3.10 Combination of Resistors — Series and Parallel

The current through a single resistor \(R\) across which there is a potential difference \(V\) is given by Ohm’s law \(I=V / R\). Resistors are sometimes joined together and there are simple rules for calculation of equivalent resistance of such combination.

Combination of Resistances

A number of resistances can be connected and all the complicated combinations can be reduced to two different types, namely

• Series and
• Parallel.

Steps to Calculate Equivalent Resistance

• Step 1: Identify the terminal across which you need to calculate the equivalent resistance (In the figure above terminal \(A-B\)).
• Step 2: Apply a voltage source \(V\) across the terminal \(A-B\).
• Step 3: Calculate the current \(I\) drawn from the voltage source \(V\).
• Step 4: The equivalent resistance of a combination is defined as \(R _{ eq }=\frac{V}{I}\)

Resistances in Series

When the resistances (or any type of elements) are connected end to end then they are said to be in series. The current through each element is the same in a series circuit. Two resistors are said to be in series if only one of their endpoints is joined (Fig. 3.13).

If a third resistor is joined with the series combination of the two (Fig. 3.14), then all three are said to be in series. Clearly, we can extend this definition to series combination of any number of resistors.

Equivalent resistance calculation in Series circuit

Consider two resistors \(R_1\) and \(R_2\) in series(Fig 3.13). The charge which leaves \(R_1\) must be entering \(R_2\). Since current measures the rate of flow of charge, this means that the same current \(I\) flows through \(R_1\) and \(R_2\).
By Ohm’s law:
Potential difference across \(R_1=V_1=I R_1\), and
The potential difference across \(R_2=V_2=I R_2\).
The potential difference \(V\) across the combination is \(V_1+V_2\). Hence,
\(
V=V_1+V_2=I\left(R_1+R_2\right)
\)
This is as if the combination had an equivalent resistance \(R_{e q}\), which by Ohm’s law is
\(
R_{e q} \equiv \frac{V}{I}=\left(R_1+R_2\right)
\)
If we had three resistors connected in series, then similarly
\(
V=I R_1+I R_2+I R_3=I\left(R_1+R_2+R_3\right) .
\)
This is as if the combination had an equivalent resistance \(R_{e q}\), which by Ohm’s law is
\(
R_{e q} \equiv \frac{V}{I}=\left(R_1+R_2+R_3\right)
\)
This obviously can be extended to a series combination of any number n of resistors \(R_1, R_2 \ldots \ldots, R_{ n }\). The equivalent resistance \(R_{e q}\) is
\(
R_{e q}=R_1+R_2+\ldots+R_{ n }=\sum_{j=1}^n R_j \quad(n \text { resistances in series }) .
\)

About Series combination

• The effective resistance appearing across the battery (or between the terminals \(A\) and \(B\) ) is \(R=R_1+R_2+R_3+\ldots \ldots \ldots \ldots+R_n\) (this means \(R_{e q}\) is greater then any resistor) and \(\quad V = V _1+ V _2+ V _3+\ldots \ldots \ldots \ldots+ V _{ n }\)
• If \(n\) identical resistors of resistance \(R\) each, are connected in series, then equivalent resistance, \(R_{\text {eq }}=n R\) and potential difference across each resistor is \(V^{\prime}=V / n\).
• The potential difference across a resistor is proportional to the resistance. Power in each resistor is also proportional to the resistance
  \(
  \because \quad V = IR \text { and } P=I^2 R
  \)
  where \(I\) is same through any of the resistor.
• \(
  V _1=\frac{R_1}{R_1+R_2+\ldots \ldots .+R_n} V ; V _2=\frac{R_2}{R_1+R_2+\ldots \ldots \ldots+R_n} V ; \quad \text { etc }
  \)

Example 27: Find the current in the circuit.

Solution: \(R_{e q}=1+2+3=6 \Omega\) the given circuit is equivalent to

current \(i =\frac{v}{R_{e q}}=\frac{30}{6}=5 A\)

Example 28: In the figure shown \(B_1, B_2\) and \(B_3\) are three bulbs rated as \((200 V, 50 W),(200 V, 100 W)\) and ( \(200 V, 25 W)\) respectively. Find the current through each bulb and which bulb will give more light?

