Lesson Explainer: Parallel Circuits Physics • 9th Grade

In this explainer, we will learn how to calculate the potential difference, current, and resistance at different points within simple parallel circuits.

The diagram below shows a circuit consisting of a cell and a resistor. The potential difference provided by the cell is 𝑉, the current in the circuit is 𝐼, and the resistance of the resistor is 𝑅.

In the circuit above, there is only one path that electrons can flow along to move from one terminal of the cell to the other. In the circuit below, however, there are two paths that the electrons could flow along to move from one terminal of the cell to the other.

The path between the two terminals of the cell splits into two branches at point A. The paths join together again at point B. Each path has a resistor on it.

When there are multiple different paths that the electrons could take between the terminals of the cell like this, we say that that part of the circuit is in parallel. If a circuit has components both in series and in parallel, then it is called a combination circuit. When all the components in a circuit are only connected by parallel connections, it is called a parallel circuit.

Let’s look at an example.

Example 1: Identifying Components Connected in Parallel

The diagram shows four circuits. Which circuit contains two resistors in parallel?

Answer

Components are connected in parallel if they are on separate paths of a circuit.

Circuit a has two resistors, but they are on the same path, not different ones. Electrons flowing from one terminal of the cell to the other must pass through both resistors. Hence, they are in series, not in parallel.

Circuit b is a very similar diagram; the only difference is that the components have been placed on different sides of the diagram. This does not affect whether the components are in series or in parallelβ€”these components are also connected in series.

In circuit d, there are two possible paths for electrons to flow along between the terminals of the cell, but one path has a light bulb and the other has the two resistors in series. The resistors are in parallel with the light bulb but not with each other.

In circuit c, there are also two possible paths for electrons to flow along, with each path containing a resistor. These resistors are in parallel.

The correct answer is circuit c.

Identifying where exactly a parallel circuit begins and ends can be tricky. Consider the following parallel circuits.

Despite appearing to start and end at different spots, these circuits are equivalent. The circled area of both is exactly the same: a split, with π‘…οŠ§ on one path and π‘…οŠ¨ on the other, then both paths coming together at the end.

As long as the paths that the currents take are the same, the circuits are equivalent. The diagram below has two more circuits that are equivalent.

The orientation of the components does not matter so long as the circuit can be described the same wayβ€”in this case, a split into two paths, one containing π‘…οŠ§ and the other π‘…οŠ¨, that then comes together again.

The diagram below shows four completely equivalent circuits, with two resistors and a light bulb.

These circuits are equivalent because the components are in parallel in the same way. They can all be described as a split in the path, with one containing two resistors and the other containing a light bulb and the paths then coming back together. The figure below shows this description visually, using two of the circuits.

Let’s look at an example.

Example 2: Identifying Equivalent Circuit Diagrams

The diagram shows four circuits with components connected in parallel. Which two circuits are equivalent?

Answer

Let’s consider the paths in these circuits. Circuit a has a split, with the 10 Ξ© resistor on one path and a light bulb and a 20 Ξ© resistor on the other.

Circuit b has a split with the 20 Ξ© resistor on one path and the light bulb and 10 Ξ© resistor on the other.

Circuit c first has a 20 Ξ© resistor, before splitting into two paths with a light bulb on one and a 10 Ξ© resistor on the other. Notably, it is not a true parallel circuit, but rather a combination circuit, since it has components both in series and in parallel.

Circuit d has a split, with the 10 Ξ© resistor on one path and a light bulb and a 20 Ξ© resistor on the other.

If the current follows the same path through the same components, the circuits are equivalent. The two that are described the same are a and d.

In a parallel circuit, the potential difference across each parallel branch will be the same. No matter the values of resistance the components along those paths have, the potential difference across each will be the same. The diagram below shows two voltmeters measuring the potential difference across two resistors.

If π‘…οŠ§ is 1 Ξ© and π‘…οŠ¨ is 1β€Žβ€‰β€Ž000 Ξ©, it does not matter. The potential difference across both is 20 V.

