Capacitors
13.0 Method of Finding Equivalent Capacitance
13.0 Method of Finding Equivalent Capacitance
Method of Finding equivalent capacitance is similar as finding equivalent resistance. We know that in series combination, $$R_{{eq}} = {R_1} + {R_2} + {R_3} + ... + {R_n}$$$${\frac{1}{C}_{eq}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + ... + \frac{1}{{{C_n}}}$$
In parallel combination, $${\frac{1}{R}_{eq}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + ... + \frac{1}{{{R_n}}}$$$${C_{eq}} = {C_1} + {C_2} + {C_3} + ... + {C_n}$$
In some circuit capacitors and resistances are connected in such a manner that the circuit become too much complected, so to solve such circuits here some methods are described below.
1. Wheatstone's bridge circuits: In Wheatstone's bridge circuit in case of capacitors, if
$$\frac{{{C_1}}}{{{C_2}}} = \frac{{{C_3}}}{{{C_4}}}$$ bridge is said to be balanced and in this case $${V_E} = {V_D}$$or $${V_{ED}} = 0$$
2. Infinite series problems: In this case, we have a infinite series of identical loops. To find $C_{eq}$ of such a series first we have to consider by our self a value of $C_{eq}$(let $x$).
Now, we have to break the chain in such a manner that only one loop is left with us and in place of remaining portion which we are removing from the circuit, should connect a capacitor of capacitance $x$. By this we will calculate equivalent capacitance and will put it's valve equal to $x$.
And hence we will get a quadratic equation in $x$, by solving it we will get the value of $C_{eq}$.
Example: An infinite ladder is constructed with $1\mu F$ and $2\mu F$ capacitor as shown in the figure. Find the equivalent capacitance between points $A$ and $B$.
Solution: Let consider that the equivalent capacitance between $A$ and $B$ is $x$. Now the we will break the circuit from the points where from identical chains repeat. Now the circuit will be as shown in the figure.
Hence we can write,$$\begin{equation} \begin{aligned} {C_{AB}} = \frac{{(1)(2 + x)}}{{1 + (2 + x)}} \\ x = \frac{{2 + x}}{{3 + x}} \\ x(3 + x) = 2 + x \\ {x^2} + 2x - 2 = 0 \\ x = \frac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times ( - 2)} }}{2} \\ x = - 1 \pm \sqrt 3 \\\end{aligned} \end{equation} $$Ignoring the negative value, we have $${C_{AB}} = x = \sqrt 3 - 1$$
3. Method of same potential: In this method we will give any arbitrary potential (say $V_1$, $V_2$,...) to all terminal of the capacitor and we have to notice that the potential of the points connected to the conducting wires should be same.
Example: Five identical conducting plates, $1,2,3,4$ and $5$ are fixed parallel plate equidistant from each other (see figure). A conductor connected plates $2$ and $ 5$ while another conductor joins $1$ and $3$. The junction of $1$ and $3$ and the plate $4$ are connected to a source of constant e.m.f. $V$. Find the effective capacitance of the system between the terminals of the source.
Solution: To solve this problem first there are two surfaces of each plate (i.e., upper surface and lower surface). Now, we should connect these two surfaces by conducting wires as shown in figure. Now we can detect each and all capacitors 
easily and let potential of plate $4$ is zero and of plate$2$ is $x$. It is given that the area of plates is $A$ and distance between the plates is $d$. So capacitance, $$C = \frac{{A{\varepsilon _ \circ }}}{d}$$
easily and let potential of plate $4$ is zero and of plate$2$ is $x$. It is given that the area of plates is $A$ and distance between the plates is $d$. So capacitance, $$C = \frac{{A{\varepsilon _ \circ }}}{d}$$Now we can make the following table.
| Potential Difference | Capacitance |
| $V$ | $C$ |
| $V-x$ | $2C$ |
| $x$ | $C$ |
Now corresponding to the table we can make a simple circuit diagram as given below.
Now the equivalent capacitance between the terminals of the source ($C_{eq}$) is,$$\begin{equation} \begin{aligned} {C_{eq}} = C + \frac{{(C)(2C)}}{{C + 2C}} \\ {C_{eq}} = \frac{5}{3}C \\ {C_{eq}} = \frac{5}{3}\frac{{A{\varepsilon _ \circ }}}{d} \\\end{aligned} \end{equation} $$
4. Connection removal method: This method is useful when the circuit diagram is symmetric except for the fact that the input and output are reversed. That is the flow of current is a mirror image between input and output above a particular axis. In such cases some junctions are unnecessarily made. Even if we remove that junction there is no difference in the remaining circuit or current distribution. But after removing the junction, the problem becomes very simple. This concept will be more clear by upcoming example.
Example: Find the equivalent capacitance between the points $A$ and $B$.