Solution:

\(
\begin{aligned}
& R _1=\frac{(200)^2}{50} ; & R _2=\frac{(200)^2}{100} ; \quad R _3=\frac{(200)^2}{25}
\end{aligned}
\)
the current following through each bulb is
\(
=\frac{200}{R_1+R_2+R_3}=\frac{200}{(200)^2\left[\frac{2+1+4}{100}\right]}=\frac{100}{200 \times 7}=\frac{1}{14} A
\)
Since \(\quad R_3>R_1>R_2\)
\(\therefore \quad\) Power consumed by bulb \(=i^2 R\)
\(\therefore \quad\) if the resistance is of higher value then it will give more light.
\(\therefore \quad\) Here Bulb \(B_3\) will give more light.

Resistances in Parallel

Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends are joined together (Fig. 3.15).

Consider now the parallel combination of two resistors (Fig. 3.15). The charge that flows in at A from the left flows out partly through \(R_1\) and partly through \(R_2\). The currents \(I, I_1, I_2\) shown in the figure are the rates of flow of charge at the points indicated. Hence,
\(
I=I_1+I_2
\)
The potential difference between A and B is given by Ohm’s law applied to \(R_1\)
\(
V=I_1 R_1
\)
Also, Ohm’s law applied to \(R_2\) gives
\(
\begin{aligned}
& V=I_2 R_2 \\
& \therefore I=I_1+I_2=\frac{V}{R_1}+\frac{V}{R_2}=V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)
\end{aligned}
\)
If the combination was replaced by an equivalent resistance \(R_{e q}\), we would have, by Ohm’s law
\(
I=\frac{V}{R_{e q}}
\)
Hence,
\(
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}
\)

We can easily see how this extends to three resistors in parallel (Fig. 3.16).

\(
I=I_1+I_2+I_3
\)
and applying Ohm’s law to \(R_1, R_2\) and \(R_3\) we get,
\(
V=I_1 R_1, V=I_2 R_2, V=I_3 R_3
\)
So that
\(
I=I_1+I_2+I_3=V\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)
\)
An equivalent resistance \(R_{e q}\) that replaces the combination, would be such that
\(
I=\frac{V}{R_{e q}}
\)
and hence
\(
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
\)

We can reason similarly for any number of resistors in parallel. The equivalent resistance of \(n\) resistors \(R_1, R_2 \ldots, R_{ n }\) is
\(
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots+\frac{1}{R_{ n }}=\sum_{j=1}^n \frac{1}{R_{ n }} \quad(n \text { resistances in series }) .
\)

Few examples are given below.

In the figure (a) and (b) all the resistors are connected between points A and B so they are in parallel.

About parallel combination

• The potential difference across each resistor is the same
• \(I = I _1+ I _2+ I _3+\ldots \ldots \ldots I _{ n }\).
• Effective resistance (R) then \(\frac{l}{R}=\frac{l}{R_1}+\frac{l}{R_2}+\frac{l}{R_3}+\ldots \ldots .+\frac{l}{R_n}\). ( \(R\) is less than each resistor).
• If \(n\) identical resistors of resistance \(R\) each, are connected in parallel, then equivalent resistance, \(R_{ eq }=\frac{R}{n}\) and current through each resistor is \(I^{\prime}=\frac{I}{n}\).
• Current in different resistors is inversely proportional to the resistance.
  \(
  I _1: I _2: \ldots \ldots \ldots I _{ n }=\frac{1}{R_1}: \frac{1}{R_2}: \frac{1}{R_3}: \ldots \ldots . . \frac{1}{R_n} .
  \)
  \(
  I _1=\frac{G_1}{G_1+G_2+\ldots \ldots \ldots+G_n} I, I _2=\frac{G_2}{G_1+G_2+\ldots \ldots .+G_n} I \text {, etc. }
  \)
  \(
  \text { where } G =\frac{1}{ R }=\text { Conductance of a resistor. [Its unit is } \Omega^{-1} \text { or } \mho \text { (mho)] }
  \)

Example 29: When two resistors are in parallel combination then determine \(i_1\) and \(i_2\), if the combination carries a current \(i\).

