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Capacitors in parallel - class-XII

Description: capacitors in parallel
Number of Questions: 39
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Tags: physics electrostatics capacitance electrostatic potential and capacitance electricity and magnetism
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A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plate are $+\sigma$ and $-\sigma$ respectively. In the region between the plates the magnitude of electric field is:

  1. $\dfrac { \sigma }{ 2{ \varepsilon } _{ 0 } } $

  2. $\dfrac { \sigma }{ { \varepsilon } _{ 0 } } $

  3. 0

  4. none of these


Correct Option: B
Explanation:

The magnitude of the electric field due to a charged plate is given as:

$\vec E=\dfrac{\sigma}{2\varepsilon _0}$

The charge density on the upper plate is positive whereas on the lower plate it is negative.

Therefore, the electric field in the region between the two plates is the difference of the two fields.
It is given as:
$E _{net}=\dfrac{\sigma}{2\varepsilon _0}-\dfrac{-\sigma}{2\varepsilon _0}$

$E _{net}=\dfrac{\sigma}{\varepsilon _0}$

If the voltage across the parallel combination in increases which capacitor will undergo breakdown first?

  1. $C$

  2. $2C$

  3. Both at same moment

  4. None of these


Correct Option: C

A capacitors of $2 \mu F$ is required is an electric circuit across a potential difference of 1.0kv. A large number of $1 \mu F$ capacitors are available which can with stand a potential difference of not more than 300V The minimum number of capacitors required to achieve this is:

  1. 24

  2. 32

  3. 2

  4. 16


Correct Option: B

Two capacitors were charged to potentials 80 and 30 V. Then they connected in parallel. The potential difference across both condensers is 60 V. The ratio of the capacitances of the capacitors is

  1. 3 : 2

  2. 1 : 3

  3. 1 : 4

  4. 2 : 3


Correct Option: A

The plates of a parallel plate capacitor are $4$cm apart, the first plate is at $300$V and the second plate at $-100$V. The voltage at $3$cm from the second plate is?

  1. $200$V

  2. $400$V

  3. $250$V

  4. $500$V


Correct Option: A

Find the total capacitance for three capacitors of $10$f,$15$f and $35$f in parallel with each other?

  1. $20f$

  2. $50f$

  3. $60f$

  4. $10f$

  5. $5f$


Correct Option: C
Explanation:

Given :    $C _1 = 10$ f                  $C _2 = 15$  f                    $C _3 = 35$  f

Equivalent capacitance for parallel combination          $C _{eq} = C _1 + C _2 + C _3$
$\therefore$   $C _{eq} = 10 + 15 + 35  = 60$  $f$

Two capacitors of capacity $C _1$ and $C _2$ are connected in parallel, then the equivalent capacity is:

  1. $C _1+C _2$

  2. $C _1C _2/(C _1+C _2)$

  3. $C _1/C _2$

  4. $C _2/C _1$


Correct Option: A
Explanation:

$C _1, C _2$ are connected in parallel then equivalent capacitance is calculated as
$V=V _1=V _2$.....(1)
$q=q _1+q _2$
$\therefore CV=C _1V _1+C _2V _2$
From (1) $C=C _1+C _2$

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

  1. stored energy

  2. capacitance

  3. intensity of electric field

  4. potential difference


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 

  1. $100\ V$

  2. $200\ V$

  3. $300\ V$

  4. $150\ V$


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:

  1. $5.50A$ 

  2. $ 5.56 \times 10^2 A $

  3. $ 5.56 \times 10^3 A $

  4. $ 2.28 \times 10^4 A $


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:

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

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

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

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


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-

  1. 1 : 4

  2. 4 : 1

  3. 1 : 2

  4. 2 : 1


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

  1. 2

  2. 4

  3. 16

  4. 32


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

  1. mica

  2. tin

  3. copper

  4. stainless steel


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:

  1. potential difference

  2. electric field

  3. stored energy

  4. E.M.F. of battery


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-

  1. $(n-1)C$

  2. $(n+1)C$

  3. $C$

  4. $nC$


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:

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

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

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

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


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

  1. $6.9\mu A$

  2. $2.3\mu A$

  3. $9.2\mu A$

  4. $4.6\mu A$


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 

  1. $2$

  2. $4$

  3. $5$

  4. $9$


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?

  1. $10V$

  2. $30V$

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

  4. $60V$


Correct Option: C

Two capacitors of capacitance $2\, \mu F$ and $4\, \mu F$ are charge to $200\, V$ and $100\, V$ respectively. They are then connected in parallel to each other. What is the potential across each capacitor ?

