Tag: capacitance

Quizzes (4) Questions (92)

Questions Related to capacitance

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

Inside a hollow charged spherical conductor, the electric field is found to be.

  1. Proportional to the distance from the centre

  2. A function of the area of the sphere

  3. Zero

  4. A function of the charge density of the sphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Gauss's Law, the electric field inside a hollow charged conductor is zero because there is no enclosed charge.

Multiple choice capacitance of an isolated spherical conductor capacitance of isolated bodies capacitance physics

Of the following about capacitive reactance which is correct

  1. the reactance of the capacitor is directly proportional to its ability to store charge

  2. capacitive reactance is inversely proportional to the frequency of the current

  3. capacitive reactance is me sured in farad

  4. the reactance of a capacitor in an A.C circuit is similar to the resistance of a capacitor in a D.C circuit
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

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. $120 volt$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

$Q=cv=1\mu fv=80\Rightarrow v=80v$
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

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 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

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. $\cfrac{1}{C} = C _{1} + C _{2} +$ ........
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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. $454.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $0.1\,\mu F\,\,\,is\,\,\,454.5\,V$ 

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics capacitance capacitors in series combination of capacitors capacitors in parallel and series

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. $9 \times 10^{-4} m$
Reveal answer Fill a bubble to check yourself
B Correct answer