Rule: Potential Difference across Components in Parallel

The potential difference across each branch of a parallel circuit is the same: 𝑉=𝑉=𝑉=….

Let’s look at an example.

Example 3: Finding Potential Differences across Components in Parallel

The diagram shows two resistors connected in parallel to a cell. If the potential difference across the 3 Ξ© resistor is 18 V, what is the potential difference across the 6 Ξ© resistor?

Answer

The potential difference across parallel components is given by the rule 𝑉=𝑉=𝑉=….

For this circuit with two components, the rule looks like 𝑉=𝑉.

We are given the value of potential difference across the 3 Ξ© resistor, π‘‰οŠ§, to be 18 V. The potential difference across the 6 Ξ© resistor must then be the same: 18 V.

Unlike the potential difference across different branches, which is the same for each branch, the current in each branch may be different.

Recall that resistance is the opposition to the flow of electric charge. If one branch has a greater resistance than another, it will be harder for charge to flow along it, and so the current in it will be weaker and the current in the other branch will be stronger.

However, all of the electrons from a cell must pass along one of the branches. Therefore, the total current through all the branches will be equal to the current before the path of the circuit split.

In the diagram below, the total current 𝐼total is equal to the sum of the currents in the individual branches, 𝐼 and 𝐼.

Rule: Total Current in Parallel Paths of a Circuit

The total current of components in parallel, 𝐼total, is equal to the sum of the current in each path: 𝐼=𝐼+𝐼+𝐼+…,total where 𝐼, 𝐼, 𝐼, and so on are the currents in a specific path.

The exact way that the current splits up is dependent on the resistance of each path. The diagram below shows a circuit that splits into three paths, with a current in each.

We are given the total current, 𝐼total, and the current in two of the paths, along the resistors π‘…οŠ¨ and π‘…οŠ©. We can use these to find the current through the leftmost path, 𝐼.

The total current in the circuit is 9 A. We can call the middle path 𝐼 and the rightmost path 𝐼, which are 5 A and 1 A respectively. We can use the rule for total current along parallel paths to represent this as an equation: 𝐼=𝐼+𝐼+𝐼.total

Putting in our known values of the current, the equation becomes (9)=𝐼+(5)+(1)9=𝐼+6.AAAAA

To solve for 𝐼, we need to subtract 6 A from both sides: 9βˆ’6=𝐼+6βˆ’63=𝐼.AAAAA

So, the current along the leftmost path, 𝐼, is 3 A.

Suppose now that we are not given the current along a path, like in the diagram below.

We can determine how much current is going through each path by comparing the values of resistance. In the above diagram, the resistors have the same value of resistance 𝑅. This means the current will be split evenly between the two paths: 𝐼=𝐼.

So, looking at the total current, 𝐼=𝐼+𝐼.total

Substituting all currents for 𝐼, since they are the same, gives us 𝐼=𝐼+(𝐼)𝐼=2𝐼.totaltotal

To find 𝐼, we just divide both sides by 2: 𝐼2=2𝐼2𝐼2=𝐼.totaltotal

So, the individual paths have one-half the total current. The total current is 8 A, so 𝐼2=4.totalA

Since 𝐼 and 𝐼 are equal to each other, they are both 4 A each.

Let’s look at some example questions.

Example 4: Finding Currents through Components Connected in Parallel

The circuit shown in the diagram consists of two resistors connected in parallel to a cell. The value of 𝐼total is equal to 30 A. What is the value of 𝐼?

Answer

The total current in a circuit with parallel components is given by the rule 𝐼=𝐼+𝐼+𝐼+….total

For just two components on separate paths, this rule becomes 𝐼=𝐼+𝐼.total

The two resistors along the paths in this circuit are the same, so the current is being split evenly between them. This means that the current 𝐼 is the same as 𝐼: 𝐼=𝐼.

Substituting this relation into the rule for the total current gives 𝐼=(𝐼)+𝐼𝐼=2𝐼.totaltotal

Now to solve for 𝐼, we divide both sides by 2: 𝐼2=2𝐼2.total

This causes the 2 on the right side of the equation to cancel, leaving 𝐼2=𝐼.total

The total current is 30 A. Substituting in this value gives the value of 𝐼: (30)2=15.AA

So, the value of 𝐼 is 15 A.

