Resistors in Series and Parallel Combinations With Solved Examples

Physics Topics are often described using mathematical equations, making them precise and quantifiable.

Analysis Techniques for Series and Parallel Resistor Circuits

Apart from potential difference, current in a circuit depends on resistance of the circuit. So, in the electrical circuits of radio, television and other similar things, it is usually necessary to combine two or more resistances to get the required current in the circuit. We can combine the resistances lengthwise (called series) or we can put the resistances parallel to one another.

Thus, the resistances can be combined in two ways : (i) in series, and (ii) in parallel. If we want to increase the total resistance, then the individual resistances are connected in series, and if we want to decrease the resistance, then the individual resistances are connected in parallel. We will study these two cases in detail, one by one.

When two (or more) resistances are connected end to end resistances (or resistors) . These can be consecutively, they are said to be connected in series. Above Figure shows connected in series or parallel combinations, shows two resistances R₁ and R₂ which are connected in series.

On the other hand, when two (or more) resistances are connected between the same two points, they are said to be connected in parallel (because they become parallel to one another). In above Figure, the two resistances R₁ and R₂ are connected in parallel arrangement between the same two points A and B. In the above examples, we have shown only two resistances (or resistors) connected in series and parallel combinations. We can, however, connect any number of resistors in these two arrangements.

Resistances (Or Resistors) In Series

The combined resistance (or resultant resistance) of a number of resistances or resistors connected in series is calculated by using the law of combination of resistances in series. According to the law of combination of resistances in series : The combined resistance of any number of resistances connected in series is equal to the sum of the individual resistances. For example, if a number of resistances R₁, R₂, R₃ ……………………etc., are connected in series, then their combined resistance R is given by : R = R₁ + R₂ + R₃ + ……………..

Suppose that a resistance R₁ of 2 ohms and another resistance R₂ of 4 ohms are connected in series and we want to find out their combined resistance R.
We know that : R = R₁ + R₂
So, R = 2 + 4
And, Combined resistance, R = 6 ohms

Thus, if we join two resistances of 2 ohms and 4 ohms in series, then their combined resistance (or resultant resistance) will be 6 ohms which is equal to the sum of the individual resistances. Before we derive the formula for the resultant resistance of a number of resistances connected in series, we should keep in mind that :

(i) When a number of resistances connected in series are joined to the terminals of a battery, then each resistance has a different potential difference across its ends (which depends on the value of resistance). But the total potential difference across the ends of all the resistances in series is equal to the voltage of the battery. Thus, when a number of resistances are connected in series, then the sum of the potential differences across all the resistances is equal to the voltage of the battery applied.

(ii) When a number of resistances are connected in series, then the same current flows through each resistance (which is equal to the current flowing in the whole circuit).

1. Resultant Resistance of Two Resistances Connected in Series

We will now derive a formula for calculating the combined resistance (equivalent resistance or resultant resistance) of two resistances connected in series.

Above Figure shows two resistances R₁ and R₂ connected in series. A battery of V volts has been applied to the ends of this series combination. Now, suppose the potential difference across the resistance R₁ is V₁ and the potential difference across the resistance R₂ is V₂. We have applied a battery of voltage V, so the total potential difference across the two resistances should be equal to the voltage of the battery.
That is : V = V₁ + V₂ ……………. (1)
We have just seen that the total potential difference due to battery is V. Now, suppose the total resistance of the combination be R, and the current flowing through the whole circuit be I. So, applying Ohm’s law to the whole circuit, we get :
\(\frac{V}{I}\) = R
or V = I × R …………….. (2)
Since the same current / flows through both the resistances R₁ and R₂ connected in series, so by applying Ohm’s law to both the resistances separately, we will get :
V₁ = I × R₁ ………… (3)
and V₂ = I × R₂ ………….. (4)
Now, putting the values of V, V₁ and V₂ from equations (2), (3) and (4) in equation (1), we get:
I × R = I × R₁ + I × R₂
or I × R = I × (R₁ + R₂)
Cancelling I from both sides, we get:
Resultant resistance (combined resistance or equivalent resistance),
R = R₁ + R₁

