Tag: capacitors in parallel

Questions Related to capacitors in parallel

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

Show answer
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.

A Parallel platecapacitor made of circular plates each of radius $R=6.0cm$ has a capacitance 100$\mathrm { pF }$ is connected to 230$\mathrm { V }$ of $\mathrm { AC }$ supply of 300 rad/sec.frequency. The rms value of displacement current

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

  1. $6.9\mu A$

  2. $2.3\mu A$

  3. $9.2\mu A$

  4. $4.6\mu A$

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Correct Option: A
Explanation:

Given$:-$

$R=6.0cm$
$C = 100pF$
$ = 100 \times {10^{ - 12}}F$
$w = 300\,rad/s$
${I _{rms}} = 230 \times 300 \times 100 \times {10^{ - 12}}$
$ = 6.9 \times {10^{ - 9}}$
$ = 6.9\mu A$
Hence, 
option $(A)$ is correct answer.

Two parallel plates have equal and opposite charge. When the space between them is evacuated. the electric field between the plates $2 \times {10^5}\,V/m.$ When the space is filed with dielectric the electric field becomes ${10^5}\,V/m$ The dielectric constant of he dielectric material is 

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

  1. $2$

  2. $4$

  3. $5$

  4. $9$

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Correct Option: A
Explanation:

Dielectric constant$:-$

$K = \dfrac{{{E _0}}}{E}$
$ = \dfrac{{2 \times {{10}^5}}}{{1 \times {{10}^5}}} = 2$
Hence,
option $(A)$is correct answer

3 identical capacitor are joined in parallel and are charged with a battery of 10V. now the battery is removed and they are joined in series with each other. in this condition what would be the potential difference between the free plates in combination?

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

  1. $10V$

  2. $30V$

  3. $\frac {10}{3}$

  4. $60V$

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Correct Option: C