Physics

Capacitors and Capacitance

186 Questions

Capacitors and capacitance are crucial topics in physics, covering the storage of electric charge and energy. These concepts explore series and parallel combinations, dielectric materials, and capacitive reactance. Questions frequently appear in various competitive engineering and medical entrance exams.

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Capacitors and Capacitance Questions

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

A capacitor $ C _1 = 4 \mu F $ is connected in series with another capacitor $ C _2 = 1 \mu F $. the combination is connected across a d.c. source of voltage 200 V. the ration of potential across $ C _1 $ and $C _2 $ is-

  1. 1 : 4

  2. 4 : 1

  3. 1 : 2

  4. 2 : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For capacitors in series, the charge Q on each is the same. Potential is V = Q/C, meaning potential is inversely proportional to capacitance. For C1 = 4 uF and C2 = 1 uF, the ratio of potential V1 / V2 = C2 / C1 = 1 / 4.

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

From a supply of identical capacitors rated $8\;\mu F, 250 \;V$ the minimum number of capacitors required to form a composite of $16\;\mu F, 1000 \;V$ is

  1. 2

  2. 4

  3. 16

  4. 32
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The number of capacitance to be connected in series $\displaystyle n=\frac{voltage \  rating \ required}{voltage\  rating \ of \ a \ capacitor \ given}=\frac{1000}{250}=4$
Equivalent capacitance, $\displaystyle C _{eq}=(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})^{-1}=(\frac{4}{8})^{-1}=2$
Number of rows required $\displaystyle =\frac{capacitance \ required}{capacitance \  of \ each \  row}=\frac{16}{2}$
Thus the minimum number of capacitors to be required $=4\times 8=32$

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

In order to increase the capacity of parallel plate condenser one should introduce between the plates, a sheet of

  1. mica

  2. tin

  3. copper

  4. stainless steel
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

mica as it is having higher conductivity$.$

So$,$ increase the capacity of parallel plate condenser one should introduce between the plates$,$ a sheet of $mica.$
Hence,
option $(A)$ is correct answer.

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

A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $C$, then the resultant capacitance is-

  1. $(n-1)C$
  2. $(n+1)C$
  3. $C$
  4. $nC$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n$ plates connected alternately give rise to $\left(n – 1\right)$ capacitors connected in parallel $\therefore$, Resultant capacitance $=\left(n – 1\right)C$.

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

A parallel plate condenser has plates of area $200\mathrm { cm } ^ { 2 }$ and separation $0.05\mathrm { cm } .$ The space between plates have been filled with a dielectric having $\mathrm { k } = 8$ and then charged to $300$ volts. The stored energy:

  1. $121.5 \times 10 ^ { - 6 } \mathrm { J }$
  2. $28 \times 10 ^ { - 6 } \mathrm { J }$
  3. $112.4 \times 10 ^ { - 5 } \mathrm { J }$
  4. $1.6 \times 64 \times 10 ^ { - 5 } \mathrm { J }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$C = \dfrac{{KA{ \in _0}}}{d} = \dfrac{{3 \times 200 \times {{10}^{ - 4}} \times 8.85 \times {{10}^{ - 12}}}}{{5 \times {{10}^{ - 4}}}}$

$ = 27 \times {10^{ - 10}}F$
$E = \dfrac{1}{2}C{V^2} = \frac{1}{2} \times 27 \times {10^{ - 10}} \times 300$
$ = \dfrac{{243}}{2} \times {10^{ - 6}}$
$ = 121.5 \times {10^{ - 6}}J$
Hence,
option $(A)$ is correct answer.

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

A Parallel platecapacitor made of circular plates each of radius $R=6.0cm$ has a capacitance 100$\mathrm { pF }$ is connected to 230$\mathrm { V }$ of $\mathrm { AC }$ supply of 300 rad/sec.frequency. The rms value of displacement current

  1. $6.9\mu A$
  2. $2.3\mu A$
  3. $9.2\mu A$
  4. $4.6\mu A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given$:-$

$R=6.0cm$
$C = 100pF$
$ = 100 \times {10^{ - 12}}F$
$w = 300\,rad/s$
${I _{rms}} = 230 \times 300 \times 100 \times {10^{ - 12}}$
$ = 6.9 \times {10^{ - 9}}$
$ = 6.9\mu A$
Hence, 
option $(A)$ is correct answer.

