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Electricity always flows from higher potential to lower potential in a circuit. A regular circuit contains conductors, resistors, 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.

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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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

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

Using this formula, we will try to solve some questions on resistors in series and parallel.

We say that the resistors are in parallel when 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. We will now try to derive the resistors in 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 circuit's resistance.

(image will be uploaded soon)

The image shows the circuit of resistors placed in parallel.

According to Kirchhoff’s junction rule we have,

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

I = I1 +I2

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

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

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

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.

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 = V/Rp = 3V/0.333Ω = 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.

FAQ (Frequently Asked Questions)

Question 1) How to find Total Resistance in a Parallel Circuit?

Answer 1) The formula to calculate the total resistance in a parallel circuit is as follows:

Rₚ = (1/R₁ + 1/R₂ + 1/R₃)⁻¹.

The potential drop will remain equal for each component of the parallel circuit. Whereas the current will be divided proportionally across the resistors.

If the resistances are equal, this simplifies to,

Rₚ = Rᵢ/n, Where Rₚ is the equivalent resistance, Rₚ is the resistance of each resistor, and n is the number of resistors.

Question 2) What is the resistance in a series circuit?

Answer 2) The formula to calculate the total resistance in a series circuit is as follows:

R₅ = R₁ + R₂ + R₃ + . . . + R_{N-1} + R_{N}

If the resistances are equal, this simplifies to

R₅ = n*R₁, Where R₅ is the equivalent resistance, R₁ is the resistance of each resistor, and n is the number of resistors.