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Equivalent Capacitance Of Complex Circuits

Equivalent Capacitance Of Complex Circuits :- In electrical circuits, when we use capacitors, they are often connected to each other in various combinations. When these combinations contain two or more capacitors and it becomes difficult to represent them in a simple form, such circuits are called complex circuits. For these complex circuits, we need to calculate the equivalent capacitance.

(1). Equivalent capacitance of complex circuits using nodal analysis

When two or more capacitors are connected in a way that makes it difficult to classify them as being purely in series or parallel, we solve such circuits using nodal analysis. In this method, we assume the potentials at the various junction points (nodes) (where more than two branches meet) of the circuit, apply Kirchhoff’s Current Law (KCL) at these nodes, and then analyze the circuit by solving the resulting equations.

Main principles and useful formulas :

(a). Charge on the capacitor :

Where and are the potentials at the ends of the capacitor.

(b). Kirchhoff’s current law (KCL) :

Total current entering a node = Total current leaving the node

Since, in the steady state, the current flowing through a capacitor is zero, therefore from dQ/dt = 0 we get Q = constant.

Therefore, the charge of all bodies connected at any junction point will remain constant.

Procedure of Nodal Analysis :

Understand the circuit and identify all its nodes.
Choose one node as zero potential (ground).
Assign potentials to the remaining nodes.
Apply KCL (Kirchhoff’s Current Law) at each node — that is, the law of conservation of charge.
Solve the equations to determine the potential at all nodes.
For the two points between which the equivalent capacitance is to be found, determine the potential difference (ΔV) and the total charge of the system (Q_total).
Finally, use :

$\displaystyle {{C}_{eq}}=\frac{{{Q}_{total}}}{\Delta V}$

Example 1.

Find the equivalent capacitance between points A and B in the following circuit diagram.

Solution :

Here, between points A and B, we must assume that a battery is connected. Therefore, let us assume that a battery of V volts is connected between points A and B, with its positive terminal connected to point A and its negative terminal connected to point B. In the following figure, the potentials at the junction points (nodes) P and Q are assumed to be x and y respectively.

All the capacitor plates connected to nodes P and Q constitute an isolated system; therefore, the net charge on them is zero. Hence,

$\displaystyle \left( x-V \right){{C}_{1}}+\left( x-0 \right){{C}_{2}}+\left( x-y \right){{C}_{3}}=0$

$\displaystyle \Rightarrow \left( x-V \right){{C}_{1}}+x{{C}_{2}}+\left( x-y \right){{C}_{3}}=0$ …..(1)

Similarly,

$\displaystyle \left( y-V \right){{C}_{2}}+\left( y-x \right){{C}_{3}}+\left( y-0 \right){{C}_{1}}=0$

$\displaystyle \Rightarrow \left( y-V \right){{C}_{2}}+\left( y-x \right){{C}_{3}}+y{{C}_{1}}=0$ …..(2)

Adding equations (1) and (2),

$\displaystyle \left[ \left( x-V \right){{C}_{1}}+x{{C}_{2}}+\left( x-y \right){{C}_{3}} \right]+\left[ \left( y-V \right){{C}_{2}}+\left( y-x \right){{C}_{3}}+y{{C}_{1}} \right]=0$

$\displaystyle \Rightarrow \left[ \left( x-V \right){{C}_{1}}+x{{C}_{2}} \right]+\left[ \left( y-V \right){{C}_{2}}+y{{C}_{1}} \right]=0$

$\displaystyle \Rightarrow x{{C}_{1}}-V{{C}_{1}}+x{{C}_{2}}+y{{C}_{2}}-V{{C}_{2}}+y{{C}_{1}}=0$

$\displaystyle \Rightarrow \left( {{C}_{1}}+{{C}_{2}} \right)x+\left( {{C}_{1}}+{{C}_{2}} \right)y=\left( {{C}_{1}}+{{C}_{2}} \right)V$

$\displaystyle \Rightarrow x+y=V$

$\displaystyle \Rightarrow x=V-y$ …..(3)

Using equation (3) in equation (1),

$\displaystyle \left( V-y-V \right){{C}_{1}}+\left( V-y \right){{C}_{2}}+\left( V-y-y \right){{C}_{3}}=0$

$\displaystyle y=\left[ \frac{{{C}_{2}}+{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right]V$ …..(4)

and

$\displaystyle x=V-\left[ \frac{{{C}_{2}}+{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right]V$

