Tag: capacitors in series

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Questions Related to capacitors in series

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

Two identical capacitors are connected in series with a source of potential V. If Q is the charge on one of the capacitors, the capacitance of each capacitor is: 

  1. Q/2V

  2. 2Q/V

  3. Q/V

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In series connection charge on each capacitor would be constant also equivalent capacitance in series $c'=\dfrac{C}{2}$ [ following $\dfrac{1}{c'}=\dfrac{1}{c _1}+\dfrac{1}{c _2}$] and voltage $V$ is applied across it so,from capacitive law,$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$

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

Two capacitors of $4\ \mu F$and $2\ \mu F$ are connected in series with the battery. If total potential difference across the two capacitors is $200$ volts then the ratio  of potential difference across one capacitor to another is

  1. $1:2$

  2. $2:1$

  3. $1:4$

  4. $4:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In series, charge Q is constant. V = Q/C. Therefore, V1/V2 = C2/C1. With C1=4uF and C2=2uF, V1/V2 = 2/4 = 1/2.

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 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 $C$
$C _{Result}=\displaystyle\frac{3C\times C}{3C+C}=3.75$
$\Rightarrow C=5\mu F$.