Tag: combination of capacitors

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

The equivalent capacitance of capacitors $6\mu F$ and $3\mu F$ connected in series is ______.

  1. $3\mu f$

  2. <span>$2\mu f$</span>

  3. <span>$4\mu f$</span>

  4. <span>$6\mu f$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know the equivalent capacitance of capacitors connected in series can be found by using

$\dfrac{1}{C _{eq}}$$=\dfrac{1}{C _{1}}$$+\dfrac{1}{C _{2}}$$+\dfrac{1}{C _{3}}+...$

$\dfrac{1}{C _{eq}}$$=\dfrac{1}{6}$$+\dfrac{1}{3}$

$\Rightarrow C _{eq} = \dfrac{3\times 6}{3+6} = 2\mu F $
Therefore, B is correct option.

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

When two capacitors of capacities of $3\mu F$ and $6\mu F$ are connected in series and connected to $120\ V$, the potential difference across $3\mu F$ is:

  1. $40\ V$

  2. $60\ V$

  3. $80\ V$

  4. $180\ V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equivalent capacitance is C

$\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{6}$, So $C=2\mu f$ 
Now $Q=VC=120\times 2=240\mu F$
 Now potential across $3\mu f$ is $V=\dfrac{Q}{3}=240/3=80V$

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

Three capacitors, $3\mu F, 6\mu F$ and $6\mu F$ are connected in series to a source of 120V. The potential difference, in volts, across the $3\mu F$ capacitor will be

  1. 24

  2. 30

  3. 40

  4. 60
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equivalent capacitance of the two $6\mu F$ and $6\mu F$ capacitors in series is $3\mu F$.

Hence the potential across the two capacitors, original $3\mu F$ capacitor and the equivalent $3\mu F$ capacitor is divided equally. 
Hence voltage across each of the capacitors is half of the external applied voltage, $60V.$

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

A capacitor of capacitance ${ C } _{ 1 }=1\mu F$ can with stand maximum voltage ${ V } _{ 1 }=6kV$ (kilo-volt) and another capacitor of capacitance ${ C } _{ 2 }=3\mu F$ can withstand maximum voltage ${ V } _{ 2 }=4kV$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of:

  1. $4kV$

  2. $6kV$

  3. $8kV$

  4. $10kV$
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

Complete the following statements with an appropriate word /term be filled in the blank space(s).


The equivalent capacitance C for the series combination of three capacitance $C _1,C _2$ and $C _3$ is given by $\cfrac{1}{C} =$..............

  1. $C _1+C _2+C _3$

  2. $\left ( \cfrac{1}{C _{1}+C _{2}+C _{3}} \right )$

  3. $\left ( \cfrac{1}{\cfrac{1}{C _{1}}+\cfrac{1}{C _{2}}+\cfrac{1}{C _{3}}}\right )$

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

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

The equivalent capacitance of the pair of capacitors is $C = \cfrac{Q}{V}$
$\cfrac{1}{C} = \cfrac{V}{Q} = \cfrac{(v _1 + v _2+ v _3)}{ Q }=\cfrac{v _1}{Q} + \cfrac{v _2}{Q}+ \cfrac{v _3}{Q} = \cfrac{1}{C _1} + \cfrac{1}{C _2}+\cfrac{1}{C _3}$

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

Which one of the following gives the resultant capacitor when capacitors are joined in series?

  1. The sum of the individual capacitors

  2. The reciprocal of the sum of the reciprocals of the individual capacitors

  3. The reciprocal of the sum of the capacitors

  4. The sum of the reciprocals of the individual capacitors
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The resultant capacitor when capacitors are joined in series is the reciprocal of the sum of the reciprocals of the indivisual capacitors.

$\cfrac{1}{c _{eq}}=$$\cfrac{1}{c _{1}}$+$\cfrac{1}{c _{2}}$

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

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.

  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) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an RC series circuit, the phase angle phi between voltage and current is given by tan(phi) = Xc / R. Since Xc = 1 / (omega * C), the expression is tan(phi) = 1 / (omega * C * R). Thus, phi = tan^-1(1 / (omega * C * R)).

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

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?

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

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

  3. $C$

  4. $2C$

  5. $4C$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 , $C _{12}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{2C\times 2C}{2C+2C}=C$