$\displaystyle x=\left[ \frac{{{C}_{1}}+{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right]V$ …..(5)

The total charge supplied to the circuit by the battery :

$\displaystyle Q={{Q}_{1}}+{{Q}_{2}}={{C}_{1}}\left( V-x \right)+{{C}_{2}}\left( V-y \right)$

[The battery only charges the capacitors that are directly connected to it. The charge that appears on is due to the potentials created by the charge stored in C1 and C2, not directly from the battery. Therefore, the total charge is stored only in and . To understand this, imagine a tank with two pipes coming out of it (like and ), and an internal connection between these two pipes (like ). Water from the tank will flow only into the two pipes directly connected to it. However, the connection between the pipes () does not take water directly from the tank—it only facilitates the exchange of water between the two pipes internally.]

$\displaystyle Q={{C}_{1}}\left( V-\left[ \frac{{{C}_{1}}+{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right]V \right)+{{C}_{2}}\left( V-\left[ \frac{{{C}_{2}}+{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right]V \right)$

$\displaystyle \frac{Q}{V}=\left( \frac{{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right)+\left( \frac{{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}} \right)$

$\displaystyle \Rightarrow \frac{Q}{V}=\frac{2{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{3}}+{{C}_{2}}{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}}$

Now the equivalent capacitance of the system between points A and B,

$\displaystyle {{C}_{eq}}=\frac{Q}{V}=\frac{2{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{3}}+{{C}_{2}}{{C}_{3}}}{{{C}_{1}}+{{C}_{2}}+2{{C}_{3}}}$

Determining the charge on various capacitors using nodal analysis.

Example 2.

Find the charge on each capacitor in the following circuit diagram.

Solution :

In the given circuit diagram, let the potentials at the junction points (nodes) P and Q be x and y, respectively. Then the voltage division will be as follows :-

All the capacitor plates connected to nodes P and Q form an isolated system :-

Before charging, the total charge of the isolated system was zero. Therefore, even after charging, the total charge on all the capacitor plates connected to nodes P and Q will remain zero. Hence,

$\displaystyle \left( x-\text{10} \right)2\mu +\left( x-y \right)4\mu +\left( x-0 \right)2\mu =0$

$\displaystyle 2x-y=5$ …..(6)

Similarly,

$\displaystyle \left( y-x \right)4\mu +\left( y-15 \right)4\mu +\left( y-5 \right)4\mu =0$

$\displaystyle 3y-x=20$ …..(7)

Adding equations (6) and (7),

$\displaystyle x+2y=25$

$\displaystyle \Rightarrow x=25-2y$ …..(8)

Using equation (8) in equation (6),

$\displaystyle y=9Volt$

And similarly,

$\displaystyle x=25-2y=25-2\times 9=7Volt$

Therefore, the values of the charge on each capacitor will be as follows :-

(2). Solving problems based on the circuit and the charge that flows through the switch/key (after the switch is shorted/closed).

[JEE]

In such questions, an electric circuit is given that contains some capacitors and a switch. After the switch is closed (shorted), it is asked which parts of the circuit will have a flow of charge and in which direction the charge will flow !

There are two steps to solving these types of problems :

(1). Determine the charge on each capacitor in both states of the switch (when the switch is open and later when it is closed).

(2). Now, the amount of charge flowing from a given point/branch will be equal to the difference between the initial and final values of charge on the capacitors.

$\displaystyle q=\left| \sum{{{q}_{i}}-}\sum{{{q}_{f}}} \right|$

Example 3.

In the circuit shown below, find the amount of charge that flows through the switch after closing the switch S.

Solution :

Before closing the switch, determine the total charge on the two upper capacitors (4 μF and 2 μF) or the two lower capacitors (8 μF and 2 μF). Let us first calculate the charge on the two upper capacitors (4 μF and 2 μF) before closing the switch.

Before closing the switch S, the 4 μF and 2 μF capacitors are connected in series, and since in a series combination the charge on each capacitor is the same, initially the total charge is zero (the negative charge on the right plate of the 4 μF capacitor is equal in magnitude to the positive charge on the left plate of the 2 μF capacitor). Therefore

q_i = 0

Now let’s close the switch and find the final value of the charge.