Solution:
\(
\begin{aligned}
& i _1 R _1= i _2 R _2 \\
& \frac{i_1}{i_2}=\frac{R_2}{R_1} \\
& i _1=\frac{R_2 i}{R_1+R_2} \quad \Rightarrow \quad i _2=\frac{R_1 i}{R_1+R_2},
\end{aligned}
\)

Example 30: Find the current passing through the battery and each resistor.

Solution: It is easy to see that the potential difference across each resistor is 30 V.
\(\therefore\) current is each resistors are \(\frac{30}{2}=15 A, \frac{30}{3}=10 A\) and \(\frac{30}{6}=5 A\)
\(\therefore\) Current through battery is \(=15+10+5=30 A\).

Example 31: Find the current which is passing through the battery.

Solution: Here potential difference across each resistor is not 30 V
\(\because\) battery has internal resistance. Here the concept of combination of resistors is useful.
\(
\begin{aligned}
& R_{\text {eq }}=1+1=2 \Omega \\
& i =\frac{30}{2}=15 A .
\end{aligned}
\)

Example 32: Find equivalent Resistance

Solution: Here all the Resistance are connected between the terminals \(A\) and \(B\)and they are in parallel.

\(R _{ eq }=\frac{R}{3}\)

Mixed grouping 

When a group of resistors is neither combined in parallel nor in series grouping, the combination is called as mixed grouping. The equivalent resistance corresponding to mixed grouping can be found by one of the following methods:

Method of successive reduction

This method is based on simplification of the circuit by successive reduction of series and parallel combinations. For eg., Let us take a circuit as shown in the figure.

Method of symmetry 

A line drawn across the circuit perpendicular to the line joining the two points between which resistance is to be found, which divides a circuit in two mirror images is a line of symmetry. This line is an equipotential line. By removing a resistor on equipotential line or by placing a conductor on this circuit is modified without altering its electrical properties. Now the modified circuit is solved by method of successive reduction. A point of symmetry is a false junction. Current does not get redistributed at this point. By separating the elements carrying the same current, the circuit is made simple & is then solved by method of successive reduction.

Example 33: Find the equivalent resistance between A and B.

Solution: Here \(V _{ A }= V _{ C }\) and \(V _{ B }= V _{ D }\)

\(
\frac{1}{ R _{ eq }}=\frac{3}{8 R }+\frac{1}{2 R }+\frac{3}{8 R } \quad \Rightarrow \quad R _{ eq }=\frac{4 R }{5}
\)

Example 34: Find equivalent resistance across points A and B.

Solution: Due to symmetry we can redraw the circuit as shown below. The voltage at points C, D & E are the same so the current across CD & DE is 0 A. We can remove those resistors from the circuit.

Example 35: Find the current in \(2 \Omega\) resistance

Solution: 

\(
\begin{aligned}
& 2 \Omega, 1 \Omega \text { in series }=3 \Omega \\
& 3 \Omega, 6 \Omega \text { in parallel }=\frac{18}{9}=2 \Omega \\
& 2 \Omega, 4 \Omega \text { in series }=6 \Omega \\
& 6 \Omega, 3 \Omega \text { in parallel }=2 \Omega \\
& R _{ eq }=4+4+2=10 \Omega \\
& i =\frac{120}{10}=12 A
\end{aligned}
\)
So current in \(2 \Omega\) Resistance \(=\frac{8}{3} A\)

Example 36: A wire of resistance \(6 R\) is bent in the form of \(a\) circle. What is the effective resistance between the ends of the diameter?

Solution: 

As shown in the figure, the two resistances of value \(3 R\) each are in parallel with each other. So, the resistance between the ends \(A\) and \(B\) of a diameter is
\(
R^{\prime}=\frac{R_1 \times R_2}{R_1+R_2}=\frac{3 R \times 3 R}{3 R+3 R}=\frac{9 R^2}{6 R}=\frac{3}{2} R
\)

Example 37: A letter ‘A’ consists of a uniform wire of resistance 0.2 ohm per cm. The sides of the letter are each 20 cm long and the cross-piece in the middle is 10 cm long while the apex angle is \(60^{\circ}\). Find the resistance of the letter between the two ends of the legs.

