161. Work done in charging a capacitor is ____________
A. QV
B. 1⁄2QV
C. 2QV
D. QV2
Answer: B
We know that work done = Q2/2C.
Substituting C as Q/V, we get work done = Q/2V.
162. Energy stored in a 2000mF capacitor charged to a potential difference of 10V is?
A. 100J
B. 200J
C. 300J
D. 400J
Answer: A
From the expression of Capacitor energy
WD = CV2/2 = 100J.
163. When do we get maximum energy from a set of capacitors?
A. When they are connected in parallel
B. When they are connected in series
C. Both in series and parallel
D. Insufficient information provided
Answer: A
We get maximum energy when capacitors are connected in parallel because the equivalent capacitance is larger than the largest individual capacitance when connected in parallel. The relation between capacitance and energy is:
Energy = CV2/2, hence as the capacitance increases, the energy stored in it also increases.
164. If the charge stored in a capacitor is 4C and the value of capacitance is 2F, calculate the energy stored in it.
A. 2 J
B. 4 J
C. 8 J
D. 16 J
Answer: B
The expression for finding the value of capacitor energy is:
U = Q2/2C = 4 × 4/(2 × 2) = 4J.
165. If the charge in a capacitor is 4C and the energy stored in it is 4J, find the value of capacitance.
A. 2F
B. 4F
C. 8F
D. 16F
Answer: A
The expression for finding the value of capacitance energy is:
U = Q2/2C.
Substituting the values of U and Q, we get C = 2F.
166. If the charge in a capacitor is 4C and the energy stored in it is 4J, calculate the voltage across its plates.
A. 2V
B. 4V
C. 8V
D. 16V
Answer: A
The expression for finding the value of capacitor energy is:
U = Q2/2C.
Substituting the values of U and Q, we get C = 2F.
V = Q/C, hence V = 4/2 = 2V.
167. Calculate the energy in the 2F capacitor in the given circuit below.
A. 8.6kJ
B. 64kJ
C. 64J
D. 6.4kJ
Answer: D
From the expression value of capacitor energy is
WD = CV2/2
= 2 × 802/2 = 6400J = 6.4kJ.
168. Calculate the energy in the 4F capacitor in the given circuit below.
A. 128kJ
B. 1.28kJ
C. 12.8kJ
D. 1280J
Answer: C
From the expression of capacitor energy
WD = CV2/2
= 4 × 802/2 = 12800J = 12.8kJ.
169. Calculate the energy stored in the combination of the capacitors.
A. 192 kJ
B. 1.92 kJ
C. 19.2 kJ
D. 1920 J
Answer: C
The equivalent capacitance is: Ceq = 4+2 = 6F.
From the expression of capacitor energy
WD = CV2/2
= 6 × 802/2 = 19200J = 19.2kJ.
171. Which among the following is the correct expression for the force between the plates of a parallel plate capacitor?
A. F = ε × A × (V/x)2/2
B. F = ε × A × (V/x)2/3
C. F = ε (V/x)2/2
D. F = ε (V/x)2/3
Answer: A
The force is proportional to the square of the potential gradient and the area.
Hence the force F = ε × A × (V/x)2/2.
172. When the area of a cross-section of the plate increases, what happens to the force between the plates?
A. Increases
B. Decreases
C. Remains the same
D. Becomes zero
Answer: A
The force of attraction between the two plates of the capacitor is directly proportional to the area of a cross-section of the plates, hence an area of cross-section increases, and the force of attraction also increases.
173. When the potential gradient increases, what happens to the force between the plates?
A. Increases
B. Decreases
C. Remains the same
D. becomes zero
Answer: A
The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential gradient, hence as a potential gradient increases, the force of attraction also increases.
174. In which of the following mediums, will the force of attraction between the plates of a capacitor be greater?
A. Air
B. Water
C. Does not depend on the medium
D. Cannot be determined
Answer: B
The absolute permittivity(ε) of water is greater than that of air.
The expression relating F and ε is F = ε × A × (V/x)2/2.
From this expression, we can see that as ε increases, the force of attraction also increases.
175. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in the air. Calculate the force on each plate if the potential difference between the plates is 1kV.
A. 350 N
B. 0.035 kN
C. 0.035 N
D. 3.35 kN
Answer: C
From the given data:
A = π × d2/4 = 0.007854m2
Potential gradient = V/x = 106V/m
F = ε × A × (V/x)2/2
Therefore, F = 0.035N.
176. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in the air. The force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
A. 1 m
B. 1 cm
C. 10 cm
D. 1 mm
Answer: D
From the given data:
A = π × d2/4 = 0.007854m2
Potential gradient = V/x = 1000/a
F = ε × A × (V/x)2/2
Substituting the given values, we find a = 1mm.
177. A metal parallel plate capacitor has ‘an a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in the air. The force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.
A. 10 mm
B. 100 mm
C. 1000 m
D. 1000 cm
Answer: B
From the given data:
A = π × d2/4 = π × a2/4
Potential gradient = V/x = 106V/m
F = ε × A × (V/x)2/2
Substituting the given values, we get d = 100 mm.
178. A metal parallel plate capacitor has a 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in the air. Calculate the potential difference between the plates if the force on each plate is 0.035N.
A. 1 kV
B. 1 V
C. 2 kV
D. 2 V
Answer: A
From the given data:
A = π × d2/4 = 0.007854m2
Potential gradient = V/x = 1000 × V
F = ε × A × (V/x)2/2
Substituting the given values in the above expression, we get V = 1kV.
179. What happens to the force of attraction between the capacitors when the potential difference between the plates decreases?
A. Increases
B. Decreases
C. Remains the same
D. Becomes zero
Answer: B
The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential difference between the plates, hence as the potential difference decreases, the force of attraction also decreases.
180. What happens to the force of attraction between the capacitors when the distance of separation between the plates increases?
A. Increases
B. Decreases
C. Remains the same
D. Becomes zero
Answer: B
The force of attraction between the two plates of the capacitor is inversely proportional to the square of the distance between the plates, hence as distance increases, the force of attraction decreases.
