Tag: combination of capacitors

Questions Related to combination of capacitors

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:

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

  1. $40 volt$

  2. $60 volt$

  3. $80 volt$

  4. $120 volt$

Show answer
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:

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

  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

Show answer
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

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

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

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

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

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

Show answer
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 

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

  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 } } $

Show answer
Correct Option: A

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

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

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

Show answer
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 :

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

  1. $50$

  2. $500$

  3. $45.5$

  4. $454.5$

Show answer
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:

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

  1. 2d/3

  2. 3d/2

  3. 3d/4

  4. 8d/5

Show answer
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:-

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

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

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

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

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

Show answer
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)

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

  1. 2 KV

  2. 4 KV

  3. 6 KV

  4. 9 KV

Show answer
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

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

  1. True

  2. False

Show answer
Correct Option: A