From the above figure, x = 15 volts; therefore, the final charge on the right plate of the 4 μF capacitor and the left plate of the 2 μF capacitor is,

$\displaystyle {{q}_{f}}=(x-20)\times 4\mu +(x-0)2\mu =(15-20)\times 4\mu +(15-0)2\mu$

$\displaystyle \Rightarrow {{q}_{f}}=-20\mu C+30\mu C$

$\displaystyle \therefore {{q}_{f}}=+10\mu C$

Hence, the charge flowing through the key,

$\displaystyle q=\left| \sum{{{q}_{i}}-}\sum{{{q}_{f}}} \right|=\left| 0-10\mu C \right|=+10\mu C$

Note :- The above question can also be solved using the concepts of series and parallel combinations, without performing nodal analysis.

Example 4.

Determine the heat produced in the capacitors after closing the switch in the above question (Example 3).

Solution :

We know that the heat generated :

$\displaystyle H=\sum\limits_{k=1}^{N}{\frac{{{\left( {{Q}_{fk}}-{{Q}_{ik}} \right)}^{2}}}{2{{C}_{k}}}}=\sum\limits_{k=1}^{N}{\frac{{{\left( \Delta {{Q}_{k}} \right)}^{2}}}{2{{C}_{k}}}}$

Values of charges on all capacitors before and after closing the key S :-

Hence heat generated :

$\displaystyle H=\sum\limits_{k=1}^{N}{\frac{{{\left( \Delta {{Q}_{k}} \right)}^{2}}}{2{{C}_{k}}}}=\frac{\text{1}}{2}\left[ \frac{{{\left( \Delta {{q}_{1}} \right)}^{2}}}{{{C}_{1}}}+\frac{{{\left( \Delta {{q}_{2}} \right)}^{2}}}{{{C}_{2}}}+\frac{{{\left( \Delta {{q}_{3}} \right)}^{2}}}{{{C}_{3}}}+\frac{{{\left( \Delta {{q}_{4}} \right)}^{2}}}{{{C}_{4}}} \right]$

$\displaystyle \Rightarrow H=\frac{\text{1}}{2}\left[ \frac{{{\left( \frac{20\mu }{3} \right)}^{2}}}{4\mu }+\frac{{{\left( \frac{10\mu }{3} \right)}^{2}}}{2\mu }+\frac{{{\left( 8\mu \right)}^{2}}}{20\mu }+\frac{{{\left( 2\mu \right)}^{2}}}{20\mu } \right]$

$\displaystyle \therefore H=13.33\mu Joule$

Another way to solve

Generated heat (H) = | Total energy stored in the system before closing the switch – Total energy stored in the system after closing the switch |

Equivalent capacitance before closing the switch,

$\displaystyle {{\left( {{C}_{eq.}} \right)}_{i}}=\frac{\left( 4\mu \times 2\mu \right)}{\left( 4\mu +2\mu \right)}+\frac{\left( 8\mu \times 2\mu \right)}{\left( 8\mu +2\mu \right)}=\frac{44}{15}\mu F$

Initial energy,

$\displaystyle {{U}_{i}}=\frac{1}{2}{{\left( {{C}_{eq.}} \right)}_{i}}{{V}^{2}}=\frac{1}{2}\times \left( \frac{44}{15}\mu \right)\times {{\left( 20 \right)}^{2}}=586.67\mu J$

Equivalent capacitance after closing the switch,

$\displaystyle {{\left( {{C}_{eq.}} \right)}_{f}}=\frac{\left( 4\mu +8\mu \right)\times \left( 2\mu +2\mu \right)}{\left( 4\mu +8\mu \right)+\left( 2\mu +2\mu \right)}=3\mu F$

Final energy,

$\displaystyle {{U}_{f}}=\frac{1}{2}{{\left( {{C}_{eq.}} \right)}_{f}}{{V}^{2}}=\frac{1}{2}\times \left( 3\mu \right)\times {{\left( 20 \right)}^{2}}=600\mu J$

Hence the heat generated (H),

$\displaystyle \therefore H=\left| 586.67\mu J-600\mu J \right|=13.33\mu J$

Example 5.

In the following diagram, after closing switch S, determine the values of the charges flowing in sections 1 and 2 of the circuit in the directions indicated by the arrows.

