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Capacitors in series - class-XII

Description: capacitors in series
Number of Questions: 31
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Tags: capacitance electrostatics physics electrostatic potential and capacitance electricity and magnetism
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Two capacitors of $1\mu F$ and $2\mu F$ are connected in series and this combination is changed upto a potential difference of $120$ volt. What will be the potential difference across $1 \mu F$ capacitor:

  1. $40 volt$

  2. $60 volt$

  3. $80 volt$

  4. $120 volt$


Correct Option: C
Explanation:
Given, $c _1=1\mu f,c _2=2\mu f,PD=120v$

$C _{eq}=\dfrac{c _1c _2}{c _1+c _2}=\dfrac{2\times1}{2+1}=\dfrac{2}{3}\mu f$

We know,  $Q=cv$ Where Q is the charge, C is the capacitance of the capacitor and v is the potential difference.

Now, $Q _{net}$ in circuit is equivalent capacitance of capacitors attached in the circuits is multiplied by PD

$Q _{net}=12\times\dfrac{2}{3}=80$

$Q=cv=1\mu fv=80\Rightarrow v=80v$

Three long concentric cylindrical shells have radii R, 2R and $2\sqrt{2}R$. Inner and outer shells are connected to each other. The capacitance across middle and inner shells per unit length is:

  1. $\dfrac{\dfrac{1}{3}\epsilon _0}{ln 2}$

  2. $\dfrac{6\pi \epsilon _0}{ln 2}$

  3. $\dfrac{\pi \epsilon _0}{2 ln 2}$

  4. None


Correct Option: B

Six identical square metallic plates are arranged as shown in figure length of each plate is l the capacitance of this arrangement should be

  1. $3\epsilon _0l^2/d$

  2. $4\epsilon _0l^2/d$

  3. $3\epsilon _0l^2/2d$

  4. $2\epsilon _0l^2/d$


Correct Option: A

Two capacitors of caacity ${ C } _{ 1 }$ and ${ C } _{ 2}$ are connected in series and potential difference V is applied across it. Then the potential difference across${ C } _{ 1 }$ will be 

  1. $V\frac { { C } _{ 2 } }{ { C } _{ 1 } } $

  2. $V\frac { { C } _{ 1 }+{ C } _{ 2 } }{ { C } _{ 1 } } $

  3. $V\frac { { C } _{ 2 } }{ { C } _{ 1 }+{ C } _{ 2 } } $

  4. $V\frac { { C } _{ 1 } }{ { C } _{ 1 }+{ C } _{ 2 } } $


Correct Option: A

For capacitors in the series combination, the total capacitance C is given by

  1. $C=(\cfrac{1}{C _1}+\cfrac{1}{C _2} + ......)$

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

  3. $\cfrac{1}{C}=(\cfrac{1}{C _1}+\cfrac{1}{C _2}+.....)$

  4. $\cfrac{1}{C} = C _{1} + C _{2} +$ ........


Correct Option: C
Explanation:

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

A series combination of two capacitances of value $0.1\ mu F$ and $1\mu F$ is connected with a source of voltage $500\ volts$. The potential difference in volts across the capacitor of value $0.1\ muF$ will be :

  1. $50$

  2. $500$

  3. $45.5$

  4. $454.5$


Correct Option: D
Explanation:

Given,

Capacitance, ${{C} _{1}}=0.1\,\mu F\,\,and\,\,{{C} _{2}}=1\,\mu F$

In series charge is equal

$ Q={{C} _{1}}{{V} _{1}}={{C} _{2}}{{V} _{2}} $

$ {{V} _{2}}=\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}} $

In series total potential difference is sum of all paternal difference

$ V={{V} _{1}}+{{V} _{2}} $

$ V={{V} _{1}}+\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}}={{V} _{1}}\left( \dfrac{{{C} _{2}}+{{C} _{1}}}{{{C} _{2}}} \right) $

$ {{V} _{1}}=\dfrac{{{C} _{2}}V}{{{C} _{2}}+{{C} _{1}}}=\dfrac{1\times 500}{1+0.1}=454.54\,V $

Hence, Potential difference across $0.1\,\mu F\,\,\,is\,\,\,454.5\,V$ 

Two parallel plate capacitors are connected in series. Each capacitor has a plate area A and a separation d between the plates. The dielectric constant of the medium between their plates are 2 and 4 . The separation between the plates of a single air capacitors of plate area A which effectively replaces the combination is:

  1. 2d/3

  2. 3d/2

  3. 3d/4

  4. 8d/5


Correct Option: B

A capacitor comprises of two parallel circular plates. Diameter of each of plates is equal to $6 cm$. If capacitance of above system is equivalent to capacitance of sphere, whose diameter is equal to $200 cm$. Distance between two plates will be:-

