Calculating the Capacitance MCQ Quiz in मल्याळम - Objective Question with Answer for Calculating the Capacitance - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT: 

• A capacitor is an arrangement made to store electric charge and electrical energy, using two parallel metal plates that are separated by a dielectric medium.
• The capacitance of a capacitor is the amount of charge which is required to be supplied to both the metal plates of a capacitor in order to increase the potential difference by unity and is given by

Where C = Capacitance, Q= Charge , V = Potential difference

• The charge on a body is quantized and is given by

⇒ q = ne

Where q = Charge n = number of charges and e =  charge of the electron 

CALCULATION : 

Given: n = 4 × 10⁶, e = 1.6× 10^-19 C, and V = 8V

• According to quantization, the total charge transferred to both the metal plates is given by

⇒ q = ne = 4 × 10⁶ × 1.6 × 10^-19 = 6.4 × 10^-13 = 0.64 × 10^-12 = 0.64pC  

• Since the charges where equally distributed the charges on each of the metal conductor is 

⇒ q = 0.32 pC

• The capacitance of the capacitor is given by

Substituting the given values in the above equation 

• Hence option 2 is the answer

Concept:

Capacitance:

The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q = CV

where C = capacitance.

Energy stored in capacitor:

The work done in charging the capacitor is stored as its electrical potential energy.

The energy stored in the capacitor is:

where Q = charge stored on the capacitor, U = energy stored in the capacitor, C = capacitance of the capacitor and V = Electric potential difference.

Calculation:

Given:

The capacitance of both the condensers are C and  both are connected in parallel with the battery at V volt.

Charge stores in both capacitors are q₁ and q₂ respectively

⇒ q₁ = CV

⇒ 

Now total energy stored or work done in both Capacitors is U and it is the summation of work done in the first capacitor U₁ and the second Capacitor U₂.

​∴ U = U₁+ U₂

​∴ 

​∴ 

​∴ 

⇒ So the work done in charging fully both the condensers is .

CONCEPT:

• Capacitance: The capacitance tells that for a given voltage how much charge the device can store.

Q = CV

where Q is the charge in the capacitor, V is the voltage across the capacitor and C is the capacitance of it.

EXPLANATION:

Given that Q = 100 μC; V = 10V.

We know that for a capacitance Q = CV

C = Q/V = 100 / 10 = 10 μF

So the correct answer is option 3.

CONCEPT:

• In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.
• And in the Series circuit, the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.

Ceq = C1 + C2 + C3 +......  (In parallel)

1/Ceq = 1/C1 + 1/C2 + 1/C3 +...... (in series)

CALCULATION:

• Here First three capacitors are in parallel combination with each other and the next three are in parallel with each other.
• And both are in series with each other

For parallel combination:

C_net = Ceq = C1 + C2 + C3 = C + C + C = 3 C

C^'_net =  Ceq = C1 + C2 + C3 = C + C + C = 3 C

For Series combination:

Now 1/C_resultant = 1/3C + 1/3C = 2/3C 

C_resultant = 3C/2 = 1.5 C

So the correct answer is option 3

Additional Information

• Wheatstone bridge: The Wheatstone bridge works on the principle of null deflection, i.e. the ratio of the resistances in the circuit is equal and there is no current flow in the circuit.

• It works on the principle of null deflection.
  • Under normal conditions, when the bridge is in an unbalanced condition the current flows through the galvanometer.
  • When the bridge is in a balanced condition, there will be no current flow through the galvanometer.

CONCEPT:

• The device that stores electrical energy in an electric field is called a capacitor.
• The capacity of a capacitor to store electric charge is called capacitance.
• When combination, where two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference across them, is called the parallel combination of a capacitor.
• Equivalent capacitance (Ceq) for parallel combination:

Ceq = C1 + C2

• When two or more capacitors are connected end to end and have the same electric charge on each is called a series combination of the capacitor.
• Equivalent capacitance (Ceq) in series combination:

Where C1 and C2 are two capacitors in the circuit.

CALCULATION:

Let the two capacitors with capacitances are C and C;

 if series equivalent capacitance will be C_eq, then the two capacitances are

∴ From the above example, it is clear that the equivalent of a series of capacitances is always less than the least capacitance in the series.

CONCEPT:

• The capacitance of a capacitor (C): The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V), i.e.

C = Q/V

• The unit of capacitance is the farad, (symbol F).

Parallel Plate Capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates of area A and separated by a small distance d.
•  Mathematical expression for the capacitance of the parallel plate capacitor is given by

Where C = capacitance, A = area of the two plates, εo = permittivity of free space and d = separation between the plates,

EXPLANATION:

• The capacitance of parallel conducting wires is given by:

where R is the radius of the wire and d is the distance between conductors.

• Therefore capacitance between two long parellel conducting wires depends upon the diameter of wires and the distance between them. Therefore option 3 is correct.

Calculation: 

Initially = C₀ = 4πε₀R_1........(Capacitance of the isolated sphere)

finally

⇒  = nC0 = 4πε0nR1

⇒  = n

⇒ 

⇒ 

⇒ 

∴ The ratio of their radii (R2/R1) is 

Concept:

The capacitance for a parallel plate capacitor with a dielectric constant ϵ is defined as:

d = separation between the plates

ϵ = dielectric constant

A = Area of the plates

Also, for two capacitors in series, the net capacitance is given by:

Application:

Given capacitance of the capacitor = C, i.e.

∴      …1)

Now, the dielectric slab of thickness one-fourth of the plate is inserted. This can be represented as:

This is nothing but two capacitors connected in series, as shown:

Using equation (1), we can write:

The capacitance C₂ of the dielectric will be:

Using equation (1), we can write:

C₂ = 4 KC

The equivalent capacitance will be:

CONCEPT:

The capacitance of the capacitor is written as;

   ---- (1)

where C is the capacitance, A is the area and d is the distance between the plates.

CALCULATION:

The capacitance of the capacitor of the air as the medium, C_O = 6 µF

using equation (1) we have;

 µF ----(2)

Similarly, the capacitance of the capacitor for dielectric material is written as;

 µF ---(3)

Here, K is the dielectric constant of the given material.

Now, dividing equation (3) by equation (2), we get

 ⇒K = 5

Now, the permittivity of the medium (ε), we have;

⇒ ϵ = ϵ_o K

Putting ϵ_o and K in the above equation we get;

ϵ = 5 × 8.85 × 10-12

⇒ ϵ = 0.44 × 10-10 C2 N-1 m^-2

Hence, option (1) is the correct answer.

CONCEPT: 

• Capacitance: The capacitance tells that for a given voltage how much charge the device can store.

Q = CV

where Q is the charge in the capacitor, V is the voltage across the capacitor and C is the capacitance of it.

• In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.
• And in the Series circuit, the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.

Ceq = C1 + C2 + C3 +......  (In parallel)

1/Ceq = 1/C1 + 1/C2 + 1/C3 +...... (in series)

CALCULATION:

First we will make a 600 V capacitor to the combination of 200V.

For this three 200V capacitor have to connect in series.

Equivalent capacitance 1/Ceq = 1/C1 + 1/C2 + 1/C3

1/C = 1/6 + 1/6 + 1/6

C = 2μF, and 600V

Now To make 18μF, 9 similar sets of branches had to be connected in parallel.

Ceq = C1 + C2 + C3 +......  (In parallel)

Ceq = 2 + 2 + 2 + .......... 9 times 

Ceq = 18μF, 600V

The voltage will be the same 600 V, Since voltage does not change in parallel.

Total needed capacitor = 3 × 9 = 27