Solution :

Before closing switch S, capacitors C₁ and C₂ are connected in series, therefore the charge on both will be the same. The equivalent capacitance of the circuit :

$\displaystyle {{C}_{eq.}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}$

Initial values of the charges,

$\displaystyle {{q}_{1}}={{q}_{2}}={{C}_{eq.}}\varepsilon =\left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon$

After closing switch S, the values of the electrical potentials and charges at the various junction points (nodes) are :

Now the potential difference across the capacitor of capacitance C₁ is zero, hence the new value of charge on C₁ is,

$\displaystyle q_{\text{1}}^{'}={{C}_{1}}\left( \varepsilon -\varepsilon \right)=0$

Similarly, the new value of charge on the capacitor of capacitance C₂,

$\displaystyle q_{\text{2}}^{'}={{C}_{2}}\left( \varepsilon -0 \right)=\varepsilon {{C}_{2}}$

Earlier, the charge on the right plate of capacitor C₁ was $\displaystyle \left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon$ and after closing the switch , the charge on its right plate becomes zero. This means that a charge of $\displaystyle -\left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon$ has flowed in the direction shown in section 2, so that the net charge on this plate is now zero.

Therefore, after closing the switch , the magnitude of the charge flowing in the direction shown in section 2 is $\displaystyle -\left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon$ .

Now, the initial charge on the upper plate of capacitor C₂ was $\displaystyle \left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon$ and after closing the switch S, the charge on its upper plate becomes $\displaystyle \varepsilon {{C}_{2}}$ . The increase in charge on the upper plate of C₂ is $\displaystyle \left[ \varepsilon {{C}_{2}}-\left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon \right]=+\left( \frac{\varepsilon C_{2}^{2}}{{{C}_{1}}+{{C}_{2}}} \right)$ .

From charge conservation,

$\displaystyle \left[ \left( \frac{\varepsilon C_{2}^{2}}{{{C}_{1}}+{{C}_{2}}} \right)+\left( \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)\varepsilon \right]=\varepsilon {{C}_{2}}$

Thus, after closing the switch S, the magnitude of the charge flowing in the direction shown in section 1 is $\displaystyle +\varepsilon {{C}_{2}}$ .

(3). Finding the equivalent capacitance of complex circuits using symmetry

Some circuits exhibit symmetry, much like the reflection of an object in a mirror. To find the equivalent resistance of these circuits, we need to identify the equipotential points in the circuit and connect them. The identification of equipotential points is based on the fact that the circuit is symmetric with respect to these points.

In a given circuit, there can be two axes of symmetry :-

(a) Normal Symmetric Axis :This axis (YY’) is perpendicular to the direction of charge flow. The electric potential is the same at all points located on this axis. In the above figure :-

V₆ = V₅

V₂ = V₀ = V₄

V₇ = V₈

(b) Parallel Symmetric Axis : This axis (XX’) lies in the direction of the flow of charge. The electric potential at points lying on this axis is never the same. As we move in the direction of electric current, the electric potential decreases along this axis. In the above figure, if point A is connected to the positive terminal of the battery and point B to the negative terminal, then :

V₁ > (V₆ = V₅) > (V₂ = V₀ = V₄) > (V₇ = V₈) > V₃

If the circuit is folded along the axis of symmetry (XX’), then, due to the equal electric potential of the overlapping points [(5 and 6), (2, 0 and 4) व (7 and 8)], the circuit can be represented as follows :-

And on solving further :-

Therefore, equivalent capacitance between A and B :

C_AB = 2C/3

Note :- The above circuit can also be considered as being divided into two parts with respect to the parallel axis of symmetry (XX’), and the capacitance of each part will be C’ = C/3, as shown in the figure below :-

Now, if the equivalent capacitance of both parts is determined (both parts will be considered to be connected in parallel), then :-

Note :- Sometimes, it is easy to calculate the equivalent capacitance in capacitor circuits using symmetry, but when the symmetry is not obvious or difficult to identify, this method can be confusing. In such situations, nodal analysis is a better method. In this method, potential values are assigned to the various nodes (junction points) of the circuit, and equations are formed using the principle of charge conservation (Kirchhoff’s current law). These equations are then solved using the principle Q = CV. This method can be computationally intensive, but it is applicable to all types of circuits and helps avoid errors.

Let’s solve the above problem using nodal analysis !

Assume that a battery of 100 volts (you can assume a battery of any voltage, as this does not affect the equivalent capacitance of the circuit) is connected between points A and B. Now, the potential at various nodes (junction points) and the equivalent capacitance are determined as shown in the following figure :-

Example 6.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = 1.3 C

Example 7.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = 11C/8

Example 8.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = 7C/5

Example 9.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = 4C/3

Example 10.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = C

Example 11.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Therefore, the equivalent capacitance between points A and B,

C_AB = 32C/11

(4). The equivalent capacitance of the Wheatstone bridge circuit

A Wheatstone bridge is a circuit consisting of four capacitors, with two capacitors of capacitance C₁and C₃ connected at the top and two more capacitors of capacitance C₂ and C₄ connected at the bottom. An external voltage source is connected between the opposite ends (A and B), and a fifth capacitor of capacitance C₅ is connected across the intermediate branch (bridge arm).