  1. $2.25 \times 10^{-4} m$

  2. $4.5 \times 10^{-4} m$

  3. $6.75 \times 10^{-4} m$

  4. $9 \times 10^{-4} m$


Correct Option: B

A capacitor 1 mF withstands a maximum voltage of 6KV while another capacitor 2 mF withstands a maximum voltage of 4 KV. If the capacitors are connected in series, the system will withstand a maximum voltage of (MNR)

  1. 2 KV

  2. 4 KV

  3. 6 KV

  4. 9 KV


Correct Option: D

what is the series combination of condenses and $\dfrac { 1 }{ c } =\dfrac { 1 }{ { c } _{ 1 } } +\dfrac { 1 }{ { c } _{ 2 } } +\dfrac { 1 }{ { c } _{ 3 } } $ farad

  1. True

  2. False


Correct Option: A

Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is: 

  1. Q/2V

  2. 2Q/V

  3. Q/V

  4. None of these


Correct Option: B
Explanation:

In series connection charge on each capacitor would be constant also equivalent capacitance in series $c'=\dfrac{C}{2}$ [ following $\dfrac{1}{c'}=\dfrac{1}{c _1}+\dfrac{1}{c _2}$] and voltage $V$ is applied across it so,from capacitive law,$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$

Two capacitors of $4\ \mu F$and $2\ \mu F$ are connected in series with the battery. If total potential difference across the two capacitors is $200$ volts then the ratio  of potential difference across one capacitor to another is

  1. $1:2$

  2. $2:1$

  3. $1:4$

  4. $4:1$


Correct Option: A

A capacitor of capacitance $ 1 \mu F $ withstands a maximum voltage of 6 kilovolt while another capacitor of $ 2 \mu F $ withstands a maximum voltage 4 kilovolt . if the two capacitor are connected in series, the system will withstand a maximum of:

  1. 2kV

  2. 4kV

  3. 6kV

  4. 9kV


Correct Option: D

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

  1. $\dfrac{C}{3}, \dfrac{V}{3}$

  2. $3C, \dfrac{V}{3}$

  3. $\dfrac{C}{3}, 3V$

  4. $3C, 3V$


Correct Option: C

When two condensers of capacitance $1\mu F$ and $2\mu F$ are connected is series then the effective capacitance will be :

  1. $\dfrac{2}{3}\mu F$

  2. $\dfrac{3}{2}\mu F$

  3. $3\mu F$

  4. $4\mu F$


Correct Option: A
Explanation:

When two condenser are in series , the equivalent capacitance $C _{eq}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{1\times2}{1+2}=\dfrac{2}{3} \mu F$

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

  1. 6 F

  2. 5 F

  3. 2/3 F

  4. 3/2 F


Correct Option: C
Explanation:

Let the resultant capacitor is $C _{R}$
For series combination of three capacitors , $\dfrac{1}{C _R}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2} $ F
$\therefore C _R=\dfrac{2}{3}F$

A resistor $ ^{\prime} R^{\prime}  $ and $2  \mu F  $ capacitor in series is connected through a switch to $200  \mathrm{V}  $ direct supply. Across the capacitor is a neon bulb that lights up at $120  \mathrm{V} $ Calculate the value of $  R  $ to make the bulb light up $5  s  $ after the switch has been closed. $ \left(\log _{10} 2.5=0.4\right) $

  1. $2.7 \quad 10^{6} \Omega $

  2. $3.3 \quad 10^{7} \Omega $

  3. $1.3 \quad 10^{4} \Omega $

  4. $1.7 \quad 10^{5} \Omega $


Correct Option: A

Two capacitors of capacitances $4\mu F$ and $6\mu F$ are connected across a 120 V battery in series with each other. What is the potential difference across the $4\mu F$ capacitor?

  1. 40V

  2. 48V

  3. 60V

  4. 72V


Correct Option: B

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}}$

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 , $C _{12}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{2C\times 2C}{2C+2C}=C$

$4\ \mu F$ and $6\ \mu F$ capacitors are joined in series and $500\ v$ are applied between the outer plates of the system. What is the charge on each plate ?

  1. $1\cdot 2\times 10^{3}\ C$

  2. $6\cdot 0\times 10^{3}\ C$

  3. $5\cdot 0\times 10^{-3}\ C$

  4. $2\cdot 0\times 10^{-3}\ C$


Correct Option: A
Explanation:

Since the capacities are connected in series

$\dfrac{1}{C _{eq}}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C _{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$
$C _{eq}=\dfrac{12\mu}{5}$
$C=\dfrac{Q}{V}$
$\dfrac{12\mu}{5}=\dfrac{Q}{500}$
$Q=1.2\times 10^{3}$

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