2. Resultant Resistance of Three Resistances Connected in Series

Figure shows three resistances R₁, R₂ and R₁ connected in series. A battery of V volts has been applied to the ends of this series combination of resistances. Now, suppose the potential difference across the resistance R₁ is V₁, the potential difference across the resistance R₂ is V₂ and that across resistance R₃ is V₃. We have applied a battery of voltage V, so the total potential difference across the three resistances should be equal to the voltage of the battery applied. That is,

V = V₁ + V₁ + V₁ …………. (1)
We have just seen that the total potential difference due to battery is V. Now, let the total resistance (or resultant resistance) of the combination be R. The current flowing through the whole circuit is I. So, applying Ohm’s law to the whole circuit, we get:
\(\frac{V}{I}\) = R
or V = I × R …………….. (2)
Since the same current I flows through all the resistances Ry R2 and R3 in series, so by applying Ohm’s law to each resistance separately, we will get:
V₁ = I × R₁ …………….. (3)
V₂ = I × R₂ ………….. (4)
and V₃ = I × R₃ …………… (5)
Putting these values of V, V₁, V₂ and V₃ in equation (1), we get:
I × R = I × R₁ + I × R₂ + I × R₃
or I × R = I × (R₁ + R₂ + R₃)
Cancelling I from both sides, we get:
R = R₁ + R₂ + R₃
Thus, if three resistors R₁, R₂, and R₃ are connected in series then their total resistance R is given by the formula :
R = R₁ + R₂ + R₃
Similarly, if there are four resistors R₁, R₂, R₃ and R₄ connected in series, then their resultant resistance R is given by the formula : R = R₁ + R₂ + R₃ + R₄ and so on.
We will now solve some problems based on the combination of resistances in series.

Example Problem 1.
If four resistances, each of value 1 ohm, are connected in series, what will be the resultant resistance ?
Solution:
Here we have four resistances, each of 1 ohm, connected in series. These are shown in the Figure below.

Now, if we have four resistances R₁, R₂, R₃ and R₄ connected in series, then their resultant resistance R is given by : R = R₁ + R₂ + R₃ + R₄
Here R₁ = 1 Ω, R₂ = 1 Ω, R₃ = 1 Ω, R₄ = 1 Ω
So, Resultant resistance, R = 1 + 1 + 1 + 1
or R = 4Ω
Thus, the resultant resistance is equal to 4 ohms.
We will now solve some problems by applying Ohm’s law to the circuits having resistances in series.

Example Problem 2.
A resistance of 6 ohms is connected in series with another resistance of 4 ohms. A potential difference of 20 volts is applied across the combination. Calculate the current through the circuit and potential difference across the 6 ohm resistance.
Solution:
The first step in solving such problems based on current electricity is to draw a proper circuit diagram. For example, in this problem we have two resistances of 6 ohms and 4 ohms which are connected in series.

So, first of all we have to draw these two resistances on paper as shown in Figure alongside. Now, a potential difference of 20 volts has been applied across this combination of resistances. So, we draw a cell or a battery of 20 volts and complete the circuit as shown in Figure alongside. Suppose the current flowing in the circuit is I amperes.

We will now find out the value of current I flowing through the circuit. To do this we should know the total resistance R of the circuit. Here we have two resistances of 6 Ω and 4 Ω connected in series. So,
Total resistance, R = R₁ + R₂
R = 6 + 4
R = 10 ohms
Now, Total resistance, R = 10 ohms
Potential difference, V = 20 volts
and, Current in the circuit, I = ? (To be calculated)
So, applying Ohm’s law to the whole circuit, we get:
\(\frac{V}{I}\) = R
So that, \(\frac{20}{I}\) = 10
And, 10I = 20
I = \(\frac{20}{10}\)
So, Current, I = 2 amperes (or 2 A)
Thus, the current flowing through the circuit is 2 amperes.