Types of Wheatstone bridge circuits :-

(a) Unbalanced Wheatstone bridge

(b) Balanced Wheatstone bridge

(a) Unbalanced Wheatstone bridge : यदि C₁/C₂ ≠ C₃/C₄ अथवा C₁ . C₄ ≠ C₂ . C₃ तब व्हीटस्टोन सेतु असंतुलित अवस्था में कहा जाता है। इस अवस्था में मध्य संधारित्र (bridge arm) C₅ सक्रिय होता है और इसमें भी आवेश संचित होता है।

If C₁/C₂ ≠ C₃/C₄ or C₁ . C₄ ≠ C₂ . C₃, then the Wheatstone bridge is said to be unbalanced. In this state, the central capacitor (bridge arm) C₅ is active and also stores charge.

Equivalent capacitance of an unbalanced Wheatstone bridge circuit by nodal analysis :

Assume that a 1-volt battery is connected between junction points A and B, and the potentials at the ends (junction points) of capacitor C₅ are x and y, as shown in the figure below :-

Nodal equation for x :

$\displaystyle (x-1){{C}_{1}}+(x-0){{C}_{3}}+(x-y){{C}_{5}}=0$

$\displaystyle \Rightarrow x({{C}_{1}}+{{C}_{3}}+{{C}_{5}})-y{{C}_{5}}={{C}_{1}}$ …..(1)

Nodal equation for y :

$\displaystyle (y-1){{C}_{2}}+(y-x){{C}_{5}}+(y-0){{C}_{4}}=0$

$\displaystyle \Rightarrow y({{C}_{2}}+{{C}_{4}}+{{C}_{5}})-x{{C}_{5}}={{C}_{2}}$ …..(2)

From equation (1),

$\displaystyle y=\frac{x({{C}_{1}}+{{C}_{3}}+{{C}_{5}})}{{{C}_{5}}}-\frac{{{C}_{1}}}{{{C}_{5}}}$ …..(3)

Substituting the value of y from equation (3) in equation (2),

$\displaystyle \left[ \frac{x({{C}_{1}}+{{C}_{3}}+{{C}_{5}})}{{{C}_{5}}}-\frac{{{C}_{1}}}{{{C}_{5}}} \right]({{C}_{2}}+{{C}_{4}}+{{C}_{5}})-x{{C}_{5}}={{C}_{2}}$

$\displaystyle \frac{x({{C}_{1}}+{{C}_{3}}+{{C}_{5}})}{{{C}_{5}}}({{C}_{2}}+{{C}_{4}}+{{C}_{5}})-\frac{{{C}_{1}}}{{{C}_{5}}}({{C}_{2}}+{{C}_{4}}+{{C}_{5}})-x{{C}_{5}}={{C}_{2}}$

$\displaystyle \Rightarrow x\left[ \frac{({{C}_{1}}+{{C}_{3}}+{{C}_{5}})({{C}_{2}}+{{C}_{4}}+{{C}_{5}})}{{{C}_{5}}}-\frac{{{C}_{5}}}{1} \right]=\frac{{{C}_{2}}}{1}+\frac{{{C}_{1}}}{{{C}_{5}}}({{C}_{2}}+{{C}_{4}}+{{C}_{5}})$

$\displaystyle \Rightarrow x\left[ \frac{{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+C_{5}^{2}+{{C}_{4}}{{C}_{5}}-C_{5}^{2}}{{{C}_{5}}} \right]=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)}{{{C}_{5}}}$

$\displaystyle \Rightarrow x\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)=\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)$

$\displaystyle \Rightarrow x=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}$ …..(4)

On substituting the value of x from equation (4) in equation (3), we get

$\displaystyle y=\frac{x({{C}_{1}}+{{C}_{3}}+{{C}_{5}})}{{{C}_{5}}}-\frac{{{C}_{1}}}{{{C}_{5}}}=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}\times \frac{({{C}_{1}}+{{C}_{3}}+{{C}_{5}})}{{{C}_{5}}}-\frac{{{C}_{1}}}{{{C}_{5}}}$

