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

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. $3C, 3V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For capacitors in series, the equivalent capacitance is C_eq = C/n = C/3. Since the voltage divides equally across identical capacitors in series, each capacitor drops V/3, meaning the total voltage the combination can withstand is 3V.

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

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. $4\mu F$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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 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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two capacitors of 4 microfarad and 6 microfarad are in series, the equivalent capacitance is (4 * 6) / (4 + 6) = 2.4 microfarad. The total charge is Q = C_eq * V = 2.4 microfarad * 120 V = 288 microcoulombs. The potential difference across the 4 microfarad capacitor is V1 = Q / C1 = 288 microcoulombs / 4 microfarad = 72 V.

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.