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

Two capacitors of capacitance $2\, \mu F$ and $4\, \mu F$ are charge to $200\, V$ and $100\, V$ respectively. They are then connected in parallel to each other. What is the potential across each capacitor ?

  1. $116\, V$
  2. $133\, V$
  3. $148\, V$
  4. $164\, V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Common potential V = (C1V1 + C2V2) / (C1 + C2) = (2*200 + 4*100) / (2 + 4) = (400 + 400) / 6 = 800 / 6 = 133.33 V.

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

A capacitor is charged by a battery. the battery is removed and another identical uncharged capacitor is connected in parallel. the total electromagnetic energy of resulting system

  1. Decrease by a factor of 2

  2. remains the same

  3. increase by a factor of 2

  4. increase by a factor 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial energy U1 = 0.5 * C * V^2. When connected to an uncharged identical capacitor, charge Q redistributes such that Q' = Q/2 and V' = V/2. Final energy U2 = 2 * (0.5 * C * (V/2)^2) = 2 * (0.5 * C * V^2 / 4) = 0.5 * (0.5 * C * V^2) = U1 / 2. The energy decreases by a factor of 2.

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

A parallel plate air capacitor has capacity 'C', a distance of separation between plate is 'd' and potential difference 'V' is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is:

  1. $\dfrac{{C{V^2}}}{{2d}}$
  2. $\dfrac{{C{V^2}}}{{d}}$
  3. $\dfrac{{{C^2}{V^2}}}{{2{d^2}}}$
  4. $\dfrac{{{C^2}{V^2}}}{{{d^2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force between plates is F = Q^2 / (2 * e0 * A) = (CV)^2 / (2 * e0 * A). Since C = e0 * A / d, then e0 * A = Cd. Substituting this, F = (C^2 * V^2) / (2 * Cd) = CV^2 / (2d).

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

The distance between the plates of a parallel plate capacitor is $1\ mm$. What must be the area of the plate of the capacitor if the capacitance is to be $1.0\mu F$?

  1. $102.5\ m^{2}$
  2. $205.9\ m^{2}$
  3. $112.9\ m^{2}$
  4. $302.9\ m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the formula for a parallel plate capacitor, C = epsilon_0 * A / d. Rearranging for area gives A = C * d / epsilon_0. Plugging in C = 1.0 * 10^-6 F, d = 1 * 10^-3 m, and epsilon_0 = 8.85 * 10^-12 F/m yields A = (10^-9) / (8.85 * 10^-12) = 112.9 m^2.

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

Two parallel plate capacitor of capacitances C and 2C are conncected in parallel and changed to a potential difference V.If  the bsttery is disconnected and the space between the plate of the capacitor of cpacince c is cpmpletely  filled with a metrial of dielectric constant K, then the potential difference a cross the capacitor will be come

  1. $3V\left( {K + 2} \right)$
  2. $\left( {\frac{{K + 2}}{{3V}}} \right)$
  3. $\left( {\frac{{3V}}{{K + 2}}} \right)$
  4. $\frac{{3\left( {K + 2} \right)}}{V}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial charge on C is CV, on 2C is 2CV. Total charge Q = 3CV. When battery is disconnected, Q is constant. New capacitance of first capacitor is KC. Total capacitance C_new = KC + 2C = C(K+2). New potential V_new = Q / C_new = 3CV / (C(K+2)) = 3V / (K+2).

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

A parallel plate capacitor has circular plates of $8.0\ cm$ radius and are separated by $1.0\ mm$. Calculate the capacitance.

  1. $120\ pF$
  2. $140\ pF$
  3. $160\ pF$
  4. $180\ pF$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

C = e0 * A / d = e0 * pi * r^2 / d. C = (8.85e-12 * 3.14 * 0.08^2) / 0.001 = 8.85e-12 * 3.14 * 0.0064 / 0.001 = 1.77e-10 F = 177 pF. The closest option is 180 pF.