Solution:  Clearly it is given that,
\(
\text { and } \begin{aligned}
A B & =B C=C D=D E=B D=10 cm \\
R_1 & =R_2=R_3=R_4=R_5=2 \Omega
\end{aligned}
\)
As \(R_2\) and \(R_3\) are in series, their combined resistance
\(
=2+2=4 \Omega
\)
This combination is in parallel with \(R_5(=2 \Omega)\).
Hence, resistance between points of \(B\) and \(D\) is given by
\(
\begin{aligned}
\frac{1}{R} & =\frac{1}{4}+\frac{1}{2} \\
\frac{1}{R} & =\frac{6}{8} \\
R & =\frac{4}{3} \Omega
\end{aligned}
\)

Now, resistance \(R_1, R, R_4\) form a series combination. So, resistance between the ends \(A\) and \(E\) is
\(
R^{\prime}=4+\frac{4}{3}+4=9.34 \Omega
\)

Example 38: Find the equivalent resistance between A and B.

Solution: (i) The points connected by a conducting wire are at the same potential. Then, redraw the diagram, by placing the points of the same potential at one place and then solve for equivalent resistance.

Hence, from the new figure \(A\) and \(Y\) are at the same potential; \(B\) and \(X\) are at the same potential.
\(
\begin{aligned}
\Rightarrow \quad \frac{1}{R_{\text {eq }}} & =\frac{1}{6 R}+\frac{1}{9 R}+\frac{1}{12 R}=\frac{13}{36 R} \\
R_{\text {eq }} & =\frac{36 R}{13}=2.77 R
\end{aligned}
\)

(ii) Similarly, placing the points of the same potential at one place, then the equivalent resistance is

In parallel, \(R_{\text {eq }}=\frac{2 R}{4}=0.5 R\)

(iii) 

\(
\text { In parallel, } \frac{1}{R^{\prime}}=\frac{1}{6 R}+\frac{1}{9 R}+\frac{1}{12 R}=\frac{13}{36 R} \Rightarrow R^{\prime}=\frac{36}{13} R
\)
\(
\text { In series, } R_{\text {eq }}=R+\frac{36 R}{13}=\frac{49}{13} R=3.77 R
\)

Example 39: Calculate the current shown by the ammeter A in the circuit shown in the figure.

Solution: The given circuit can be redrawn as

From the above figure, the two \(8 \Omega\) resistances are connected in parallel, so equivalent resistance, \(R_{\text {eq }}=\frac{8 \times 8}{8+8}=4 \Omega\).
These two combinations are connected in series, so equivalent resistance \(=4+4=8 \Omega\).
Now, we have resistances of \(4 \Omega, 8 \Omega\) and \(4 \Omega\) connected in parallel, so
\(\Rightarrow \quad \frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{4}=\frac{5}{8}\) or \(R=\frac{8}{5} \Omega\)
Also, \(V=12 V\) (given)
\(\therefore \quad\) Current, \(I=\frac{V}{R}=\frac{12}{8 / 5}=\frac{15}{2}=7.5 A\)

Example 40: Find the equivalent resistance between P and Q.

Solution: (i) It can be seen that this diagram is symmetrical about \(P Q\), so points on the perpendicular bisector of \(P Q\), i.e. \(X, Y\) and \(Z\) are at the same potential. So, in this type of diagram, to calculate the equivalent resistance, we can remove the resistances at the same potential, i.e. the resistances between \(X\) and \(Y, Y\) and \(Z\), are redundant and can be removed.

All resistances in the circuit are in parallel,
\(
\begin{aligned}
\quad & \frac{1}{R_{\text {eq }}}=\frac{1}{8 R}+\frac{1}{4 r}+\frac{1}{8 R}=\frac{R+r}{4 R r} \\
\Rightarrow \quad & R_{\text {eq }}=\frac{4 R r}{R+r} \Omega
\end{aligned}
\)

(ii) Similarly as in (i), we see that there is symmetry about \(P Q\) and \(X, Y\) and \(Z\) are at same potential. So, remove resistances between \(X\) and \(Y ; Y\) and \(Z\).

In parallel, \(R_{\text {eq }}=\frac{4 R}{3} \Omega\)

Example 41: Find the equivalent resistance between A and B.

Solution: Here, we have infinite pairs of \(R\) and \(2 R\). Suppose, the equivalent resistance is \(R_0\) between \(C\) and \(D\), i.e. excluding one pair near \(A B\) (since, pairs are infinite, equivalent resistance will remain same, if we include pair near \(A B\) ).

