Tag: capacitance

Quizzes (4) Questions (92)

Questions Related to capacitance

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

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When two capacitors charged to V are connected in parallel, they have charge Q = CV each. If they are disconnected and reconnected in series, the total potential across the combination is V + V = 2V. Thus, the potential on the free plates is doubled.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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. $\dfrac {C _2^2}{C _1^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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.
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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   $C' _{eq} = C + C = 2C$
$\therefore$   $\dfrac{C _{eq}}{C' _{eq}}  = \dfrac{C/2}{2C}  =\dfrac{1}{4}$

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

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. $\left ( \cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}\right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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 :

  1. 120 V

  2. 60 V

  3. 180 V

  4. 150 V
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=Q/C = 120V$

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

$= 10 + 5 + 5$

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