0

Capacitors in Series and Parallel

Description: Engineering Entrance Physics
Number of Questions: 15
Created by:
Tags: Engineering Entrance Physics Capacitors in Series and Parallel Capacitors in Parallel
Attempted 0/15 Correct 0 Score 0

What is the C.G.S. unit of capacity?

  1. Stat faraday

  2. Dyne/stat coulomb

  3. Erg/stat coulomb

  4. erg

  5. Erg/sec


Correct Option: A
Explanation:

C.G.S. unit of capacity

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


Correct Option: B
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 

Find out the common potential of two spheres of radius 30 cm and 60 cm, which are charged to potential 60 volt to 55 volt, and are connected by a wire.

  1. 44.3

  2. 17 x 10-3

  3. 35.29 x 10-2

  4. 28.3 x 10-1

  5. 56.66


Correct Option: E
Explanation:

Common potential v = (c1v+ c2v2)/c+ c2 Putting the values in above formula V = 56.66

A Leyden jar has a radius of 20 cm the height of the tin foils is 10 cm and thickness of glass is 3 cm. If state-independent contextuality (SIC) of glass is 6.4, then find out its capacity.

  1. 4.74 x 10-10

  2. 4.26

  3. 0

  4. 6144 x 10-5

  5. 6.82 x 10-9


Correct Option: A
Explanation:

C = 4π єo єr[ {r2+2rh}/4t] t = thickness 4π єo =1/9 x 109 єr = 6.4, r = 0.2, h = 0.1, t = 0.03 Putting  the values in above formula, C = 4.74 x 10-10

If five identical condensers are connected in series, then their equivalent capacity is 6 microfarad. Find out the equivalent capacity when they are connected in parallel.

  1. 150 µF

  2. 6 µF

  3. 55 µF

  4. 5 µF

  5. 16 µF


Correct Option: A
Explanation:

C= C1 / n Cs  = 6 µF n = 5 C1 = Cs x  n      = 30 µF Cp=n x C1     = 5 x 30     = 150 µF

Find the capacity of parallel plate condenser consisting of two plates of area 100 cm. The plates are separated by a mica sheet of relative permittivity 6 and thickness 1 mm.

  1. 53.1 x 10-11

  2. C = 53.1 x 10-17

  3. 53.1 x 10-7

  4. 53.1 x 10-13

  5. 1.475 x 10-17


Correct Option: A
Explanation:

C = єoєr A/d C = {(8.85  x 10-12) x 6 x (100 x 10-4)}/1 x 10-3

C = 53.1 x 10-11

There is a model of a red blood cell capacitor. The radius of the blood cell is about 6 µm and the membrane wall of 0.1 µm is considered to be a dielectric material with a dielectric constant of the cell is ________.

  1. 0.2 x 10-12 F

  2. 0.18 x 10-2 F

  3. 0.18 x 10-9 F

  4. 0.2 x 10-19 F

  5. 0.36 x 10-3 F


Correct Option: A
Explanation:

  C =( 4π є0єrRr)/(R-r) = 0.2 x 10-12 F

  1. 0.18 x 10-2 F

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.


Correct Option: D
Explanation:

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

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


Correct Option: A
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

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


Correct Option: A
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 

If 125 identical drops each of capacity C combines to form a big drop, what is the capacity of big drop?

  1. 125 C

  2. 5/C

  3. 25/C

  4. 25 C

  5. 5 C


Correct Option: E
Explanation:

C1 = n1/3C

     = (53)1/3C

     = 5 C 

When a battery of emf 12 volt is connected to a conductor, it gets a charge of 60 µC. Find out the capacity of the conductor.

  1. 72 µF

  2. 720 µF

  3. 5 µF

  4. 48 µF

  5. 360 µF


Correct Option: C
Explanation:

C = q/v     = 60/12       = 5 µF

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


Correct Option: B
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.

A dielectric slab of thickness 4 mm is placed between the plates of a parallel plate condenser. When the distance between the plates is increased by 3.5 mm, then the capacity of the condenser remains same. Find the dielectric of the medium.

  1. 8

  2. 30 x 10-6

  3. 2 x 10-6

  4. 1.87

  5. 12.5 x 10-2


Correct Option: A
Explanation:

Initially the capacity, 

C1 = єoA/d

Let the distance is increased by x.

When a slab of thickness d is introduced, then capacity

C2 =  єoA/ (x+d/єr) since C1 = C2

Therefore, єoA/d = єoA/ (x + d/єr)

Or d = (x+d/єr)

єr = d/(d – x)

    = 4 x10-3/(4 x10-3 – 3.5 x10-3)

    = 4/0.5

 = 8

Find out the heat generated in the wire of 8 µF capacitor, which is charged to 200 volt and its plates are connected to a resistance wire.

  1. 16 x 10-2 joule

  2. 8 x 10-4 joule

  3. 32 x 10-2 joule

  4. 4 x 10-8 joule

  5. 16 x 10-4 joule


Correct Option: A
Explanation:

E = (1/2) c x v2 Putting the values in above formula E = 16 x 10-2 joule

- Hide questions