$\displaystyle \Rightarrow y=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)({{C}_{1}}+{{C}_{3}}+{{C}_{5}})-{{C}_{1}}\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right){{C}_{5}}}$

$\displaystyle \Rightarrow y=\frac{C_{1}^{2}{{C}_{2}}+C_{1}^{2}{{C}_{4}}+C_{1}^{2}{{C}_{5}}+{{C}_{1}}{{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{3}}{{C}_{4}}+{{C}_{1}}{{C}_{3}}{{C}_{5}}+{{C}_{2}}{{C}_{3}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}{{C}_{5}}+{{C}_{1}}C_{5}^{2}+{{C}_{2}}C_{5}^{2}-C_{1}^{2}{{C}_{2}}-C_{1}^{2}{{C}_{5}}-C_{1}^{2}{{C}_{4}}-{{C}_{1}}{{C}_{2}}{{C}_{3}}-{{C}_{1}}{{C}_{3}}{{C}_{5}}-{{C}_{1}}{{C}_{3}}{{C}_{4}}-{{C}_{1}}{{C}_{2}}{{C}_{5}}-{{C}_{1}}{{C}_{4}}{{C}_{5}}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right){{C}_{5}}}$

$\displaystyle \Rightarrow y=\frac{{{C}_{1}}{{C}_{2}}{{C}_{5}}+{{C}_{2}}{{C}_{3}}{{C}_{5}}+{{C}_{1}}C_{5}^{2}+{{C}_{2}}C_{5}^{2}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right){{C}_{5}}}$

$\displaystyle \Rightarrow y=\frac{{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{5}}+{{C}_{2}}{{C}_{5}}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}$ …..(5)

The total charge supplied to the system by the battery,

$\displaystyle Q=x{{C}_{3}}+y{{C}_{4}}$

Using the values of x and y from equations (4) and (5),

$\displaystyle Q=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}\times {{C}_{3}}+\frac{{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{5}}+{{C}_{2}}{{C}_{5}}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}\times {{C}_{4}}$

Capacitance,

$\displaystyle C=\frac{Q}{V}=\frac{Q}{1}=Q$

$\displaystyle \Rightarrow C=Q=\frac{\left( {{C}_{2}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}} \right)}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}\times {{C}_{3}}+\frac{{{C}_{1}}{{C}_{2}}+{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{5}}+{{C}_{2}}{{C}_{5}}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}\times {{C}_{4}}$

$\displaystyle \Rightarrow C=\frac{{{C}_{2}}{{C}_{3}}{{C}_{5}}+{{C}_{1}}{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{3}}{{C}_{5}}+{{C}_{1}}{{C}_{3}}{{C}_{4}}+{{C}_{1}}{{C}_{2}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}{{C}_{4}}+{{C}_{1}}{{C}_{4}}{{C}_{5}}+{{C}_{2}}{{C}_{4}}{{C}_{5}}}{\left( {{C}_{1}}{{C}_{2}}+{{C}_{1}}{{C}_{5}}+{{C}_{1}}{{C}_{4}}+{{C}_{2}}{{C}_{3}}+{{C}_{3}}{{C}_{5}}+{{C}_{3}}{{C}_{4}}+{{C}_{2}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}$

$\displaystyle \therefore C=\frac{{{C}_{1}}{{C}_{2}}({{C}_{3}}+{{C}_{4}})+\left( {{C}_{1}}+{{C}_{2}} \right)\left( {{C}_{3}}{{C}_{4}}+{{C}_{3}}{{C}_{5}}+{{C}_{4}}{{C}_{5}} \right)}{\left( {{C}_{2}}+{{C}_{4}} \right)\left( {{C}_{1}}+{{C}_{3}}+{{C}_{5}} \right)+\left( {{C}_{1}}+{{C}_{3}} \right){{C}_{5}}}$ …..(6)

Equation (6) is the required expression for the equivalent capacitance of an unbalanced Wheatstone bridge circuit.

(b) Balanced Wheatstone bridge : If C₁/C₂ = C₃/C₄ or C₁ . C₄ = C₂ . C₃ then the Wheatstone bridge is said to be balanced. In this state, there is no charge on the middle capacitor (bridge arm) C₅, hence it can be considered separate from the circuit.

