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

Capacity of a parallel plate capacitor is $2\mu F$. The two plates of the capacitor are given $400\mu C$ and $-200\mu C$charges respectively. The potential difference between the plates is 

  1. $100\ V$

  2. $200\ V$

  3. $300\ V$

  4. $150\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The potential difference between plates is V = (Q1 - Q2) / (2C) is incorrect. The potential difference is determined by the charge on the inner surfaces. For a capacitor with charges Q1 and Q2, the charge on the inner faces is (Q1 - Q2)/2. Here, (400 - (-200))/2 = 300uC. V = Q/C = 300uC / 2uF = 150V.

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

A parallel plate capacitor consist of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of $5 \times {10^{13}}V{s^{ - 1}}$ , then the value of displacement current is:

  1. $5.50A$ 

  2. $ 5.56 \times 10^2 A $

  3. $ 5.56 \times 10^3 A $

  4. $ 2.28 \times 10^4 A $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Displacement current Id = e0 * d(Phi_E)/dt = e0 * A * dE/dt = e0 * A * (1/d) * dV/dt = (e0 * A / d) * dV/dt = C * dV/dt. C = e0 * pi * r^2 / d. C = (8.85e-12 * 3.14 * 0.02^2) / 0.0001 = 1.11e-10 F. Id = 1.11e-10 * 5e13 = 5.55 A.

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

When $n$ identical capacitors are connected in series their effective capacity is $C _s$ and when they are connected in parallel their effective capacity is $C _p$. The relation between $C _p$ and $C _s$ is:

  1. $C _p = n \,C _s$

  2. $C _p = \dfrac{C _s}{n}$

  3. $C _p = n^2 \,C _s$

  4. $C _p = \dfrac{C _s}{n^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For n identical capacitors of capacity C, Cs = C/n and Cp = nC. Therefore, Cp = n * (n * Cs) = n^2 * Cs.

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.