The second part of this problem is to find out the potential difference across the ends of the 6 ohm resistance. To do this we will have to apply Ohm’s law to this resistance only. We know that the current flowing through the 6 ohm resistance will also be 2 amperes. Now,

Potential difference (across 6 Ω resistance), V = ? (To be calculated)
Current (through 6 Ω resistance), I = 2 amperes
And, Resistance, R = 6 ohms
So, applying Ohm’s law to the 6 Ω resistance only, we get:
\(\frac{V}{I}\) = R
or \(\frac{V}{2}\) = 6
So, Potential difference, V = 12 volts
Thus, the potential difference across the 6 ohm resistance is 12 volts.

Here is an exercise for you. Find out the potential difference across the 4 ohm resistance yourself. The answer will be 8 volts. Remember that the same current of 2 amperes flows through the 4 ohm resistance.

Example Problem 3.
(a) Draw the diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor and a 12 Ω resistor, and a plug key, all connected in series.
(b) Redraw the above circuit putting an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Solution:
(a) In this problem, we have a battery of 3 cells of 2 V each, so the total potential difference (or voltage) of the battery will be 3 × 2 = 6 V. The circuit consisting of a battery of three cells of 2 V each (having a voltage of 6 V), resistors of 5 Ω, 8 Ω, 12 Ω and a plug key, all connected in series is given in Figure alongside.

(b) The above circuit can be redrawn by including an ammeter in the main circuit and a voltmeter across the 12 Ω resistor, as shown in Figure alongside. Please note that the ammeter has been put in series with the circuit but the voltmeter has been put in parallel with the 12 Ω resistor. We will now calculate the current reading in the ammeter and potential difference reading in the voltmeter.

(i) Calculation of current flowing in the circuit. The three resistors of 5 Ω, 8 Ω and 12 Ω are connected in series. So,
Total resistance, R = 5 + 8 + 12 = 25 Ω
Potential difference, V = 6 V And,
And, Current, I = ? (To be calculated)
Now, \(\frac{V}{I}\) = R
So, \(\frac{6}{I}\) = 25
I = \(\frac{6}{25}\)
I = 0.24 A
Now, since the current in the circuit is 0.24 amperes, therefore, the ammeter will show a reading of 0.24 A.

(ii) Calculation of potential difference across 12 Ω resistor. We have just calculated that a current of 0.24 A flows in the circuit. The same current of 0.24 A also flows through the 12 Ω resistor which is connected in series. Now, for the 12 Ω resistor :
Current, I = 0.24 A (Calculated above)
Resistance, R = 12 Ω (Given)
And, Potential difference, V = ? (To be calculated)
We know that, \(\frac{V}{I}\) = R
So, \(\frac{V}{0.24}\) = 12
Ans V = 0.24 × 12
V = 2.88 V
Thus, the potential difference across the 12 Q resistor is 2.88 volts. So, the voltmeter will show a reading of 2.88 V.

Resistances (Or Resistors) In Parallel

The combined resistance (or resultant resistance) of a number of resistances or resistors connected in parallel can be calculated by using the law of combination of resistances in parallel. According to the law of combination of resistances in parallel: The reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

For example, if a number of resistances, R₁, R₂, R₃ ……………… etc., are connected in parallel, then their combined resistance R is given by the formula :
\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots \ldots\)
Suppose that a resistance R₁ of 6 ohms and another resistance R₂ of 12 ohms are connected in parallel and we want to find out their combined resistance R.
We know that : \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)
= \(\frac{1}{6}\) + \(\frac{1}{12}\)
= \(\frac{2+1}{12}\)
= \(\frac{3}{12}\)
Now, \(\frac{1}{R}\) = \(\frac{1}{4}\)
So, Combined resistance, R = 4 ohms

This means that if we join two resistances of 6 ohms and 12 ohms in parallel then their combined resistance is only 4 ohms which is less than either of the two individual resistances (of 6 ohms and 12 ohms). Thus, when a number of resistances are connected in parallel then their combined resistance is less than the smallest individual resistance.

This is due to the fact that when we have two or more resistances joined parallel to one another, then the same current gets additional paths to flow and the overall resistance decreases. Before we derive a formula for the resultant resistance of a number of resistances connected in parallel, we should keep in mind that :

(i) When a number of resistances are connected in parallel, then the potential difference across each resistance is the same which is equal to the voltage of the battery applied.

