Tag: combination of capacitors

Questions Related to combination of capacitors

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

Show answer
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. $\dfrac {C _2^2}{C _1^2}$

Show answer
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.

Show answer
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. $4$

Show answer
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

Show answer
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:

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

$= 10 + 5 + 5$

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

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?

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

  1. 0.07$A$

  2. 1.39$A$

  3. 13.9$\mathrm { A }$

  4. 139$\mathrm { A }$

Show answer
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 :

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

  1. $2E$

  2. $E$

  3. $E/2$

  4. zero

Show answer
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 $\displaystyle =\frac{total \  charge }{total \  capacity}=\frac{CE}{2C}=\frac{E}{2}$