Tag: combination of capacitors

Questions Related to combination of capacitors

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.

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