Physics

Capacitors and Capacitance

189 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 series combination of capacitors capacitors in parallel and series

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

  1. 6 F

  2. 5 F

  3. 2/3 F

  4. 3/2 F
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the resultant capacitor is $C _{R}$
For series combination of three capacitors , $\dfrac{1}{C _R}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2} $ F
$\therefore C _R=\dfrac{2}{3}F$

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

A resistor $ ^{\prime} R^{\prime}  $ and $2  \mu F  $ capacitor in series is connected through a switch to $200  \mathrm{V}  $ direct supply. Across the capacitor is a neon bulb that lights up at $120  \mathrm{V} $ Calculate the value of $  R  $ to make the bulb light up $5  s  $ after the switch has been closed. $ \left(\log _{10} 2.5=0.4\right) $

  1. $2.7 \quad 10^{6} \Omega $
  2. $3.3 \quad 10^{7} \Omega $
  3. $1.3 \quad 10^{4} \Omega $
  4. $1.7 \quad 10^{5} \Omega $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The voltage across a charging capacitor is Vc = V0 * (1 - exp(-t/RC)). Given Vc = 120, V0 = 200, t = 5, and C = 2 * 10^-6, we have 120 = 200 * (1 - exp(-5/(R * 2 * 10^-6))). Simplifying gives 0.6 = 1 - exp(-5/(2 * 10^-6 * R)), so 0.4 = exp(-5/(2 * 10^-6 * R)). Taking the natural log, ln(0.4) = -5/(2 * 10^-6 * R). Using log10(2.5) = 0.4, we find R is approximately 2.7 * 10^6 ohms.

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

Two capacitor of capacity $C _{1}$ and $C _{2}$ are connected in series. The combined capacity $C$ is given by

  1. $C _{1} + C _{2}$
  2. $C _{1} - C _{2}$
  3. $\dfrac {C _{1}C _{2}}{C _{1} + C _{2}}$
  4. $\dfrac {C _{1} + C _{2}}{C _{1}C _{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For two capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances: 1/C = 1/C1 + 1/C2. Solving for C gives C = (C1 * C2) / (C1 + C2).

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

Three condenser of capacitance $C(\mu F)$ are connected in parallel to which a condenser of capacitance $C$ is connected in series. Effective capacitance is $3.75$, then capacity of each condenser is

  1. $4\mu F$
  2. $5\mu F$
  3. $6\mu F$
  4. $8\mu F$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The effective capacitance of three condenser connected in parallel$=3C$.
When $3C$ is connected in series to $C$
$C _{Result}=\displaystyle\frac{3C\times C}{3C+C}=3.75$
$\Rightarrow C=5\mu F$.

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. $2\mu f$
  3. $4\mu f$
  4. $6\mu f$
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

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 , $C _{12}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{2C\times 2C}{2C+2C}=C$

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

$4\ \mu F$ 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 ?

  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. $2\cdot 0\times 10^{-3}\ C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the capacities are connected in series

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