\(
\begin{gathered}
R_{\text {eq }}=R_0=R+\frac{2 R R_0}{2 R+R_0} \\
\left(R_0-R\right)\left(2 R+R_0\right)=2 R R_0 \\
2 R R_0-2 R^2+R_0^2-R R_0=2 R R_0 \\
R_0^2-R R_0-2 R^2=0 \\
R_0=\frac{R \pm \sqrt{R^2+8 R^2}}{2}=\frac{R \pm 3 R}{2}=2 R \text { or }-R
\end{gathered}
\)
Equivalent resistance between \(A\) and \(B=2 R(\because\) equivalent resistance cannot be negative).

Example 42: Resistances \(R, 2 R, 4 R, 8 R \ldots \infty\) are connected in parallel. What is their resultant resistance?

Solution: 

\(
\frac{1}{ R _{ eq }}=\frac{1}{ R }+\frac{1}{2 R }+\frac{1}{4 R }+\frac{1}{8 R } \ldots \ldots \ldots \infty=\frac{1}{ R }\left[1+\frac{1}{2}+\frac{1}{4}+\ldots \infty\right]=\frac{1}{ R }\left[\frac{1}{1-\frac{1}{2}}\right]=\frac{2}{ R } \Rightarrow R _{ eq }=\frac{ R }{2}
\)

Example 43: A wire of \(\rho_{ L }=10^{-6} \Omega / m\) is turned in the form of a circle of diameter \(2 m\), A piece of the same material is connected as a diameter \(A B\). Then find the resistance between \(A\) and \(B\).

Solution: 

\(
\begin{aligned}
& R =\rho_{ L } \times \text { length } \\
& R _1=\pi \times 10^{-6} \Omega, R _2=2 \times 10^{-6} \Omega, R _3=\pi \times 10^{-6} \Omega
\end{aligned}
\)
\(
\frac{1}{ R _{ AB }}=\frac{1}{\pi \times 10^{-6}}+\frac{1}{2 \times 10^{-6}}+\frac{1}{\pi \times 10^{-6}} ; \quad R _{ AB }=0.88 \times 10^{-6} \text { ohm. }
\)

Example 44: Give the nature of the V-I graph for
(a) ohmic
(b) non-ohmic
circuit elements. Give one example of each type.

Solution: Linear graph for ohmic. Non-linear graph for non-ohmic circuit elements.
eg: Ohmic-electrical resistance, Non-ohmic-diode.

Example 45: Twelve identical resistances each of resistance \(R\) are connected as in the figure. Find the net resistance between \(X\) and \(Y\).

Solution: Given circuit can be modified according to the following figures :

\(
\frac{1}{ R _{ XY }}=\frac{1}{2 R }+\frac{3}{2 R }+\frac{1}{2 R }=\frac{5}{2 R } \Rightarrow R _{ XY }=\frac{2 R }{5}
\)

Example 46: 12 wires, each of resistance \(r\) ohms are connected to form a cube. What is the equivalent resistance of the cube when the current enters through one corner and leaves from diagonally opposite corner?

Solution:

Let the current \(6 I\) enter at point \(A\) and leave at \(G\) at the opposite end of the diagonal
On applying Kirchhoff’s Law to a path ABFG between AG, we get
\(
\begin{aligned}
& 2 I r + Ir +2 I r= V =6 I R \\
& 5 I r=6 IR \\
& R =\frac{5}{6} r
\end{aligned}
\)

Example 47: Find the equivalent resistance between A and B.

Solution:

In this case, the circuit has symmetry in the two branches AC and AD at the input
\(\therefore\) current in them are same but from input and from exit the circuit is not similar
( \(\because\) on left R and on right 2 R )
\(\therefore\) on both sides the distribution of current will not be similar.
Here \(V _{ c }= V _{ d }\)
hence \(C\) and \(D\) are same point
the circuit can be simplified that as shown
Now it is a balanced wheat stone bridge
\(
\begin{aligned}
R _{ eq } & =\frac{3 R \times \frac{3 R}{2}}{3 R+\frac{3 R}{2}} \\
& =\frac{\frac{9}{2} R}{\frac{9}{2}}= R .
\end{aligned}
\)