Tag: capacitance

Questions Related to capacitance

The current in a contining a capacitance C and a resistance R in series over the applied voltage of frequency $\cfrac { \omega  }{ 2\pi  } $ by.

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

  1. ${ tan }^{ -1 }\left( \frac { 1 }{ \omega CR } \right) $

  2. ${ tan }^{ -1 }\left( \omega CR \right) $

  3. ${ tan }^{ -1 }\left( \omega \frac { 1 }{ R } \right) $

  4. ${ cos }^{ -1 }\left( \omega CR \right) $

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Correct Option: A

A very thin metal sheet is inserted halfway between the parallel plates of an air-gap capacitor. The sheet is thin compared to the distance between the plates, and it does not touch either plate when fully inserted. The system had capacitance, $C$, before the plate is inserted.
What is the equivalent capacitance of the system after the sheet is fully inserted?

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

  1. $\cfrac{1}{4}C$

  2. $\cfrac{1}{2}C$

  3. $C$

  4. $2C$

  5. $4C$

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Correct Option: C
Explanation:

Initially (before metal sheet inserted) the capacitance of a parallel plate capacitor is $C=\dfrac{A\epsilon _0}{d}$ where A be the area of plates and d be the separation between parallel plates.
When a metal sheet inserted fully halfway between the parallel plates, the capacitance will be divided into two capacitors $C _1, C _2$ and they are in series.
Thus, $C _1=\dfrac{A\epsilon _0}{(d/2)}=2C$ and $C _2=\dfrac{A\epsilon _0}{(d/2)}=2C$
The equivalent capacitance , $C _{12}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{2C\times 2C}{2C+2C}=C$

$4\ \mu F$ and $6\ \mu F$ capacitors are joined in series and $500\ v$ are applied between the outer plates of the system. What is the charge on each plate ?

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

  1. $1\cdot 2\times 10^{3}\ C$

  2. $6\cdot 0\times 10^{3}\ C$

  3. $5\cdot 0\times 10^{-3}\ C$

  4. $2\cdot 0\times 10^{-3}\ C$

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Correct Option: A
Explanation:

Since the capacities are connected in series

$\dfrac{1}{C _{eq}}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C _{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$
$C _{eq}=\dfrac{12\mu}{5}$
$C=\dfrac{Q}{V}$
$\dfrac{12\mu}{5}=\dfrac{Q}{500}$
$Q=1.2\times 10^{3}$

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plate are $+\sigma$ and $-\sigma$ respectively. In the region between the plates the magnitude of electric field is:

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

  1. $\dfrac { \sigma }{ 2{ \varepsilon } _{ 0 } } $

  2. $\dfrac { \sigma }{ { \varepsilon } _{ 0 } } $

  3. 0

  4. none of these

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Correct Option: B
Explanation:

The magnitude of the electric field due to a charged plate is given as:

$\vec E=\dfrac{\sigma}{2\varepsilon _0}$

The charge density on the upper plate is positive whereas on the lower plate it is negative.

Therefore, the electric field in the region between the two plates is the difference of the two fields.
It is given as:
$E _{net}=\dfrac{\sigma}{2\varepsilon _0}-\dfrac{-\sigma}{2\varepsilon _0}$

$E _{net}=\dfrac{\sigma}{\varepsilon _0}$

If the voltage across the parallel combination in increases which capacitor will undergo breakdown first?

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

  1. $C$

  2. $2C$

  3. Both at same moment

  4. None of these

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Correct Option: C

A capacitors of $2 \mu F$ is required is an electric circuit across a potential difference of 1.0kv. A large number of $1 \mu F$ capacitors are available which can with stand a potential difference of not more than 300V The minimum number of capacitors required to achieve this is:

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

  1. 24

  2. 32

  3. 2

  4. 16

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Correct Option: B

Two capacitors were charged to potentials 80 and 30 V. Then they connected in parallel. The potential difference across both condensers is 60 V. The ratio of the capacitances of the capacitors is

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

  1. 3 : 2

  2. 1 : 3

  3. 1 : 4

  4. 2 : 3

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Correct Option: A

The plates of a parallel plate capacitor are $4$cm apart, the first plate is at $300$V and the second plate at $-100$V. The voltage at $3$cm from the second plate is?

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

  1. $200$V

  2. $400$V

  3. $250$V

  4. $500$V

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Correct Option: A

Find the total capacitance for three capacitors of $10$f,$15$f and $35$f in parallel with each other?

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

  1. $20f$

  2. $50f$

  3. $60f$

  4. $10f$

  5. $5f$

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Correct Option: C
Explanation:

Given :    $C _1 = 10$ f                  $C _2 = 15$  f                    $C _3 = 35$  f

Equivalent capacitance for parallel combination          $C _{eq} = C _1 + C _2 + C _3$
$\therefore$   $C _{eq} = 10 + 15 + 35  = 60$  $f$

Two capacitors of capacity $C _1$ and $C _2$ are connected in parallel, then the equivalent capacity is:

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

  1. $C _1+C _2$

  2. $C _1C _2/(C _1+C _2)$

  3. $C _1/C _2$

  4. $C _2/C _1$

Show answer
Correct Option: A
Explanation:

$C _1, C _2$ are connected in parallel then equivalent capacitance is calculated as
$V=V _1=V _2$.....(1)
$q=q _1+q _2$
$\therefore CV=C _1V _1+C _2V _2$
From (1) $C=C _1+C _2$