Tag: capacitance

Questions Related to capacitance

Two capacitor of capacity $C _{1}$ and $C _{2}$ are connected in series. The combined capacity $C$ is given by

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

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

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

  4. $\dfrac {C _{1} + C _{2}}{C _{1}C _{2}}$


Correct Option: C

Three condenser of capacitance $C(\mu F)$ are connected in parallel to which a condenser of capacitance $C$ is connected in series. Effective capacitance is $3.75$, then capacity of each condenser is

  1. $4\mu F$

  2. $5\mu F$

  3. $6\mu F$

  4. $8\mu F$


Correct Option: B
Explanation:

The effective capacitance of three condenser connected in parallel$=3C$.
When $3C$ is connected in series to $C$
$C _{Result}=\displaystyle\frac{3C\times C}{3C+C}=3.75$
$\Rightarrow C=5\mu F$.

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}}$