Since C₅ is inactive, therefore,

Equivalent capacitance of the upper branch :

$\displaystyle {{C}_{up}}=\frac{{{C}_{1}}{{C}_{3}}}{{{C}_{1}}+{{C}_{3}}}$

Equivalent capacitance of the lower branch :

$\displaystyle {{C}_{down}}=\frac{{{C}_{2}}{{C}_{4}}}{{{C}_{2}}+{{C}_{4}}}$

Finally, the equivalent capacitance between A and B :

$\displaystyle {{C}_{eq.}}={{C}_{up}}+{{C}_{down}}=\frac{{{C}_{1}}{{C}_{3}}}{{{C}_{1}}+{{C}_{3}}}+\frac{{{C}_{2}}{{C}_{4}}}{{{C}_{2}}+{{C}_{4}}}$ …..(7)

Different forms of Wheatstone bridge circuit :-

Example 12.

Find the equivalent capacitance between points A and B in the following circuit.

Solution :

Given :

C₁ = 10μF , C₂ = 5μF , C₃ = 5μF , C₄ = 10μF , C₅ = 20μF

Here C₁/C₂ ≠ C₃/C₄ , Hence, this is an unbalanced Wheatstone bridge circuit. From equation (6),

$\displaystyle C=\frac{\left( 10\mu \right)\left( 5\mu \right)(5\mu +10\mu )+\left( 10\mu +5\mu \right)\left( 5\mu \times 10\mu +5\mu \times 20\mu +10\mu \times 20\mu \right)}{\left( 5\mu +10\mu \right)\left( 10\mu +5\mu +20\mu \right)+\left( 10\mu +5\mu \right)\times 20\mu }$

$\displaystyle \therefore C=\frac{80}{11}\mu F$

Note :- Although the above problem involves an unbalanced Wheatstone bridge circuit, there is still a symmetry in the circuit. The capacitors connected on the left side of the circuit (10μF and 5μF) are the same as those connected on the right side, only their order has been reversed. The charge supplied by the battery to the capacitor with capacitance C₁ will be the same as the charge supplied to the capacitor with capacitance C₄. Similarly, the charge supplied by the battery to the capacitor with capacitance C₂ will be the same as the charge supplied to the capacitor with capacitance C₃. Therefore, the potential difference across these capacitors will be equal. This problem can also be solved easily using nodal analysis, as shown below :-

The nodal equation for node x in the above diagram :

$\displaystyle \left( x-100 \right)\times 5\mu +\left( 2x-100 \right)\times 20\mu +\left( x-0 \right)\times 10\mu =0$

$\displaystyle x=\frac{500}{11}Volt$

The total charge supplied to the system by the battery :

$\displaystyle Q=\left( x-0 \right)\times 10\mu +\left( 100-x \right)\times 5\mu =\frac{500}{11}\times 10\mu +\left( 100-\frac{500}{11} \right)\times 5\mu$

$\displaystyle \Rightarrow Q=\frac{8000\mu }{11}$

Hence the capacitance of the system,

$\displaystyle C=\frac{Q}{V}=\frac{8000\mu /11}{100}=\frac{80}{11}\mu F$

(5). Equivalent capacitance of Extensive or Extended Wheatstone bridge

The extended Wheatstone bridge is larger and more complex than the standard Wheatstone bridge (which has only four branches and a central bridge arm).

The following circuit diagram shows two interconnected Wheatstone bridges. One bridge is connected between AEGHFA and the other between EGBHFE. This circuit is an extended version of the standard Wheatstone bridge, exhibiting symmetry in the ratio of capacitances on the left and right sides of the two branches EF and GH.

In the above diagram, the ratio of the capacitances of the capacitors connected in branches AE and EG is equal to the ratio of the capacitances of the capacitors connected in branches AF and FH. Therefore, branch EF can be removed from the bridge AEGHFA. Similarly, branch GH can also be removed from the bridge EGBHFE.

(6). Equivalent capacitance of complex circuits containing more than one cell

(i) Multiple cells/batteries and capacitors connected in a single loop :-

If multiple batteries/cells are connected in the same loop along with other components (capacitors, resistors, inductors, etc.), then we can use the ‘Branch Manipulation Method’ to redraw the circuit in a simplified manner.

Branch Manipulation Method :- In any circuit, when a battery and a capacitor (or resistor, inductor, etc.) are connected in series, we can rearrange the positions or order of these circuit components to simplify the circuit. In the following diagram, the two cells have been replaced by their equivalent cell, and the two capacitors have been replaced by their equivalent capacitor, and the circuit has been rearranged accordingly.

