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**5. Capacitor Circuits**

**Ex.8 Find charge on each capacitor.**

**Sol.** Charge on C_{1} = C_{1}V_{1} = 2 Ã— (20 - 5)Î¼C

= 30 Î¼C

Charge on C_{2} = C_{2}V_{2} = 2 Ã— (20 - (-10))Î¼C

= 60 Î¼C

Charge on C_{3} = C_{3}V_{3} = 4 Ã— (20 - 10)Î¼C

= 40 Î¼C

**Ex.9 Find charge on each capacitor. **

**Sol.** Charge on C_{1} = (x - 10) C_{1}

Charge on C_{2} = (x - 0) C_{2}

Charge on C_{3} = (x - 20) C_{3}

Now from charge conservation at node x

(x - 10)C_{1} (x - 0)C_{2} (x - 20)C_{3} = 0

â‡’ 2x - 20 2x 4x - 80 = 0

â‡’ x = 25 Therefore

so

**Ex.10 In the given circuit find out the charge on each capacitor. (Initially they are uncharged)**

**Sol. **Let potential at A is 0, so at D it is 30 V, at F it is 10 V and at point G potential is -25V. Now apply Kirchhoff's I^{st} law at point E. (total charge of all the plates connected to 'E' must be same as before i.e. 0)

Therefore, (x - 10) + (x - 30) 2 +(x 25) 2 = 0

5x = 20

x = 4

Final charges :

Q_{2mF} = (30 - 4) 2 = 52 mC

Q_{1mF} = (10 - 4) = 6 mC

Q_{2mF} = (4 - (-25)) 2 = 58 mC

**Ex.11**

**Find voltage across capacitor C _{1}.**

**Sol.**

Now from charge conservation at node x and y

for x

(x - 4)C_{1} + (x - 2)C_{2} + (x - y)C_{3} = 0 â‡’

2(x - 4) + 2(x - 2) (x - y) 2 = 0

6x - 2y - 12 = 0 .....(1)

For y

(y - x)C_{3} + [y -(-4)]C_{4} (y - 0)C_{5} = 0 â‡’ (y - x)2 (y 4) 2 y 2 = 0

= 6y - 2x 8 = 0 .....(2)

eq. (1) & (2)

y = - 3 Therefore

x = 7 Therefore

So potential difference = x - y =

**6. Combination of Capacitors :**

**6.1 Series Combination :**

(i) When initially uncharged capacitors are connected as shown, then the combination is called series combination

(ii) All capacitors will have same charge but different potential difference across then.

(iii) We can say that

V_{1} = potential across C_{1}

Q = charge on positive plate of C_{1}

C_{1} = capacitance of capacitor similarly

(iv) V_{1} : V_{2} : V_{3} = _{}

We can say that potential difference across capacitor is inversely proportional to its capacitance in series combination.

**Note : **In series combination the smallest capacitor gets maximum potential.

(v) , ,

Where V = V_{1 + } V_{2} + V_{3}

(vi) Equivalent Capacitance :

Equivalent capacitance of any combination is that capacitance which when connected in place of the combination stores same charge and energy as that of the combination

In series :

........................

In series combination equivalent is always less then smallest capacitor of combination.

(vii) Energy stored in the combination

U_{combination} =

U_{combination} =

Energy supplied by the battery in charging the combination

U_{battery} = Q Ã— V = Q . _{ = }

Half of the energy supplied by the battery is stored in form of electrostatic energy and half of the energy is converted into heat through resistance.

**Derivation of Formulae :**

Meaning of equivalent capacitor

Now,

Initially, the capacitor has no charge. Applying Kirchhoff's voltage law

in general

**Ex.12 Three initially uncharged capacitors are connected in series as shown in circuit with a battery of emf 30V. Find out following :**

**(i) charge flow through the battery,**

**(ii) potential energy in 3 mF capacitor. **

**(iii) U _{total} in capacitors **

**(iv) heat produced in the circuit**

**Sol. **

C_{eq} = 1 Î¼F.

(i) Q = C_{eq} V = 30 Î¼C

(ii) charge on 3Î¼F capacitor = 30 Î¼C

energy = = = 150 Î¼J

(iii) U_{total} = = 450 Î¼J

(iv) Heat produced = (30 Î¼C) (30) - 450 Î¼J = 450 Î¼J

**Ex.13 Two capacitors of capacitance 1 mF and 2mF are charged to potential difference 20 V and 15 V as shown in figure. If now terminal B and C are connected together terminal A with positive of battery and D with negative terminal of battery then find out final charges on both the capacitor.**

Now applying kirchhoff voltage law

- 40 - 2q - 30 - q = - 60

3q = - 10

Charge flow = - Î¼C.

Charge on capacitor of capacitance 1Î¼F = 20 q =

Charge on capacitor of capacitance 2Î¼F = 30 q =

**6.2 Parallel Combination :**

(i) When one plate of one capacitor is connected with one plate of the other capacitor, such combination is called parallel combination.

(ii) All capacitors have same potential difference but different charges.

(iii) We can say that :

Q_{1} = C_{1}V

Q_{1} = Charge on capacitor C_{1}

C_{1} = Capacitance of capacitor C_{1 }

V = Potential across capacitor C_{1}

(iv) Q_{1} : Q_{2} : Q_{3} : C_{1} : C_{2} : C_{3}

The charge on the capacitor is proportional to its capacitane Q Âµ C

(v)

Where Q = Q_{1} + Q_{2} + Q_{3} ..............

- Maximum charge will flow through the capacitor of largest value.

(vi) Equivalent capacitance of parallel combination

C_{eq} = C_{1} C_{2} C_{3}

- Equivalent capacitance is always greater then the largest capacitor of combination.

(vii) Energy stored in the combination :

=

- Half of the energy supplied by the battery is stored in the form of electrostatic energy and half of the energy is converted into heat through resistance.

**Formulae Derivation for parallel combination :**

Q = Q_{1} Q_{2} Q_{3}

= C_{1}V C_{2}V C_{3}V

= V(C_{1} C_{2} C_{3})

C_{eq} = C_{1} C_{2} C_{3}

In general

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