To observe how current changes with the resistance of a path, we must first look at how to find the total resistance of components in parallel.

Rule: Total Resistance of Components in Parallel

For a number of components along parallel paths, the total resistance 𝑅total is 1𝑅=1𝑅+1𝑅+1𝑅+…,total where π‘…οŠ§ is the resistance of the first component, π‘…οŠ¨ is the second, and so on.

Let’s use this equation with just two resistors, π‘…οŠ§ and π‘…οŠ¨, to demonstrate how to find 𝑅total by itself. This makes the equation 1𝑅=1𝑅+1𝑅.total

We can take the reciprocal of both sides to obtain 𝑅total on the left side: 1=1+.total

Dividing by a fraction is the same as multiplying by its denominator, so the left side becomes 1=1×𝑅1=𝑅.totaltotaltotal

This makes the equation 𝑅=1+.total

We now multiply both sides by π‘…π‘…οŠ§οŠ§; since this is just one, it will preserve the 𝑅total term on the left side: 𝑅×𝑅𝑅=1+×𝑅𝑅.total

The fraction means that the top and bottom will distribute across for the fraction on the right side: 𝑅×1=1×𝑅𝑅+οˆπ‘…=𝑅+.totaltotal

π‘…οŠ§ over π‘…οŠ§ is one, so the equation is now 𝑅=𝑅1+.total

Now, let’s multiply both sides by π‘…π‘…οŠ¨οŠ¨: 𝑅×𝑅𝑅=𝑅1+οˆΓ—π‘…π‘….total

Multiplying π‘…οŠ¨ through on both the top and bottom for the right side, we get 𝑅×1=𝑅×𝑅𝑅1+.total

The π‘…οŠ¨ distributes along the denominator: 𝑅=𝑅𝑅𝑅+.total

π‘…π‘…οŠ¨οŠ¨ in the denominator is 1, giving 𝑅=𝑅𝑅(𝑅+𝑅).total

So, when there are two resistors, this equation can be used to determine the total resistance of the circuit.

Rule: Total Resistance of a Two-Component Parallel Circuit

When two components are along parallel paths, the total resistance 𝑅total is 𝑅=𝑅𝑅(𝑅+𝑅),total where π‘…οŠ§ is the resistance of the first component and π‘…οŠ¨ is the resistance of the second component.

In the circuit below, we can calculate 𝑅total using this rule.

We take π‘…οŠ§ as 5 Ξ© and π‘…οŠ¨ as 20 Ξ©. Putting these values into the equation for total resistance gives 𝑅=(5)(20)(5)+(20)=ο€Ί100(25).totalΩΩΩΩΩΩ

The division by ohms cancels the square on the ohms term in the numerator, giving the total resistance as ο€Ί100(25)=4.ΩΩΩ

Let’s look at an example.

Example 5: Identifying How Current Changes Depending on the Number of Parallel Paths in a Circuit

A student sets up the circuit shown in the diagram. Initially, the switch is open. When the student closes the switch, will the current flowing through the circuit increase or decrease?

Answer

To determine the total current in a circuit, Ohm’s law can be put into terms of current: 𝐼=𝑉𝑅.

The potential difference in this circuit is fixed by the cell, but the resistance can change when the path below is closed. Let’s label the resistors as π‘…οŠ§ and π‘…οŠ¨ like so.

Initially, when the switch is open, there is no current in the bottom branch of this circuit. All of the electrons from the cell will only flow through π‘…οŠ§. The total resistance can thus be described as 𝑅=𝑅.total

When the switch is closed, there will be a current in the branches that contain both π‘…οŠ§ and π‘…οŠ¨. Recall that the total resistance in a circuit with two parallel resistors is given using the equation 𝑅=𝑅𝑅(𝑅+𝑅).total

Now, no matter the value of π‘…οŠ¨, it will make it so that the total resistance decreases. This is because multiplying π‘…οŠ§ by π‘…οŠ¨, then dividing by the sum will always be smaller than π‘…οŠ§ by itself: 𝑅>𝑅𝑅(𝑅+𝑅).

