Lesson Explainer: Analyzing Combination Circuits Physics • 9th Grade

In this explainer, we will learn how to determine the currents through and voltages across parts of circuits that contain resistors both in series and parallel.

Recall that resistors in series are connected in a single conductive path. The circuit diagram below shows three resistors in series.

In the example shown above, the total resistance of the circuit, 𝑅, is 𝑅=𝑅+𝑅+𝑅.

This extends to any number of resistors in series: 𝑅=𝑅+𝑅+⋯+𝑅.

On the other hand, resistors in parallel are connected along multiple conductive paths. The circuit diagram below shows three resistors connected in parallel.

In the example shown above, the total resistance of the circuit, 𝑅, is 𝑅=1𝑅+1𝑅+1𝑅.

This extends to any number of resistors in parallel: 𝑅=1𝑅+1𝑅+⋯+1𝑅.

A combination circuit contains sections of resistors in series and parallel. The circuit diagram below shows two resistors connected in series to two resistors in parallel.

In the example shown above, the section of the circuit containing 𝑅 and 𝑅 is in series, and the section of the circuit containing 𝑅 and 𝑅 is in parallel.

To analyze this circuit, each set of resistors can be converted to their equivalent resistor. The diagram below shows how the two resistors in series and the two resistors in parallel can be converted into single equivalent resistors, which can then be converted to a final equivalent resistor for the entire circuit.

The series section containing 𝑅 and 𝑅 can be converted into its equivalent resistor, 𝑅: 𝑅=𝑅+𝑅.

The parallel section containing 𝑅 and 𝑅 can be converted into its equivalent resistor, 𝑅: 𝑅=1𝑅+1𝑅.

Then, finally, the equivalent resistance of the entire circuit, 𝑅, is 𝑅=𝑅+𝑅.

Let us work through an example question of this.

Example 1: Finding the Equivalent Resistance of a Combination Circuit

The circuit shown contains both series and parallel combinations of resistors.

  1. What is the total current in the circuit shown? Give your answer to one decimal place.
  2. What is the total power dissipated by the circuit? Give your answer to one decimal place.

Answer

Part 1

In the first part of this question, we must calculate the total current in the circuit. To do this, we must find the equivalent resistance of the circuit.

We will start by labeling the components on the circuit diagram.

The first step in calculating the equivalent resistance of the circuit is to find the equivalent resistance of the parallel section, composed of 𝑅 and 𝑅. Let us denote the equivalent resistance of this parallel section by 𝑅: 𝑅=1𝑅+1𝑅.

Substituting in the known values for 𝑅=2.5Ω and 𝑅=3.2Ω gives us 𝑅=12.5+13.2𝑅=1.40.ΩΩΩ

Now we can calculate the equivalent resistance of the entire circuit, 𝑅: 𝑅=𝑅+𝑅+𝑅.

Substituting the known values for 𝑅=1.6Ω and 𝑅=1.5Ω and the value of 𝑅 that we calculated as 1.40 Ω gives us 𝑅=1.6+1.5+1.4𝑅=4.5.ΩΩΩΩ

Then, we can use Ohm’s law to calculate the total current in the circuit, with 𝑉=5.5V: 𝐼=𝑉𝑅𝐼=5.54.5.VΩ

The total current in the circuit, to one decimal place, is 𝐼=1.2.A

Part 2

The total power dissipated in the circuit can be calculated by multiplying the current in the circuit by the total potential drop across the circuit: 𝑃=𝐼𝑉𝑃=1.2×5.5.AV

The total power dissipated in the circuit, to one decimal place, is 𝑃=6.7.W

These techniques of analyzing combination circuits can also be used to analyze circuits containing components other than resistors. We will now work through an example question of this.

Example 2: Analyzing Circuits Containing Components Other Than Resistors

Current is measured by an ammeter in the circuit shown in the diagram. The ammeter has a resistance of 2.5 µΩ.

  1. What reading would be on the ammeter? Give your answer to one decimal place.
  2. What would the reading on the ammeter be if it were connected in parallel to the 3.5 Ω resistor? Give your answer to one decimal place.

Answer

Part 1

The first part of the question asks us to calculate the amount of current through the circuit. For this purpose, we can replace the ammeter with a resistor of equivalent resistance.

We can summarize this on a diagram, where each resistor is labeled. 𝑅 is the equivalent resistor for the ammeter.

The total resistance of the circuit is equal to 𝑅=𝑅+𝑅+𝑅𝑅=2.5+3.5+2.5×10𝑅=6.0000025.ΩΩΩΩ

The current through the circuit can then be calculated with Ohm’s law: 𝐼=126.0000025𝐼=1.9999991.VΩA

To one decimal place, the ammeter will read 2.0 A. Let us compare this to an ideal ammeter, which has zero internal resistance.

The total resistance of the circuit would be exactly equal to the sum of the two resistors: 𝑅=2.5+3.5𝑅=6.0.ΩΩΩ

This means that the current in the circuit is 𝐼=126.0𝐼=2.0.VΩA

So, the ammeter would give a reading of exactly 2.0 A. This highlights how important it is to have an ammeter with very low internal resistance. If the internal resistance is low enough, as it is in this example, it can be neglected in calculations.

