Resistors in Series and Parallel

Electricity always flows from higher potential to lower potential in a circuit. A regular circuit contains conductors, resistors, a switch to turn on and off the circuit, and a power source. All of these different components can be connected in multiple ways to produce a complicated network. Therefore, solving resistors in series and parallel is important. 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. Therefore, solving resistors in series and parallel is essential.

Normally, we have a combination of resistors used in all circuits. We can either have resistors in series or resistors in parallel. In this article, we will look at resistors in series and parallel problems and solutions.  

Series Combination of Resistors

We say that resistors are connected in series when the resistors are connected one after the other. The current flows through them one after the other, and Voltage will keep dropping from one resistor to another.

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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According to Kirchhoff’s loop law we have,

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

I = I1 +I2

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

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

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

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

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

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

Using this formula, we will try to solve questions on resistors in series and parallel. We will now look at some resistors in series and parallel problems and solutions. You will also be able to find some series-parallel resistance practice problems on our page.

Solved Problems

Question 1) Consider a circuit with a voltage of 9V, and consisting of five resistors with a resistance of 30Ω each. Calculate the equivalent resistance, and the current ‘I’ through the resistors.

Answer 1) Looking at the figure, we can see that the resistors are in series. 

Given: V = 9V

R1 = R2 = R3 = R4 = R5 = 30Ω

The equivalent resistance is given as,

RS = R1 + R2 + R3 + R4 + R5 = 30Ω + 30Ω + 30Ω + 30Ω + 30Ω = 150Ω

The total resistance with the correct number of significant digits is Req = 150Ω.

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

I = V/RS = 9V/150Ω = 0.06A

Therefore, we were able to find the equivalent resistance to be 150Ω and the current as 0.06A.

Question 2) Three resistors R1 = 1.00Ω, R2 = 1.00Ω, and R3 = 1.00Ω, 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 the parallel formula to calculate the equivalent resistance.

Given: V = 3V, R1 = 1.00Ω, R2 = 1.00Ω, R3 = 1.00Ω

The equivalent resistance is given as,

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

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

\[R_{p} = 0.333 \Omega\]

Therefore, we get the equivalent resistance as Req = 0.333Ω.

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

\[I = \frac{V}{R_{p}} = \frac{3V}{0.333\Omega } = 9A\]

Therefore, we were able to find the equivalent resistance to be 0.333Ω and the current as 9A.

Now that you have gone through resistors in series and parallel problems and solutions. You should easily be able to solve any questions on resistors in series and parallel.