How To Find Parallel Resistance: Detailed Insights


There are numerous techniques on How To Find parallel resistance that we shall elucidate in this article. Unlike series resistors, parallel joined resistors have different equivalent resistance computation methods.

Suppose, we have two resistors R1 and R2 as shown in image 1. We know that the total current in a parallel circuit = sum of branch currents. 

Therefore, [Latex] i = \frac{V} {R_{1}}+ \frac{V} {R_{2}} [/Latex] ( potential of A and B are same)

Or, [Latex] i = V\left ( \frac{1} {R_{1}}+ \frac{1} {R_{2}} \right ) [/Latex]

Now, total current i = Voltage / equivalent resistance = V/Req

So, [Latex] \frac{V} {R_{eq}} = V\left ( \frac{1} {R_{1}}+ \frac{1} {R_{2}} \right ) [/Latex] and [Latex] R_{eq} = \left ( \frac{1} {R_{1}}+ \frac{1} {R_{2}} \right )^ {-1} [/Latex]

How To Find Parallel Resistance- circuit

How To Find Parallel Resistance – FAQs

How To Find Parallel Resistance for n resistors?

The method of calculating equivalent resistance for more than two resistors is similar. Image 2 depicts a circuit consisting of n resistors placed parallelly. Let us find the equivalent resistance in this case.

We know from ohm’s law, 

  1. Each branch has same voltage = V
  2. Net current [Latex]I = i_{1} + i_{2} + i_{3} + ……..+ i_{n}[/Latex] 

Net current = V/ R where R is the equivalent resistance

Hence, [Latex]\frac{V} {R} = \frac{V} {R_{1}} + \frac{V} {R_{2}} + \frac{V} {R_{3}} +………\frac{V} {R_{n}} [/Latex]

Or [Latex]R = \left (\frac{1} {R_{1}} + \frac{1} {R_{2}} + \frac{1} {R_{3}} +………\frac{1} {R_{n}} \right ) ^{-1} [/Latex]

We can replace the values according to circuit requirement and get the desired equivalent resistance.

What are the features of parallel resistance?

Parallel resistances have several properties in a circuit. The most important feature of parallel resistance is – The reciprocal equivalent resistance is the sum of all individual reciprocal resistances.

The other features of parallel resistance are-

  1. All the resistors share the same voltage and it is equal to the node voltage
  2. The currents through the resistors sums up the net current outside the entire parallel connection.
  3. The equivalent resistance value is less than any resistor present in the circuit.

Read more on….Is Current The Same In Parallel: Complete Insights and FAQs

How does parallel resistance affect voltage and current?

We are aware of the fact that equivalent resistance in a parallel circuitry is obtained by summing up the inverse of all resistances and again reciprocating them. This resistance determines the current in the circuit.

Suppose, we construct an electrical circuit with parallel connection of resistors RA and RB with a voltage source of V. The source voltage will be shared by both resistors and the voltage drop across both of them will be V. Current in the path of RA will be V/ RA and Current in the path of RA will be V/ RB

Read more on….Is Voltage The Same In Parallel: Complete Insights and FAQs

Why is the equivalent resistance in parallel is less than the individual resistances?

In parallel, the charge flowing from the source when it arrives at the node has the option of moving to any branch. So a large number of charges flow from the source. Therefore, the current increases.

From ohm’s law we know, V = IR

The voltage will be the same for all branches in parallel. As such, the current grows with the growth of the branches (i.e. connecting more resistance) .The only way in which the voltage can remain unchanged when the resistance decreases. Therefore, that is why resistance is reduced.

