles. (Note : we use in this case and not the expression ,

, as both C and V is changing ).

QX = (5 × 10−6) × (8)

QX = 40 × 10−6

QY = (20 × 10−6) × (8)

QY = 160 × 10−6

12

CV 2

12

(5 × 10−6) × (40)2 = 4 × 10−3J

12

(5 × 10−6) × (8)2 +12

× (20 × 10−6) × (8)2 = .8 × 10−3J

C =εAd

C = QVC V

W =Q2

2CC Q

W =Q2

2CW =

12

CV 2

7 Mr. P. Moktan S.A.S

CAPACITORS - 2 Saturday, 18 July 2020Battery Connected

When the capacitor is connected to a battery the potential difference across the

capacitor is constant. In this case it is useful to make use of . Thus,

since is halved and is constant, the energy stored by the capacitor halves .

When the capacitor is isolated work has to be done in order to put the positive charge on one plate away from the negative charge on the other . The work which is done is equal to the increase in the electric energy stored by the capacitor.

When the capacitor is connected to a battery the decrease in capacitance results in a decrease in the amount of charge stored by the capacitor. This charge is returned to the battery. The decrease in energy is as a result of the capacitor discharging.

W =12

CV 2

C V

8 Mr. P. Moktan S.A.S