(ii) When a number of resistances connected in parallel are joined to the two terminals of a battery, then different amounts of current flow through each resistance (which depends on the value of resistance). But the current flowing through all the individual parallel resistances, taken together, is equal to the current flowing in the circuit as a whole. Thus, when a number of resistances are connected in parallel, then the sum of the currents flowing through all the resistances is equal to the total current flowing in the circuit.

1. Combined Resistance of Two Resistances Connected in Parallel

We will now derive a formula for calculating the combined resistance (resultant resistance or equivalent resistance) of two resistors connected in parallel. In Figure, two resistances R₁ and R₂ are connected parallel to one another between the same two points A and B. A battery of V volts has been applied across the ends of this combination. In this case the potential difference across the ends of both the resistances will be the same. And it will be equal to the voltage of the battery used. The current flowing through the two resistances in parallel is, however, not the same.

Suppose the total current flowing in the circuit is I, then the current passing through resistance R₁ will be I₁ and the current passing through the resistance R₂ will be I₂ (see Figure). It is obvious that:
Total current, I = I₁ + I₂ …………………. (1)
Suppose the resultant resistance of this parallel combination is R. Then by applying Ohm’s law to the whole circuit, we get :
I = \(\frac{V}{R}\) …………….. (2)
Since the potential difference V across both the resistances R₁ and R₂ in parallel is the same, so by applying Ohm’s law to each resistance separately, we get :
I₁ = \(\frac{V}{R_1}\) …………….. (3)
and I₂ = \(\frac{V}{R_2}\) ………………. (4)
Now, putting the values of I, I₁ and I₂ from equations (2), (3) and (4) in equation (1), we get :
\(\frac{V}{R}\) =\(\frac{V}{R_1}+\frac{V}{R_2}\)
or \(V\left[\frac{1}{R}\right]\)=\(V\left[\frac{1}{R_1}+\frac{1}{R_2}\right]\)
Cancelling V from both sides, we get:
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}\)
Thus, if two resistances R( and R2 are connected in parallel, then their resultant resistance R is given by the formula :
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}\)

2. Combined Resistance of Three Resistances Connected in Parallel

In below Figure, three resistances R₁, R₂ and R₃ are connected parallel to one another between the same two points A and B. A battery of V volts has been applied across the ends of this combination. In this case the potential difference across the ends of all the three resistances will be the same. And it will be equal to the voltage of the battery used.

The current flowing through the three resistances connected in parallel is, however, not the same. Suppose the total current flowing through the circuit is I, then the current passing through resistance R₁ will be I₁, the current passing through resistance R₂ will be I₂, and that through R₃ will be I₃ (see Figure). It is obvious that:

Total current, I = I₁ + I₂ + I₃ ……….. (1)
Suppose the resultant resistance of this combination is R. Then, by applying Ohm’s law to the whole circuit, we get:
I = \(\frac{V}{R}\) ………………… (2)
Since the potential difference V across all the three resistances R₁, R₂ and R₃ in parallel is the same, so by applying Ohm’s law to each resistance separately, we get:
I₁ = \(\frac{V}{R_1}\) ……………. (3)
I₂ = \(\frac{V}{R_2}\) ………….. (4)
and I₃ = \(\frac{V}{R_3}\) ……………….. (5)
Putting these values of I, I₁, I₂ and I₃ in equation (1), we get :
\(\frac{V}{R}\) = \(\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}\)
or V \(\frac{1}{R}\) = V \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
Cancelling V from both sides, we get:
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
Thus, if three resistances R₁, R₂ and R₃ are connected in parallel, then their resultant resistance R is given by the formula :
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
Similarly, when four resistances R₁, R₂, R₃ and R₄ are connected in parallel, then their resultant resistance R is given by the formula :
\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\) and so on
Let us solve some problems now.