Also read on…What Is Voltage Drop In Parallel Circuit:How to Find, Example Problems and Detailed Facts

Numerical problems

Calculate the equivalent parallel resistance for this infinite ladder shown in image 3

For this infinite resistance ladder, we can say that the equivalent resistance Req between P and Q points is equal to that of the remaining circuit. Therefore [Latex]R_{eq} = 2+ 1|| R_{eq}[/Latex] 

So, [Latex] R_{eq} = 2+ \frac {1\times R_{eq}}{1 + R_{eq}} = \frac{ 2 + 3R_{eq} }{1 + R_{eq} } [/Latex]

Or, [Latex] R_{eq} + R_{eq}^{2} = 2 + 3R_{eq} [/Latex]

Or, [Latex] R_{eq}^{2} – 2R_{eq} -2 = 0 [/Latex]

By solving the above equation, we get, [Latex] R_{eq} = 1 \pm \sqrt{3}\; ohm [/Latex]

By neglecting the negative quantity, we can say [Latex] R_{eq} = 1 + \sqrt{3}\; ohm [/Latex]

This is the required equivalent resistance.

If the equivalent resistance for the circuit in image 4 in 15 ohm, find the missing value R.

In the first step, we shall calculate the equivalent resistance of the rightmost mesh. So, [Latex] R_{eq} = \frac{ 15R }{ 15 + R} [/Latex]. So, the circuit is now reduced to image 4.1. Now we shall calculate the next mesh of three series resistors.

Now, [Latex] R_{eq} = 4 + 4 + \frac{ 15R }{ 15 + R} = 8 + \frac{ 15R } { 15 + R} = \frac{ 120 + 23 R} { 15 + R} [/Latex]. Next we again have a parallel mesh. So the Req now is [Latex] 14\; ||\; \frac{ 120 + 23 R}{ 15 + R} = \frac{ 14 \times \frac{ 120 + 23 R}{ 15 + R} }{ 14 + \frac{ 120 + 23 R}{ 15 + R}} = \frac{ 14 \left (120 + 23 R \right )}{ 330 + 37R} [/Latex]. The final mesh is another series connection which gives Req as [Latex] 5 + 3 + \frac{ 14 \left (120 + 23 R \right )}{ 330 + 37R} = 15 [/Latex]ohm. Solving this, we get R= 10 ohm.

What will be the equivalent resistance Req for the circuit depicted in image 5.

We can redraw the above circuit as image 5. So for the mesh on the extreme right, Req = 4+6 = 10 ohm. Now, we have 3 resistors in parallel for the right mesh and 2 resistors in parallel for the top mesh shown in 5.1.

Equivalent resistance for the right mesh [Latex] = \frac{10 \times 15\times 30}{ 10\times 15 +15\times 30 + 10\times 30} = 6\; ohm [/Latex]. 

Equivalent resistance for the top mesh [Latex] = \frac{ 20\times 5}{ 20 + 5} = 4\; ohm [/Latex]. Now we have reduced the system into a simple series circuit with three resistors 1 ohm, 4 ohm and 6 ohm as shown in 5.2. So the final Req is [Latex] 1 + 4 + 6 = 11 \; ohm [/Latex].

Find the equivalent resistance in the circuit given below: VS = 12 V, R1 = 2.5 Ω, R2 = 2 Ω, R3 = 1.5 Ω, R4 = 3 Ω, R5 = 5 Ω, and R6 = 3.25 Ω.

The simplified circuit for image 6 is shown in 6.1. We shall solve for the equivalent resistance from the innermost mesh. So, Req for the mesh with R4 and R5 is [Latex] \frac{ R_{4} \times R_{5} }{ R_{4} + R_{5} } = \frac{ 5 \times 3 }{ 5 + 3 } = 1.875\; ohm [/Latex].

Now we have R3 and 1.875 ohm in series. So, [Latex] R_{eq} = 1.5+ 1.875 = 3.375\; ohm [/Latex]. This resistance is in parallel with R2. So now [Latex] R_{eq} = \frac{ 2\times 3.375}{ 2 + 3.375} = 1.25\; ohm [/Latex]. Finally we have this resistance in series with R1 and R6. Therefore, [Latex] R_{eq} = \left ( 2.5 + 3.25 + 1.25 \right ) = 7\; ohm [/Latex]. This is the equivalent resistance of the circuit.

Kaushikee Banerjee

I am an electronics enthusiast and currently devoted towards the field of Electronics and Communications . My interest lies in exploring the cutting edge technologies. I'm an enthusiastic learner and I tinker around with open-source electronics. LinkedIn ID- https://www.linkedin.com/in/kaushikee-banerjee-538321175

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