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Electricity and Magnetism - SS2 Physics Past Questions and Answers - page 2

11

The unit of capacitance is:

A

Ohms

 

B

Volts

 

C

Coulombs

 

D

Farads

correct option: d
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12

The formula for capacitance is given by:

A

C = Q / V

 

B

C = V / Q

 

C

C = ε0A / d

 

 

D

C = QV

correct option: a
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13

When the distance between the plates of a parallel-plate capacitor is halved, the capacitance:

A

Doubles

 

B

Halves

 

C

Remains the same

 

D

Becomes zero

correct option: a
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14

Capacitors in series have a total capacitance that is:

A

Equal to the sum of the individual capacitances

 

B

Less than the smallest individual capacitance

 

C

Greater than the largest individual capacitance

 

D

Inversely proportional to the individual capacitances

correct option: d
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15

The energy stored in a charged capacitor is given by:

A

U = (1/2) CV2

 

B

U = (1/2) Q2

 

C

U = (1/2) QV

 

D

U = (1/2) CV

correct option: a
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16

The dielectric material inserted between the plates of a capacitor:

A

Increases the capacitance

 

B

Decreases the capacitance

 

C

Does not affect the capacitance

 

D

Makes the capacitor non-functional

correct option: a
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17

A capacitor with a capacitance of 10 μF is connected to a voltage source of 12 V. What is the charge stored on the capacitor?

Q = CV 

substituting,

(10 μF) x (12 V) = 120 μC

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18

Two capacitors, C₁ = 5 μF and C₂ = 10 μF, are connected in series. What is the total capacitance of the combination? 

The total capacitance, Ctotal, is given by the reciprocal of the sum of the reciprocals of the individual capacitances:

1/Ctotal = 1/C₁ + 1/C₂ 

= 1/(5 μF) + 1/(10 μF) 

= (1/5 + 1/10) μF 

= 3/10 μF

Ctotal = 10/3 μF = 3.33 μF

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19

Three capacitors, C₁ = 4 μF, C₂ = 6 μF, and C₃ = 8 μF, are connected in parallel. What is the total capacitance of the combination?

The total capacitance of capacitors in parallel is the sum of the individual capacitances:

Ctotal = C₁ + C₂ + C₃ 

= 4 μF + 6 μF + 8 μF = 18 μF

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20

A capacitor is charged to a voltage of 200 V and stores a charge of 50 μC. What is the capacitance of the capacitor?

C = Q/V 

= (50 μC) / (200 V) = 0.25 μF

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