Physics

Capacitors and Capacitance

189 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

Series and parallel capacitorsParallel plate capacitorsCapacitive reactance formulasDielectrics and permittivitySpherical capacitors

Capacitors and Capacitance Questions

Multiple choice

Directions: Carry Two Marks Each.

An LC tank circuit consists of an ideal capacitor C that is connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is

  1. $\frac{1}{2 \pi \sqrt{LC}}$
  2. $\frac{1}{2 \pi \sqrt{LC}}\sqrt{1 - R^2 \frac{C}{L}}$
  3. $\frac{1}{2 \pi \sqrt{LC}}\sqrt{1 - \frac{L}{R^2 C}}$
  4. $\frac{1}{2 \pi \sqrt{LC}}\sqrt{1 - R^2 \frac{C}{L}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice

Two conductors insulated from each other are charged by transferring electrons from one conductor to another. A potential difference of 200 V was produced on transferring 6.25 x 1015 electrons from one conductor to another. The capacity of the system will be _________.

  1. 2 x 10-1 F

  2. 5 x 10-6 F

  3. 2 x 105 F

  4. 1.95 x 1036 F

  5. 51.28 x 1034 F
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

q = c x v c = q/v q = n x e c = n x e/v n = 6.25 x 1015 e = 1.6 x 10-19 v = 200 c = (6.25 x 1015) x (1.6 x 10-19)/200 c = 5 x 10-6 f 

Multiple choice

An air capacitor C, is connected to a battery of emf V. It acquires a charge Q and energy E. The capacitor is then disconnected from the battery and a dielectric slab is introduced between the plates. Which of the following is true?

  1. The values of Q and C increase but values of V and E decrease.

  2. The values of V and Q decrease but values of E and C increase.

  3. V remains unchanged but the values of Q, E and C increase.

  4. Q remains unchanged, C increases and the values of V and E decrease.

  5. The values of Q, C, V and E, increase.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The charge remains the same. Capacitance increases. Hence, the potential V = q/C decreases and the energy E = 1/2(q2/C).

Multiple choice

A capacitor of 5 µF is charged to a potential difference of 200 V. If it is discharged through two resistors of 700 and 300 ohms in series, what is the energy dissipated in each of the two resistors?

  1. 0.07 J and 0.03 J

  2. 0.03 J and 0.07 J

  3. 9 J and 49 J

  4. 21 J and 21 J

  5. 0.1 J and 0.1 J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Current in the circuit is given by I2[700+300] = 1/2  x 5 x 10-6  x (200)2 This gives I2 = 0.1 x 103 Hence, H1 = I2 R1 = 0.07 J             H2 = I2 R2 = 0.03 J

Multiple choice

The distance between plates of parallel plates condenser is 2 cm and area of each plate is 31.41 cm2. A glass slab of thickness 1 cm and area equal to the area of the plate is kept inside it. Find out the capacity if the relative permittivity of glass slab is 6.

  1. 0.238 x 10-11 F

  2. 26.92 x 10-2 F

  3. 18.5 x 10-6 F

  4. 31.41 x 10-2 F

  5. 31.41 x 10-6 F
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

C = A єo / {(d - t) + (t/єr)}   = {(31.41 x 10-4)/(36π x 109)}  x  {1/(0.02 – 0.01) + (0.01 / 6 )}   =  (31.41 x 10-4)/(36π x 109)  x  x 6 / 0.07  = 0.238 x 10-11 F 

Multiple choice

Two capacitors of capacitances C1 and C2 are connected across 200 V power supply. The potential drop across C1 is 120 V. A capacitor of capacitance 2µF is connected in parallel with C1 and the potential drop across C2 becomes 160 V. What are the values of V C1 and C2 (in µF)?

  1. 2.4 and 1.6

  2. 0.4 and 0.6

  3. 0.6 and 0.4

  4. 0.6 and 0.13

  5. 0.13 and 0.6

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initially the charge is same on both capacitors. Therefore, 120 C1 = 80 C2 In the second step, (C1 + 2) 40 = 160 C2. This gives C1 = 0.4 µF and C2 = 0.6 µF.