  1. $116\, V$

  2. $133\, V$

  3. $148\, V$

  4. $164\, V$


Correct Option: B

A parallel plate capacitor has a capacity C.If a thin metal plate (M) joins the two coating A and B of the capacitor,its new capacitance is

  1. 2 C

  2. C / 2

  3. Zero

  4. Infinity


Correct Option: C

A capacitor is charged by a battery. the battery is removed and another identical uncharged capacitor is connected in parallel. the total electromagnetic energy of resulting system

  1. Decrease by a factor of 2

  2. remains the same

  3. increase by a factor of 2

  4. increase by a factor 4


Correct Option: A

A parallel plate air capacitor has capacity 'C', a distance of separation between plate is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

  1. $\dfrac{{C{V^2}}}{{2d}}$

  2. $\dfrac{{C{V^2}}}{{d}}$

  3. $\dfrac{{{C^2}{V^2}}}{{2{d^2}}}$

  4. $\dfrac{{{C^2}{V^2}}}{{{d^2}}}$


Correct Option: A

The distance between the plates of a parallel plate capacitor is $1\ mm$. What must be the area of the plate of the capacitor if the capacitance is to be $1.0\mu F$?

  1. $102.5\ m^{2}$

  2. $205.9\ m^{2}$

  3. $112.9\ m^{2}$

  4. $302.9\ m^{2}$


Correct Option: A

Two parallel plate capacitor of capacitances C and 2C are conncected in parallel and changed to a potential difference V.If  the bsttery is disconnected and the space between the plate of the capacitor of cpacince c is cpmpletely  filled with a metrial of dielectric constant K, then the potential difference a cross the capacitor will be come

  1. $3V\left( {K + 2} \right)$

  2. $\left( {\frac{{K + 2}}{{3V}}} \right)$

  3. $\left( {\frac{{3V}}{{K + 2}}} \right)$

  4. $\frac{{3\left( {K + 2} \right)}}{V}$


Correct Option: C

A parallel plate capacitor has circular plates of $8.0\ cm$ radius and are separated by $1.0\ mm$. Calculate the capacitance.

  1. $120\ pF$

  2. $140\ pF$

  3. $160\ pF$

  4. $180\ pF$


Correct Option: D

two similar capacitor are connected to potential v in  parallel order by separating them and joining them in series

  1. the potential on fees plated will be doubled

  2. the charge on the face free plates will increase

  3. the plates in contact will lose their charge

  4. more energy will be stored in the system.


Correct Option: A

For a given potential difference V how would you connect two capacitors, to obtain greater stored charge :

  1. series

  2. parallel

  3. series and parallel combined

  4. cannot say


Correct Option: B
Explanation:

Since $q=CV$ & capacitance of two capacitor is more in parallel combination as compared to series combinations i.e., $C _P > C _S$.

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 :

  1. (n+1)C

  2. (n-1)C

  3. nC

  4. C


Correct Option: B
Explanation:

Here n plates will make $(n-1)$ capacitors with capacitance $C$ and they are in parallel combination.
Thus, the resultant capacitance $=(n-1)C$

Two capacitors of capacitance $C _1$ and $C _2$ are connected in parallel across a battery. If $Q _1$ and $Q _2$ respectively be the charges on the capacitors, then $\dfrac {Q _1}{Q _2}$ will be equal to :

  1. $\dfrac {C _2}{C _1}$

  2. $\dfrac {C _1}{C _2}$

  3. $\dfrac {C _1^2}{C _2^2}$

  4. $\dfrac {C _2^2}{C _1^2}$


Correct Option: B
Explanation:

In parallel combination the both capacitors have same potential , V (say).
So, $Q _1=C _1V$ and $Q _2=C _2V$
$\therefore \dfrac{Q _1}{Q _2}=\dfrac{C _1V}{C _2V}=\dfrac{C _1}{C _2}$

These questions consist of two statements, each printed as assertion and reason. While answering these question you are required to choose any one of the following five responses.

If three capacitors of capacitances $\displaystyle { C } _{ 1 }<{ C } _{ 2 }<{ C } _{ 3 }$ are connected in parallel then their equivalent capacitance $\displaystyle $.
Reason: $\displaystyle \frac { 1 }{ { C } _{ p } } =\frac { 1 }{ { C } _{ 1 } } +\frac { 1 }{ { C } _{ 2 } } +\frac { 1 }{ { C } _{ 3 } } $

  1. If both assertion and reason are true but the reason is the correct explanation

    of assertion.