Just to verify this, let’s say that π‘…οŠ§ is 1 Ξ© and we test π‘…οŠ¨ as 0.1 Ξ© and 1β€Žβ€‰β€Ž000 Ξ©: 𝑅=1.Ω

Substituting in 0.1 Ξ© for π‘…οŠ¨, the total resistance 𝑅total becomes 𝑅𝑅(𝑅+𝑅)(1)(0.1)((0.1)+(1))=0.099.ΩΩΩΩΩ

Now, let’s try substituting in 1β€Žβ€‰β€Ž000 Ξ© for π‘…οŠ¨: (1)(1000)((1000)+(1))=0.999.ΩΩΩΩΩ

We see that, in both these cases, the final result is less than just π‘…οŠ§.

So, if the total resistance decreases, according to Ohm’s law, the total current in the circuit must increase, since the current is inversely proportional to the resistance: 𝐼=𝑉𝑅.

The correct answer is for the current to increase when the switch is closed.

Now that we see how current, resistance, and potential difference work in parallel circuits, we can use these rules to find specific values we want in a circuit.

Let’s look at an example.

Example 6: Finding the Values of Potential Difference and Current for Components Connected in Parallel

The circuit shown in the diagram consists of two resistors connected in parallel to a cell. The value of the current given by the second ammeter, 𝐼, is 3 A. What is the value of 𝐼total?

Answer

We know that, in a parallel circuit, the total current is the sum of all the currents along the parallel paths. For this circuit, then, the total current is 𝐼=𝐼+𝐼.total

This means we have to find the current of the second path, 𝐼, so we can add it to 𝐼 and find the total current. We can do this by considering the potential differences across each path.

We are not given the potential difference of the cell, but we know that the potential difference is the same across both paths: 𝑉=𝑉.

This presents an opportunity to use Ohm’s law to relate the two paths together. Before we do this, though, let’s say that the resistor along the path of 𝐼 is π‘…οŠ§ and that along the path of 𝐼 is π‘…οŠ¨.

The potential difference along the first path, using Ohm’s law, can be written as 𝑉=πΌπ‘…οŠ§οŠ§οŠ§ and the second path as 𝑉=𝐼𝑅, but since 𝑉=π‘‰οŠ§οŠ¨, we can relate these equations as 𝐼𝑅=𝐼𝑅.

Now, we can solve for our desired value, 𝐼, by dividing both sides by π‘…οŠ§: 𝐼𝑅𝑅=𝐼𝑅𝑅.

Canceling on the left side gives 𝐼=𝐼𝑅𝑅.

𝐼 is 3 A, π‘…οŠ§ is 4 Ξ©, and π‘…οŠ¨ is 12 Ξ©. Putting these values into the equation, we get 𝐼=(3)(12)(4).AΩΩ

We now divide the units of ohms, canceling as they are in both the numerator and denominator, yielding 𝐼=(3)(3)𝐼=9.AA

Now that we have 𝐼, we just add it together with 𝐼 to get the total current: 𝐼=9+3𝐼=12.totaltotalAAA

So, the total current is 12 amperes.

Let’s summarize what we have learned in this explainer.

Key Points

  • Components are in parallel when there is more than one path for current to follow.
  • Parallel circuits are circuits in which all of the components are in parallel.
  • For a number of components along parallel paths, the total resistance 𝑅total is 1𝑅=1𝑅+1𝑅+1𝑅+…,total where π‘…οŠ§ is the resistance of the first component, π‘…οŠ¨ is the second, and so on.
  • For 2 resistors in parallel, 𝑅=𝑅𝑅(𝑅+𝑅).total
  • The total current 𝐼total in a parallel circuit is split between the different branches of the circuit such that 𝐼=𝐼+𝐼+𝐼+….total
  • The potential difference across each parallel branch of a circuit is the same: 𝑉=𝑉=𝑉=….

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