Tag: capacitance

Questions Related to capacitance

A parallel plate capacitor is connected to a battery and the distance between the plates is decreased then which quantity is same for the parallel plate capacitor

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. stored energy

  2. capacitance

  3. intensity of electric field

  4. potential difference

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Correct Option: A

Capacity of a parallel plate capacitor is $2\mu F$. The two plates of the capacitor are given $400\mu C$ and $-200\mu C$charges respectively. The potential difference between the plates is 

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $100\ V$

  2. $200\ V$

  3. $300\ V$

  4. $150\ V$

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Correct Option: D

A parallel plate capacitor consist of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of $5 \times {10^{13}}V{s^{ - 1}}$ , then the value of displacement current is:

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $5.50A$ 

  2. $ 5.56 \times 10^2 A $

  3. $ 5.56 \times 10^3 A $

  4. $ 2.28 \times 10^4 A $

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Correct Option: A

When $n$ identical capacitors are connected in series their effective capacity is $C _s$ and when they are connected in parallel their effective capacity is $C _p$. The relation between $C _p$ and $C _s$ is:

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $C _p = n \,C _s$

  2. $C _p = \dfrac{C _s}{n}$

  3. $C _p = n^2 \,C _s$

  4. $C _p = \dfrac{C _s}{n^2}$

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Correct Option: C

A capacitor $ C _1 = 4 \mu F $ is connected in series with another capacitor $ C _2 = 1 \mu F $. the combination is connected across a d.c. source of voltage 200 V. the ration of potential across $ C _1 $ and $C _2 $ is-

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. 1 : 4

  2. 4 : 1

  3. 1 : 2

  4. 2 : 1

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Correct Option: C

From a supply of identical capacitors rated $8\;\mu F, 250 \;V$ the minimum number of capacitors required to form a composite of $16\;\mu F, 1000 \;V$ is

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. 2

  2. 4

  3. 16

  4. 32

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Correct Option: D
Explanation:

The number of capacitance to be connected in series $\displaystyle n=\frac{voltage \  rating \ required}{voltage\  rating \ of \ a \ capacitor \ given}=\frac{1000}{250}=4$
Equivalent capacitance, $\displaystyle C _{eq}=(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})^{-1}=(\frac{4}{8})^{-1}=2$
Number of rows required $\displaystyle =\frac{capacitance \ required}{capacitance \  of \ each \  row}=\frac{16}{2}$
Thus the minimum number of capacitors to be required $=4\times 8=32$

In order to increase the capacity of parallel plate condenser one should introduce between the plates, a sheet of

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. mica

  2. tin

  3. copper

  4. stainless steel

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Correct Option: A
Explanation:

mica as it is having higher conductivity$.$

So$,$ increase the capacity of parallel plate condenser one should introduce between the plates$,$ a sheet of $mica.$
Hence,
option $(A)$ is correct answer.

A parallel plate capacitor is connected to a battery and a dielectric slab is inserted between the plates, then which quantity increase:

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. potential difference

  2. electric field

  3. stored energy

  4. E.M.F. of battery

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Correct Option: A

A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $C$, then the resultant capacitance is-

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $(n-1)C$

  2. $(n+1)C$

  3. $C$

  4. $nC$

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Correct Option: A
Explanation:

$n$ plates connected alternately give rise to $\left(n – 1\right)$ capacitors connected in parallel $\therefore$, Resultant capacitance $=\left(n – 1\right)C$.

A parallel plate condenser has plates of area $200\mathrm { cm } ^ { 2 }$ and separation $0.05\mathrm { cm } .$ The space between plates have been filled with a dielectric having $\mathrm { k } = 8$ and then charged to $300$ volts. The stored energy:

physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

  1. $121.5 \times 10 ^ { - 6 } \mathrm { J }$

  2. $28 \times 10 ^ { - 6 } \mathrm { J }$

  3. $112.4 \times 10 ^ { - 5 } \mathrm { J }$

  4. $1.6 \times 64 \times 10 ^ { - 5 } \mathrm { J }$

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Correct Option: A
Explanation:

$C = \dfrac{{KA{ \in _0}}}{d} = \dfrac{{3 \times 200 \times {{10}^{ - 4}} \times 8.85 \times {{10}^{ - 12}}}}{{5 \times {{10}^{ - 4}}}}$

$ = 27 \times {10^{ - 10}}F$
$E = \dfrac{1}{2}C{V^2} = \frac{1}{2} \times 27 \times {10^{ - 10}} \times 300$
$ = \dfrac{{243}}{2} \times {10^{ - 6}}$
$ = 121.5 \times {10^{ - 6}}J$
Hence,
option $(A)$ is correct answer.