Note :- However, the above method is only useful when the capacitors are initially uncharged and there is only one loop in the circuit.

(ii) Solving a circuit consisting of multiple cells/batteries and capacitors connected in multiple loops using nodal analysis :-

If the circuit has more than one loop or if the capacitors are already charged, then the circuit is solved using nodal analysis, as shown in Example 13 below.

Example 13.

Find the charge on each capacitor in the following circuit.

Solution :

(iii) Determination of the New Charge on the Capacitors After Closing the Switch (When the Capacitors Are Initially Charged) :-

Example 14.

In the circuit shown below, find the new values of the charges on the capacitors after closing the switch S.

Solution :

Solving the above problem using the Branch Manipulation Method will not yield the correct result, because although it appears that the capacitors are connected in series, this is not the case. Capacitors can only be considered to be connected in series when :

👉All capacitors are initially uncharged or have the same amount of charge.

Therefore, to solve this problem, we will use nodal analysis. Solution using nodal analysis :

Example 15.

In example 14 above, calculate the amount of charge that flows through the switch S after closing it.

Solution :

Let x be the charge flowing downwards from switch S, then

$\displaystyle +80\mu +x=-80\mu$

$\displaystyle \therefore x=-160\mu C$

Therefore, the charge flowing downwards through switch S is –160μC.

(7). Equivalent capacitance of an infinite ladder network

In the circuit shown above, capacitors C₁ , C₂ and C₃ are repeated; therefore, the above circuit can be simplified as follows :-

The circuit shown above appears the same when viewed from points X and Y as it does when viewed from points A and B. Therefore, the capacitance between points X and Y will be equivalent to the capacitance C_AB between points A and B.

Now, C_AB and C₃ are connected in parallel, and their resultant capacitance is connected in series with C₁ and C₂. Therefore, the equivalent capacitance between points A and B can be calculated as follows :-

$\displaystyle \frac{1}{{{C}_{AB}}}=\frac{1}{{{C}_{1}}}+\frac{1}{\left( {{C}_{3}}+{{C}_{AB}} \right)}+\frac{1}{{{C}_{2}}}$

$\displaystyle \Rightarrow \frac{1}{{{C}_{AB}}}=\frac{\left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{2}}+{{C}_{1}}{{C}_{2}}+\left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{1}}}{\left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{1}}{{C}_{2}}}$

$\displaystyle \Rightarrow \left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{1}}{{C}_{2}}=\left[ \left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{2}}+{{C}_{1}}{{C}_{2}}+\left( {{C}_{3}}+{{C}_{AB}} \right){{C}_{1}} \right]{{C}_{AB}}$

$\displaystyle \Rightarrow {{C}_{1}}{{C}_{2}}{{C}_{3}}+{{C}_{1}}{{C}_{2}}{{C}_{AB}}={{C}_{2}}{{C}_{3}}{{C}_{AB}}+{{C}_{2}}C_{AB}^{2}+{{C}_{1}}{{C}_{2}}{{C}_{AB}}+{{C}_{1}}{{C}_{3}}{{C}_{AB}}+{{C}_{1}}C_{AB}^{2}$

$\displaystyle \Rightarrow \left( {{C}_{1}}+{{C}_{2}} \right)C_{AB}^{2}+\left( {{C}_{1}}{{C}_{3}}+{{C}_{2}}{{C}_{3}} \right){{C}_{AB}}-{{C}_{1}}{{C}_{2}}{{C}_{3}}=0$

$\displaystyle \therefore {{C}_{AB}}=\frac{-\left( {{C}_{1}}{{C}_{3}}+{{C}_{2}}{{C}_{3}} \right)\pm \sqrt{{{\left( {{C}_{1}}{{C}_{3}}+{{C}_{2}}{{C}_{3}} \right)}^{2}}+4\left( {{C}_{1}}+{{C}_{2}} \right){{C}_{1}}{{C}_{2}}{{C}_{3}}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

Equivalent Capacitance Of Complex Circuits

(1). Equivalent capacitance of complex circuits using nodal analysis

(2). Solving problems based on the circuit and the charge that flows through the switch/key (after the switch is shorted/closed).

(3). Finding the equivalent capacitance of complex circuits using symmetry

(4). The equivalent capacitance of the Wheatstone bridge circuit

(5). Equivalent capacitance of Extensive or Extended Wheatstone bridge

(7). Equivalent capacitance of an infinite ladder network

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