Tag: capacitors in parallel and series

Questions Related to capacitors in parallel and series

The equivalent capacitance of capacitors $6\mu F$ and $3\mu F$ connected in series is ______.

  1. $3\mu f$

  2. $2\mu f$

  3. $4\mu f$

  4. $6\mu f$


Correct Option: B
Explanation:

We know the equivalent capacitance of capacitors connected in series can be found by using

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

$\dfrac{1}{C _{eq}}$$=\dfrac{1}{6}$$+\dfrac{1}{3}$

$\Rightarrow C _{eq} = \dfrac{3\times 6}{3+6} = 2\mu F $
Therefore, B is correct option.

The two capacitors $2\mu F$ and $6\mu F$ are put in series, the effective capacity of the system is $\mu F$ is:

  1. $8\mu F$

  2. $2\mu F$

  3. $3/2\mu F$

  4. $2/3\mu F$


Correct Option: C
Explanation:

When connected in series 

$\dfrac{1}{C}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{6}$
$C=\dfrac{3}{2}$

When two capacitors of capacities of $3\mu F$ and $6\mu F$ are connected in series and connected to $120\ V$, the potential difference across $3\mu F$ is:

  1. $40\ V$

  2. $60\ V$

  3. $80\ V$

  4. $180\ V$


Correct Option: C
Explanation:

Equivalent capacitance is C

$\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{6}$, So $C=2\mu f$ 
Now $Q=VC=120\times 2=240\mu F$
 Now potential across $3\mu f$ is $V=\dfrac{Q}{3}=240/3=80V$

Three capacitors, $3\mu F, 6\mu F$ and $6\mu F$ are connected in series to a source of 120V. The potential difference, in volts, across the $3\mu F$ capacitor will be

  1. 24

  2. 30

  3. 40

  4. 60


Correct Option: D
Explanation:

The equivalent capacitance of the two $6\mu F$ and $6\mu F$ capacitors in series is $3\mu F$.

Hence the potential across the two capacitors, original $3\mu F$ capacitor and the equivalent $3\mu F$ capacitor is divided equally. 
Hence voltage across each of the capacitors is half of the external applied voltage, $60V.$

A capacitor of capacitance ${ C } _{ 1 }=1\mu F$ can with stand maximum voltage ${ V } _{ 1 }=6kV$ (kilo-volt) and another capacitor of capacitance ${ C } _{ 2 }=3\mu F$ can withstand maximum voltage ${ V } _{ 2 }=4kV$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of:

  1. $4kV$

  2. $6kV$

  3. $8kV$

  4. $10kV$


Correct Option: A

A capacitor of capacitance $1\mu F$ withstands a maximum voltage of $6\ kV$, while another capacitor of capacitance $2\mu F$, the maximum voltage $4\ kV$. If they are connected in series, the combination can withstand a maximum of

  1. $6\ kV$

  2. $4\ kV$

  3. $10\ kV$

  4. $9\ kV$


Correct Option: D

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


The equivalent capacitance C for the series combination of three capacitance $C _1,C _2$ and $C _3$ is given by $\cfrac{1}{C} =$..............

  1. $C _1+C _2+C _3$

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

  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: D
Explanation:

When in series, the reciprocal of the net capacitance is equal to the sum of reciprocal of individual capacitances.

The equivalent capacitance of the pair of capacitors is $C = \cfrac{Q}{V}$
$\cfrac{1}{C} = \cfrac{V}{Q} = \cfrac{(v _1 + v _2+ v _3)}{ Q }=\cfrac{v _1}{Q} + \cfrac{v _2}{Q}+ \cfrac{v _3}{Q} = \cfrac{1}{C _1} + \cfrac{1}{C _2}+\cfrac{1}{C _3}$

Which one of the following gives the resultant capacitor when capacitors are joined in series?

  1. The sum of the individual capacitors

  2. The reciprocal of the sum of the reciprocals of the individual capacitors

  3. The reciprocal of the sum of the capacitors

  4. The sum of the reciprocals of the individual capacitors


Correct Option: B
Explanation:

The resultant capacitor when capacitors are joined in series is the reciprocal of the sum of the reciprocals of the indivisual capacitors.

$\cfrac{1}{c _{eq}}=$$\cfrac{1}{c _{1}}$+$\cfrac{1}{c _{2}}$

The current in a contining a capacitance C and a resistance R in series over the applied voltage of frequency $\cfrac { \omega  }{ 2\pi  } $ by.

  1. ${ tan }^{ -1 }\left( \frac { 1 }{ \omega CR } \right) $

  2. ${ tan }^{ -1 }\left( \omega CR \right) $

  3. ${ tan }^{ -1 }\left( \omega \frac { 1 }{ R } \right) $

  4. ${ cos }^{ -1 }\left( \omega CR \right) $


Correct Option: A

A very thin metal sheet is inserted halfway between the parallel plates of an air-gap capacitor. The sheet is thin compared to the distance between the plates, and it does not touch either plate when fully inserted. The system had capacitance, $C$, before the plate is inserted.
What is the equivalent capacitance of the system after the sheet is fully inserted?

  1. $\cfrac{1}{4}C$

  2. $\cfrac{1}{2}C$

  3. $C$

  4. $2C$

  5. $4C$


Correct Option: C
Explanation:

Initially (before metal sheet inserted) the capacitance of a parallel plate capacitor is $C=\dfrac{A\epsilon _0}{d}$ where A be the area of plates and d be the separation between parallel plates.
When a metal sheet inserted fully halfway between the parallel plates, the capacitance will be divided into two capacitors $C _1, C _2$ and they are in series.
Thus, $C _1=\dfrac{A\epsilon _0}{(d/2)}=2C$ and $C _2=\dfrac{A\epsilon _0}{(d/2)}=2C$
The equivalent capacitance , $C _{12}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{2C\times 2C}{2C+2C}=C$