Capacitors in Series

For two capacitors connected in series, the charge Q is the same on both. The potential difference across C1 is V1 = Q / C1. Since equivalent capacitance is C_eq = (C1 * C2) / (C1 + C2) and total charge is Q = C_eq * V, substituting gives V1 = V * (C2 / (C1 + C2)).

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

Given,

Capacitance, ${{C} _{1}}=0.1\,\mu F\,\,and\,\,{{C} _{2}}=1\,\mu F$

In series charge is equal

$ Q={{C} _{1}}{{V} _{1}}={{C} _{2}}{{V} _{2}} $

$ {{V} _{2}}=\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}} $

In series total potential difference is sum of all paternal difference

$ V={{V} _{1}}+{{V} _{2}} $

$ V={{V} _{1}}+\dfrac{{{C} _{1}}{{V} _{1}}}{{{C} _{2}}}={{V} _{1}}\left( \dfrac{{{C} _{2}}+{{C} _{1}}}{{{C} _{2}}} \right) $

$ {{V} _{1}}=\dfrac{{{C} _{2}}V}{{{C} _{2}}+{{C} _{1}}}=\dfrac{1\times 500}{1+0.1}=454.54\,V $

Hence, Potential difference across $0.1\,\mu F\,\,\,is\,\,\,454.5\,V$ 

Two capacitors of area A and separation d with dielectric constants 2 and 4 are in series. Their capacitances are C1 = 2A e0 / d and C2 = 4A e0 / d. The equivalent capacitance C_s satisfies 1/C_s = 1/C1 + 1/C2 = d/(2A e0) + d/(4A e0) = 3d/(4A e0), so C_s = 4A e0 / (3d). Equating this to a single air capacitor C = A e0 / d_eff gives d_eff = 3d/4.

The capacitance of a parallel plate capacitor is C = epsilon_0 * A / d = epsilon_0 * pi * r^2 / d, where r = 3 cm = 0.03 m. The capacitance of a conducting sphere of radius R = 100 cm = 1 m is C_sphere = 4 * pi * epsilon_0 * R. Equating the two: pi * epsilon_0 * (0.03)^2 / d = 4 * pi * epsilon_0 * (1). Solving for d gives d = 0.0009 / 4 = 2.25 * 10^-4 m.

The formula for capacitors in series is indeed 1/C_eq = 1/C1 + 1/C2 + 1/C3.

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,$Q=c'v \Rightarrow Q=\dfrac{CV}{2} \Rightarrow C=\dfrac{2Q}{V}$

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.

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.

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

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

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.

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.

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).

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

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

When connected in series 

Equivalent capacitance is C

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

Let charge q be on the capacitors in series. The voltage across C1 is V1 = q / C1 <= V_max1 (6 kV), so q <= 1 microfarad * 6 kV = 6 micro-C. The voltage across C2 is V2 = q / C2 <= V_max2 (4 kV), so q <= 3 microfarad * 4 kV = 12 micro-C. Thus, maximum safe charge is q = 6 micro-C. At this charge, V1 = 6 kV and V2 = 6 micro-C / 3 microfarad = 2 kV. Total maximum voltage is V1 + V2 = 6 kV + 2 kV = 8 kV.

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 $C = \cfrac{Q}{V}$
$\cfrac{1}{C} = \cfrac{V}{Q} = \cfrac{(v _1 + v _2+ v _3)}{ Q }=\cfrac{v _1}{Q} + \cfrac{v _2}{Q}+ \cfrac{v _3}{Q} = \cfrac{1}{C _1} + \cfrac{1}{C _2}+\cfrac{1}{C _3}$

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

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)).

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

Since the capacities are connected in series