Tag: capacitance

Questions Related to capacitance

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

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

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Correct Option: A

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

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

  1. series

  2. parallel

  3. series and parallel combined

  4. cannot say

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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 :

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

  1. (n+1)C

  2. (n-1)C

  3. nC

  4. C

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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 :

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

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

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

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

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

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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 } } $

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

  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.

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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?

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

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

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

  3. $1$

  4. $2$

  5. $4$

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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:

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

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

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

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

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

Show answer
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 ${C} =$..............

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

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

Show answer
Correct Option: A
Explanation:

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

A $1\mu F$ capacitor is charged to 200 V and then connected in parallel (+ve to +ve) with a $4\mu F$ capacitor charged to 100 V. The resultant potential difference is :

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

  1. 120 V

  2. 60 V

  3. 180 V

  4. 150 V

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Correct Option: A
Explanation:

Total charge of the system $=C _1V _1+C _2V _2 = 600\mu C$
Since the capacitors are connected in parallel , the potential across them should be the same. Let the charge across $1\mu F$ capacitor be $q\mu C$.
$ q/1 = (600-q)/4 \Rightarrow 5q=600 \Rightarrow q =120 $
Potential across the capacitors $=Q/C = 120V$

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

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

  1. $10 \mu F $

  2. $ 5 \mu F $

  3. $ 20 \mu F $

  4. None of the above

Show answer
Correct Option: C
Explanation:

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

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

$= 10 + 5 + 5$

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