Resistors in Series and Parallel Configuration

Wherever there is a flow of charged particles, we say that it is an electric current. The flow of electricity is always stable and constant. You will find that it flows from higher potential to lower potential in a circuit. A regular circuit contains conductors (usually copper wires), resistors, switch to turn on and off the circuit, and a power source. A resistor is an electrical component that provides resistance or limits the flow of current in the circuit. For example, we can consider a tube light used in our household as a resistor.

In general use, we have a combination of resistors used in all circuits. We can either have resistors in series or resistors in parallel. In some cases, we can have both series and parallel combinations of resistors in a single circuit.

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The above figure shows a simple circuit consisting of a conductor, a resistor, and a battery.

Series Combination of Resistors

When we have resistors in series, the current flows through them one after the other. Each resistor will have the same current flowing through them. Across a series of resistors, there will always be a voltage drop.

To calculate the equivalent resistance, we need to derive the series resistance formula. To obtain the equation, we use Ohm's law. According to the law, the potential drop 'V' is given as V=IR, where 'I' is the current, and 'R' is the resistance of the circuit. 

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The image shows the circuit of resistors placed in series.

According to Kirchhoff’s loop law we have,

\[\sum_{i=1}^{N}\]   \[V_{i}\] = 0

V − V1 − V2 − V3 = 0,

V = V1 + V2 + V3

V = IR1 + IR2 + IR3

I = VR1 + R2 + R3 = VRS

Therefore, we get the series resistance formula as,

RS = R1 + R2 + R3 + . . . + RN−1 + RN

\[R_{s}\] = \[\sum_{i=1}^{N}\]  \[R_{i}\]

Parallel Combination of Resistors

When we have resistors in parallel, the electric current divides itself to travel through the different branches. The voltage drops across each resistor will be equal, unlike the resistors in series. Since there are a lot of resistors connected in parallel, we must find the total resistance. To do that we need to derive the resistors in the parallel formula. To acquire the equation, we use Ohm's law. According to the law the equation for the electric current 'I' is given as I = V/R, where 'V' is the potential drop, and 'R' is the resistance of the circuit. 

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The image shows the circuit of resistors placed in series.

According to Kirchhoff’s junction rule we have,

\[\sum\] \[I_{in}\]  = \[\sum\] \[I_{out}\]

I = I1 +I2

I =  \[\frac{V_{1}}{R_{1}}\] +  \[\frac{V_{2}}{R_{2}}\] = \[\frac{V}{R_{1}}\]  +  \[\frac{V}{R_{2}}\]

I = V  \[\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right )\]

\[R_{p}\] =  \[\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right )^{-1}\]

Therefore, we get the resistors in the parallel formula as,

\[R_{p}\] = \[\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} +  \cdots + \frac{1}{R_{N-1} + \frac{1}{R_{N}}} \right )^{-1}\]

 \[R_{p}\] = \[\left ( \sum_{i=1}^{N} \frac{1}{R_{i}} \right )^{-1}\]

Combination of Resistors In Series And Parallel

In practice, you will never find simple electrical circuits, where the resistors are only placed in series or parallel. Instead, you will find complex connections with multiple resistors connected in series and parallel at the same time. Now, just because the circuit looks complicated, it doesn't mean that it is difficult to calculate the resistance of the circuit. All you need to do is break the connections into small parts so that you can calculate the equivalent resistance easily. Your main goal is to keep decreasing the number of resistors by using the formula of resistance in series and parallel. We will now try to solve some questions on resistors in series and parallel.

Solved Problems

Question 1) Consider a circuit with a voltage of 9V, and consisting of five resistors. Calculate the equivalent resistance, and the current ‘I’ through the resistors.

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Answer 1) Looking at the figure, we can see that the resistors are in series. Therefore, we will use the series resistance formula to calculate the equivalent resistance.

Given: V = 9V

R1 = R2 = R3 = R4 = 20Ω

R5 = 10Ω

The equivalent resistance is given as,

RS = R1 + R2 + R3 + R4 + R5 = 20Ω + 20Ω + 20Ω + 20Ω + 10Ω = 90Ω

The total resistance with the correct number of significant digits is Req = 90Ω.

Using Ohm’s law, we can calculate the current in the circuit.

I = V/RS = 9V/90Ω = 0.1A

Question 2) Three resistors R1 = 1.00Ω, R2 = 2.00Ω, and R3 = 2.00Ω, are connected in parallel. The battery has a voltage of 3V. Calculate the equivalent resistance, and current ‘I’ through the circuit.

Answer 2) Since the resistors are connected in parallel, we will use the resistors in parallel formula to calculate the equivalent resistance.

Given: V = 3V

R1 = 1.00Ω

R2 = 2.00Ω

R3 = 2.00Ω

The equivalent resistance is given as,

 \[R_{p}\] =  \[\left ( \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}  \right )^{-1}\] 

 \[R_{p}\] =  \[\left ( \frac{1}{1} + \frac{1}{2} + \frac{1}{2}  \right )^{-1}\] 

\[R_{p}\]  =  0.50Ω

Therefore, we get the equivalent resistance as \[R_{eq}\] = 0.50Ω.

Using Ohm’s law, we can calculate the current in the circuit.

I = V/Rp = 3V/0.5Ω = 6A