Physics

Capacitors and Capacitance

189 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

Series and parallel capacitorsParallel plate capacitorsCapacitive reactance formulasDielectrics and permittivitySpherical capacitors

Capacitors and Capacitance Questions

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:

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

$= 10 + 5 + 5$

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

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

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$\mathrm { A }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\displaystyle =\frac{total \  charge }{total \  capacity}=\frac{CE}{2C}=\frac{E}{2}$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

A lossless coaxial cable has a capacitance of $7\times { 10 }^{ -11 }$ F and an inductance of $0.39\mu H$. Calculate characteristic impedance of the cable.

  1. 65

  2. 75

  3. 66

  4. 77
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here,$C=7\times { 10 }^{ -11 }F$,
        $L=0.39\times { 10 }^{ -6 }H$

         ${ Z } _{ o }$ As the cable is lossless,
        $\therefore { Z } _{ o }\sqrt { \dfrac { L }{ C }  } =\sqrt { \dfrac { 0.39\times { 10 }^{ -6 } }{ 7\times { 10 }^{ -11 } }  } =75ohm$

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

If $C$ is the value of the capacitance of a capacitor filled with a given dielectric and $a$ is the capacitance of an identical capacitor in a vacuum, the dielectric constant, symbolized by the Greek letter kappa, , is simply expressed as

  1. $\dfrac{C}{a}$
  2. $\dfrac{a}{C}$
  3. $C\times a$

  4. None

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

The capacitance of a parallel plate capacitor filled with the given material of dielectric constant $\kappa$ is given by:

              $C=\dfrac{\kappa\varepsilon _{0}A}{d}$ .....................eq1
And the capacitance of the same capacitor without the given material is given by:
              $a=\dfrac{\varepsilon _{0}A}{d}$  .....................eq2
Dividing eq1 by eq2:
              $C/a=\kappa$

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

The insertion of a dielectric between the plates of a parallel-plate capacitor always 

  1. increases its capacitance

  2. decreases its capacitance

  3. remains same its capacitance

  4. none of the above

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

The capacitance of a parallel plate capacitor completely filled with a given material of dielectric constant $K$ is given by:

              $C=\dfrac{K\varepsilon _{0}A}{d}$ .....................eq1
And the capacitance of the same capacitor without the given material is given by:
              $C'=\dfrac{\varepsilon _{0}A}{d}$  .....................eq2
Dividing eq1 by eq2:
              $C/C'=K$
As $K>1$ always, hence $C>C'$ i.e. the insertion of a dielectric between the plates of a parallel -plate capacitor always increases the capacitance.