Example Problem 1.
Calculate the equivalent resistance when two resistances of 3 ohms and 6 ohms are connected in parallel.
Solution:
Here we have two resistances of 3 ohms and 6 ohms which are connected in parallel. This arrangement is shown in Figure given below. Now, we want to find out their equivalent resistance or resultant resistance. We know that when two resistances R₁ and R₂ are connected in parallel, then their equivalent resistance R is given by :

\(\frac{1}{R}\) = \(\frac{1}{R_1}+\frac{1}{R_2}\)
Here, R₁ = 3 ohms
and, R₂ = 6 ohms
So, \(\frac{1}{R}\) = \(\frac{1}{3}+\frac{1}{6}\)
or \(\frac{1}{R}\) = \(\frac{2+1}{6}\)
or \(\frac{1}{R}\) = \(\frac{3}{6}\)
or \(\frac{1}{R}\) = \(\frac{1}{2}\)
and R = 2 Ω
Thus, the equivalent resistance is 2 ohms.

So far we have studied the combination of resistances in series and parallel separately. Many times, however, the practical electrical circuits involve the combination of resistances in series as well as in parallel in the same circuit. We will now solve a problem in which the resistances are connected in series as well as in parallel in the same circuit.

Example Problem 2.
In the circuit diagram given alongside, find :

(i) total resistance of the circuit,
(ii) total current flowing in the circuit, and
(iii) the potential difference across R₁
Solution:
In this problem the resistances are connected in series as well as in parallel combination. For example, the two resistances R₂ and R₃ are in parallel combination to each other but, taken together, they are in series combination with the resistance R₁.

(i) Calculation of Total Resistance. We will now find out the total resistance of the circuit. For doing this, let us first calculate the resultant resistance R of R₂ and R₃ which are connected in parallel.
Now, \(\frac{1}{R}\) = \( \frac{1}{R_2}+\frac{1}{R_3}\)
Here, R₂ = 8 Ω
and, R₃ = 12 Ω
So, \(\frac{1}{R}\) = \( \frac{1}{8}+\frac{1}{12}\)
or \(\frac{1}{R}\) = \( \frac{3+2}{24}\)
\(\frac{1}{R}\) = \( \frac{5}{24}\)
R = \(\frac{24}{5}\)
and R = 4.8 ohms

Thus, the two resistances of 8 ohms and 12 ohms connected in parallel are equal to a single resistance of 4.8 ohms. It is obvious that in the above given diagram, we can replace the two resistances R₂ and R₃ by a single resistance of 4.8 ohms. We can now draw another circuit diagram for this problem by showing a single resistance of 4.8 ohms in place of two parallel resistances. Such a circuit diagram is given alongside.

It is clear from this diagram that now we have two resistances of 7.2 ohms and 4.8 ohms which are connected in series. So,
= 12 ohms
Thus, the total resistance of the circuit is 12 ohms.

(ii) Calculation of Total Current. The battery shown in the given circuit is of 6 volts. So, Total potential difference, V = 6 volts
Total current, I = ? (To be calculated)
and Total resistance, R = 12 ohms (Calculated above)
So, applying Ohm’s law to the whole circuit, we get:
\(\frac{V}{I}\) = R
or \(\frac{6}{I}\) = 12
or 12 I = 6
I = \(\frac{6}{12}\)
I = \(\frac{1}{2}\)
So, Total current, I = 0.5 ampere (or 0.5 A)

Thus, the total current flowing in the circuit is 0.5 ampere. It should be noted that the same current flows through all the parts of a series circuit. So, the current flowing through the resistance Rx is also 0.5 ampere.

(iii) Calculation of Potential Difference Across R₁. We have now to find out the potential difference across the resistance R₁ of 7.2 ohms.
Now,Potential difference across R₁ = ? (To be calculated)
Current through R₁ = 0.5 ampere
And, Resistance of R₁ = 7.2 ohms
Applying Ohm’s law to the resistance R₁ only, we get :
\(\frac{V}{I}\) = R
or \(\frac{V}{0.5}\) = 0.72
or V = 7.2 × 0.5
or V = 3.6 volts
Thus, the potential difference across the ends of the resistance R₁ is 3.6 volts.