5. Capacitor Circuits
Ex.8 Find charge on each capacitor.
Sol. Charge on C1 = C1V1 = 2 × (20 - 5)μC
= 30 μC
Charge on C2 = C2V2 = 2 × (20 - (-10))μC
= 60 μC
Charge on C3 = C3V3 = 4 × (20 - 10)μC
= 40 μC
Ex.9 Find charge on each capacitor.
Sol. Charge on C1 = (x - 10) C1
Charge on C2 = (x - 0) C2
Charge on C3 = (x - 20) C3
Now from charge conservation at node x
(x - 10)C1 (x - 0)C2 (x - 20)C3 = 0
⇒ 2x - 20 2x 4x - 80 = 0
⇒ x = 25 Therefore
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 Ist 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 :
Q2mF = (30 - 4) 2 = 52 mC
Q1mF = (10 - 4) = 6 mC
Q2mF = (4 - (-25)) 2 = 58 mC
Find voltage across capacitor C1.
Now from charge conservation at node x and y
(x - 4)C1 + (x - 2)C2 + (x - y)C3 = 0 ⇒
2(x - 4) + 2(x - 2) (x - y) 2 = 0
6x - 2y - 12 = 0 .....(1)
(y - x)C3 + [y -(-4)]C4 (y - 0)C5 = 0 ⇒ (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
V1 = potential across C1
Q = charge on positive plate of C1
C1 = capacitance of capacitor similarly
(iv) V1 : V2 : V3 =
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 = V1 + V2 + V3
(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
Energy supplied by the battery in charging the combination
Ubattery = Q × 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
Initially, the capacitor has no charge. Applying Kirchhoff's voltage law
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) Utotal in capacitors
(iv) heat produced in the circuit
Ceq = 1 μF.
(i) Q = Ceq V = 30 μC
(ii) charge on 3μF capacitor = 30 μC
energy = = = 150 μJ
(iii) Utotal = = 450 μJ
(iv) Heat produced = (30 μC) (30) - 450 μJ = 450 μJ
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 = - μC.
Charge on capacitor of capacitance 1μF = 20 q =
Charge on capacitor of capacitance 2μF = 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 :
Q1 = C1V
Q1 = Charge on capacitor C1
C1 = Capacitance of capacitor C1
V = Potential across capacitor C1
(iv) Q1 : Q2 : Q3 : C1 : C2 : C3
The charge on the capacitor is proportional to its capacitane Q µ C
Where Q = Q1 + Q2 + Q3 ..............
(vi) Equivalent capacitance of parallel combination
Ceq = C1 C2 C3
(vii) Energy stored in the combination :
Formulae Derivation for parallel combination :
Q = Q1 Q2 Q3
= C1V C2V C3V
= V(C1 C2 C3)
Ceq = C1 C2 C3