  2. If both assertion and reason are true but the reason is not the correct explanation

    of assertion.

  3. If assertion is true but reason is false.

  4. If both the assertion and reason are false.

  5. If reason is true but assertion is false.


Correct Option: C
Explanation:

If three capacitors are joined in parallel then their equivalent capacitor will be less than the least value of capacitor so

$C _p > C _s$
$\dfrac{1}{C _p} = \dfrac{1}{C _1}+\dfrac{1}{C _2}+\dfrac{1}{C _3}$ is false.

Calculate the ratio of the equivalent capacitance of the circuit when two identical capacitors are in series to that when they are in parallel?

  1. $\dfrac{1}{4}$

  2. $\dfrac{1}{2}$

  3. $1$

  4. $2$

  5. $4$


Correct Option: A
Explanation:

Let the capacitance of each capacitor be $C$.

Series combination :  Equivalent capacitance      $\dfrac{1}{C _{eq}}=\dfrac{1}{C} +\dfrac{1}{C} $                  $\implies C _{eq} = \dfrac{C}{2}$
Parallel combination :    Equivalent capacitance   $C' _{eq} = C + C = 2C$
$\therefore$   $\dfrac{C _{eq}}{C' _{eq}}  = \dfrac{C/2}{2C}  =\dfrac{1}{4}$

Two capacitors of capacity $C _{1}$ and $C _{2}$ are connected in parallel, then the equivalent capacity is:

  1. $C _{1} + C _{2}$

  2. $C _{1}C _{2}/(C _{1} + C _{2})$

  3. $C _{1}/C _{2}$

  4. $C _{2}/C _{1}$


Correct Option: A
Explanation:

Let the potential across the capacitor be V

$q=q _1+q _2$
$CV=C _1V+C _2V$
$C=C _1+C _2$ 

Complete the following statements with an appropriate word /term be filled in the blank space(s).


The equivalent capacitance C for the parallel combination of three capacitance $C _1,C _2$ and $C _3$ is given by ${C} =$..............

  1. $C _{1}+ C _{2}+ C _{3}$

  2. $\dfrac{1}{C _{1}+ C _{2}+ C _{3}}$

  3. $\left ( \cfrac{1}{\cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}}\right )$

  4. $\left ( \cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}\right )$


Correct Option: A
Explanation:

In parallel, the net capacitance is equal to the sum of individual capacitances. In this case, $C = Q/v = Q / (v _1 +v _2+v _3) = Q/v _1 + Q/v _2+ Q/v _3 = C _1 + C _2+ C _3$

Three capacitance of capacity $10 \mu F , 5 \mu F $ are connected in parallel. The total capacity will be :

  1. $10 \mu F $

  2. $ 5 \mu F $

  3. $ 20 \mu F $

  4. None of the above


Correct Option: C
Explanation:

Equivalent capacitance if they are connected in parallel is given by:

${C _{eq}} = {C _1} + {C _2} + {C _3}$

$= 10 + 5 + 5$

$= 20\;{\rm{\mu F}}$

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^12V/ms.$ what is the displacement current?

  1. 0.07$A$

  2. 1.39$A$

  3. 13.9$\mathrm { A }$

  4. 139$\mathrm { A }$


Correct Option: A
Explanation:

$\begin{array}{l} Id={ \in _{ 0 } }\dfrac { { d\phi  } }{ { dt } } ={ \in _{ 0 } }\dfrac { { d\left( { EA } \right)  } }{ { dt } } ={ \in _{ 0 } }A\dfrac { { EA } }{ { dt } } =\in \dfrac { { \pi { r^{ 2 } }dE } }{ { dt } }  \ =8.85\times { 10^{ -12 } }\times 3.14\times 25\times { 10^{ -4 } }\times { 10^{ 12 } } \ =0.07A \end{array}$

Hence,
option $(A)$ is correct answer.

A capacitor is charged by a cell of emf $E$ and the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is :

  1. $2E$

  2. $E$

  3. $E/2$

  4. zero


Correct Option: C
Explanation:

As the battery is disconnected so total is constant. i.e $Q _t=CE$
When a identical capacitor is add in parallel so the total capacitance is $C _t=C+C=2C$.
Now the common potential $\displaystyle =\frac{total \  charge }{total \  capacity}=\frac{CE}{2C}=\frac{E}{2}$

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