Tag: physics

Questions Related to physics

A capacitor of capacitance $ 1 \mu F $ withstands a maximum voltage of 6 kilovolt while another capacitor of $ 2 \mu F $ withstands a maximum voltage 4 kilovolt . if the two capacitor are connected in series, the system will withstand a maximum of:

  1. 2kV

  2. 4kV

  3. 6kV

  4. 9kV


Correct Option: D

Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be

  1. $\dfrac{C}{3}, \dfrac{V}{3}$

  2. $3C, \dfrac{V}{3}$

  3. $\dfrac{C}{3}, 3V$

  4. $3C, 3V$


Correct Option: C

When two condensers of capacitance $1\mu F$ and $2\mu F$ are connected is series then the effective capacitance will be :

  1. $\dfrac{2}{3}\mu F$

  2. $\dfrac{3}{2}\mu F$

  3. $3\mu F$

  4. $4\mu F$


Correct Option: A
Explanation:

When two condenser are in series , the equivalent capacitance $C _{eq}=\dfrac{C _1C _2}{C _1+C _2}=\dfrac{1\times2}{1+2}=\dfrac{2}{3} \mu F$

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


Correct Option: C
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
$\therefore C _R=\dfrac{2}{3}F$

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 $


Correct Option: A

Two capacitors of capacitances $4\mu F$ and $6\mu F$ are connected across a 120 V battery in series with each other. What is the potential difference across the $4\mu F$ capacitor?

  1. 40V

  2. 48V

  3. 60V

  4. 72V


Correct Option: B

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. $\dfrac {C _{1} + C _{2}}{C _{1}C _{2}}$


Correct Option: C

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. $8\mu F$


Correct Option: B
Explanation:

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

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. $6\mu f$


Correct Option: B
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.

The two capacitors $2\mu F$ and $6\mu F$ are put in series, the effective capacity of the system is $\mu F$ is:

  1. $8\mu F$

  2. $2\mu F$

  3. $3/2\mu F$

  4. $2/3\mu F$


Correct Option: C
Explanation:

When connected in series 

$\dfrac{1}{C}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{6}$
$C=\dfrac{3}{2}$