Tag: capacitance

Quizzes (4) Questions (92)

Questions Related to capacitance

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 parallel plates have equal and opposite charge. When the space between them is evacuated. the electric field between the plates $2 \times {10^5}\,V/m.$ When the space is filed with dielectric the electric field becomes ${10^5}\,V/m$ The dielectric constant of he dielectric material is 

  1. $2$

  2. $4$

  3. $5$

  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Dielectric constant$:-$

$K = \dfrac{{{E _0}}}{E}$
$ = \dfrac{{2 \times {{10}^5}}}{{1 \times {{10}^5}}} = 2$
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

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.