This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Capacitance of a Multi Plate Capacitor”.

a) C=absolute permittivity*A/d

b) C=Actual permittivity*(n-1)*A/d

c) C=Actual permittivity*(n)*A/d

d) Actual permittivity*(n+1)*A/d

View Answer

Explanation: The correct expression is: C=Actual permittivity*(n-1)*A/d.

Where, n=number of plates, A=area of cross section of the plates, d=distance of separation between the plates.

2. What happens to the capacitance of a multi plate capacitor when the area of cross section of the plate decreases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

View Answer

Explanation: When the area of cross section decreases, the capacitance also decreases since it is related by the formula C=Actual permittivity*(n-1)*A/d. Here, we can see that the capacitance is directly proportional to the area of cross section.

3. What happens to the capacitance of a multi plate capacitor when the distance of separation between the plate increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

View Answer

Explanation: When the distance of separation between the plates decreases, the capacitance also decreases since it is related by the formula C=Actual permittivity*(n-1)*A/d. Here, we can see that the capacitance is inversely proportional to the distance of separation.

4. What happens to the capacitance of a multi plate capacitor when the number of plates increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

View Answer

Explanation: When the number of capacitors increases, the capacitance also increases since it is related by the formula C=Actual permittivity*(n-1)*A/d. Here, we can see that the capacitance is directly proportional to the number of capacitors.

^{2}and d=2m.

a) 10F

b) 20F

c) 30F

d) 40F

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity*(n-1)*A/d.

Thus, C=5*(3-1)*4/2= 20F.

6. Find the capacitance of a multi plate capacitor whose relative permittivity=5, n=3, A=4m^{2} and d=2m.

a) 1.77*10^{-10} F

b) 1.77*10^{10} F

c) 1.77*10^{-11} F

d) 1.77*10^{11} F

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Relative permittivity*absolute permittivity*(n-1)*A/d.

C= 5*8.85*10

^{-12}*(3-1)*4/2=1.77*10

^{-10}.

7. Find the number of plates in the multi plate capacitor having C=20F absolute permittivity=5F/m, A=4m^{2} and d=2m.

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity*(n-1)*A/d.

Substituting the given values in the equation, we get n=3.

8. Calculate the distance between the plates of the capacitor having C=20F, actual permittivity=F/m

n=3 and A=4m^{2}.

a) 1m

b) 2m

c) 3m

d) 4m

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity*(n-1)*A/d.

Substituting the given values in the equation, we get d=2m.

a) 1m

b) 2m

c) 3m

d) 4m

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity*(n-1)*A/d.

Substituting the given values in the equation, we get A=4m

^{2}.

10. Calculate the number of plates in the multi plate capacitor having C=1.77*10^{-10}F relative permittivity=5, A=4m^{2} and d=2m.

a) 1m

b) 2m

c) 3m

d) 4m

View Answer

Explanation: The formula for capacitance of a multi plate capacitor: C=Relative permittivity*absolute permittivity*(n-1)*A/d.

Substituting the given values in the equation, we get n=3.

**Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.**

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