Input and output circuits are mirror images of each other about the dotted line as shown. So, if a current $i$ enters from $A$ and leaves from $B$ it will distribute as shown below.
Input and output circuits are mirror images of each other about the dotted line as shown. So, if a current $i$ enters from $A$ and leaves from $B$ it will distribute as shown below.
Hence $${C_{eq}} = \frac{{7C}}{{10}}$$
5. Method of symmetry: Points having symmetrically located about initial and final points have same potentials. So, the capacitance between these points can be ignored.
Example: Twelve capacitors each of capacitance $C$ are connected together so that each lies along the edge of the cube as shown in figure. Find the equivalent capacitance between points $1$ and $4$.
Solution: Between $1$ and $4$: Points $2$ and $5$ are symmetrically located with respect to (w.r.t.) points $1$ and $4$. So, they are at same potentials. Similarly, points $3$ and $8$ are also symmetrically located w.r.t. points $1$ and $4$. So, they are again at same potentials. Now, we have $12$ capacitors each of capacitance $C$ connected across $1$ and $2$, $2$ and $3$, . . . , etc. So redrawing them with the assumption that $2$ and $5$ are at same potential and $3$ and $8$ are at same potential. The new figure is shown below.
Hence, we can find the equivalent capacitance of the circuit $${C_{eq}} = \frac{3}{2}C$$
Example: A capacitor is having area of cross-section of plates $A$ and distance between plates is $d$. Capacitance of
the capacitor is $C$. The gap between the plates is filled with two dielectric in wedge shape having dielectric constant $K_1$ and $K_2$ as shown in the figure. Find the equivalent capacitance of the resultant capacitor.
the capacitor is $C$. The gap between the plates is filled with two dielectric in wedge shape having dielectric constant $K_1$ and $K_2$ as shown in the figure. Find the equivalent capacitance of the resultant capacitor.Solution: Let the length and the width of the plates of capacitor are $l$ and $b$ and distance between the plates is $d$. let there is a elemental slab at $x$ distance of $dx$ thickness.$$QR = x\tan \theta $$And $PQ = d - x\tan \theta $ where $\tan \theta = d/l$
Capacitance of $PQ$$$\begin{equation} \begin{aligned} {C_1} = \frac{{{K_1}{\varepsilon _ \circ }(bdx)}}{{d - x\tan \theta }} = \frac{{{K_1}{\varepsilon _ \circ }(bdx)}}{{d - \frac{{xd}}{l}}} \\ {C_1} = \frac{{{K_1}{\varepsilon _ \circ }Adx}}{{d(l - x)}} \\\end{aligned} \end{equation} $$$C_2$= Capacitance of $QR$$$\begin{equation} \begin{aligned} {C_2} = \frac{{{K_2}{\varepsilon _ \circ }b\left( {dx} \right)}}{{x.\tan \theta }} \\ = \frac{{{K_2}{\varepsilon _ \circ }A\left( {dx} \right)}}{{x.d}} \\\end{aligned} \end{equation} $$$C_1$ and $C_2$ are in series, so their resultant capacitance is,
$$\begin{equation} \begin{aligned} \frac{1}{{{C_ \circ }}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} \\ \frac{1}{{{C_ \circ }}} = \frac{{d(1 - x)}}{{{K_1}{\varepsilon _ \circ }A(dx)}} + \frac{{x.d}}{{{K_2}{\varepsilon _ \circ }A(dx)}} \\ \frac{1}{{{C_ \circ }}} = \frac{d}{{{\varepsilon _ \circ }A(dx)}}\left( {\frac{{1 - x}}{{{K_1}}} + \frac{x}{{{K_2}}}} \right) \\ {C_ \circ } = \frac{{{\varepsilon _ \circ }A{K_1}{K_2}}}{{d\left. {\left\{ {{K_2}l + ({K_1} - {K_2})x} \right.} \right\}}}dx \\\end{aligned} \end{equation} $$So the capacitance of the given capacitor can be calculated by addition of small infinite number of elemental capacitance which are in parallel to each other.$$\begin{equation} \begin{aligned} {C_R} = \int\limits_{x = 0}^{x = l} {{C_ \circ }} = \int\limits_0^l {\frac{{{\varepsilon _ \circ }A{K_1}{K_2}}}{{d\left. {\left\{ {{K_2}l + ({K_1} - {K_2})x} \right.} \right\}}}dx} \\ {C_R} = \frac{{{K_1}{K_2}{\varepsilon _ \circ }A}}{{({K_2} - {K_1})}}\ln \frac{{{K_2}}}{{{K_1}}} \\ {C_R} = \frac{{C{K_1}{K_2}}}{{({K_2} - {K_1})}}\ln \frac{{{K_2}}}{{{K_1}}} \\\end{aligned} \end{equation} $$Where $C = \frac{{{\varepsilon _ \circ }A}}{d}$