Part 2

The second part of the question asks us to consider what would happen if the ammeter was connected in parallel with the 3.5 Ω resistor.

Replacing the ammeter with its equivalent resistor, we can draw in a diagram.

We must now calculate the equivalent resistance, 𝑅, of the parallel section of the circuit: 𝑅=13.5+12.5×10𝑅=(0.28+400000)𝑅=400000.28𝑅=2.499998×10.ΩΩΩ

Here, we see that the resistance of the ammeter is so low compared to the 3.5 Ω resistor that it completely dominates the equivalent resistance of the two.

The total resistance of the circuit is then 𝑅=2.5+2.499998×10𝑅=2.5000025.Ω

The current through the circuit can then be calculated using Ohm’s law: 𝐼=122.5000025𝐼=4.799995.VΩA

So, to one decimal place, the ammeter reads 4.8 A.

Let us calculate the current reading if the equivalent resistance of the parallel section is neglected in the calculation: 𝐼=122.5𝐼=4.8.VΩA

As seen, the internal resistance of the ammeter is so low that the equivalent resistance of the parallel section can be neglected.

While equivalent resistance allows us to analyze many combination circuits, there are some that cannot be solved using equivalent resistance alone.

The circuit shown in the following circuit diagram cannot be analyzed using equivalent resistance alone.

Because there are two batteries, we cannot calculate the equivalent resistance of 𝑅 and 𝑅.

To analyze circuits like this, we can use Kirchhoff’s laws.

Kirchhoff’s first law states that the current into a junction, or node, in a circuit must be the same as the current out of the junction, or node.

Definition: Kirchhoff’s First Law

Kirchhoff’s first law states that the sum of the currents into a junction/node in a circuit, 𝐼+𝐼+⋯()()inin, must be the same as the sum of the currents out of the junction/node, 𝐼+𝐼+⋯()()outout: 𝐼+𝐼+⋯=𝐼+𝐼+⋯.()()()()ininoutout

For example, suppose that the following junction in a circuit has currents 𝐼 and 𝐼 into the junction and 𝐼 out of the junction.

Kirchhoff’s first law states that the sum of the currents into the junction, 𝐼+𝐼, must equal the sum of the currents out of the junction, 𝐼: 𝐼+𝐼=𝐼.

Kirchhoff’s second law allows us to analyze the potential difference across various points in a combination circuit.

Kirchhoff’s second law states that the sum of all potential differences across components in a loop must be zero.

The following circuit diagram shows three resistors in series with a battery.

Definition: Kirchhoff’s Second Law

The sum of the potential difference across each component in a loop is equal to zero: 𝑉+𝑉+⋯+𝑉=0.

In the example circuit, the potential difference across the cell is 𝑉, and the three resistors have a potential difference across them of 𝑉, 𝑉, and 𝑉 respectively.

Kirchoff’s law states that the sum of the potential differences across all components in the loop is zero. That is, 𝑉+𝑉+𝑉+𝑉=0.

In this case, the potential difference across the cell is positive and is equal in magnitude to the total potential difference across the three resistors.

Kirchoff’s laws can be used to compare circuits. We will work through an example question of this now.

Example 3: Analyzing Multiple Similar Circuits

The circuits (a) and (b) appear very similar but are slightly different from each other. What is the difference in the total current between the circuit shown in diagram (a) and the circuit shown in diagram (b)? Give your answer to one decimal place.

Answer

Let us start by analyzing circuit (b).

We can use Kirchhoff’s second law on the loop made by the circuit. Recall that Kirchhoff’s second law states that the sum of the potential differences across each component in a loop must equal zero.

Let us label the potential difference across the 800 mΩ resistor as 𝑉 and the potential difference across the 960 mΩ resistor as 𝑉.

The two resistors can be converted to one equivalent resistor, which we will call 𝑅, with the following resistance: 𝑅=880+960𝑅=1840.mΩmΩmΩ

The potential difference across this equivalent resistor is 𝑉 and is equal to 𝑉=𝑉+𝑉.

The sum of the potential difference across each component in circuit (b) can be written as 2.5+1.4−𝑉−𝑉=0,VVV or, considering the equivalent resistor, 𝑅, it can be written as 2.5+1.4−𝑉=0𝑉=3.9.VVVV

So, the potential difference across this equivalent resistor is equal to 3.9 V.

The current through circuit (b), 𝐼, can then be calculated using Ohm’s law, converting the resistance first from milliohms to ohms: 𝐼=𝑉𝑅𝐼=3.91.84𝐼=2.12.VΩA

Now, we will analyze circuit (a).

Circuit (a) is nearly identical to circuit (b), except that the 1.4 V battery is flipped. When we apply Kirchhoff’s second law this time, this battery contributes a negative potential difference to the equation.

Combining the resistors into an equivalent resistor, as we previously did, we can write 2.5−1.4−𝑉=0𝑉=1.1.VVVV

So, the potential difference across this equivalent resistor is 1.1 V.

The current through circuit (a), 𝐼, can then be calculated using Ohm’s law, converting the resistance first from milliohms to ohms: 𝐼=𝑉𝑅𝐼=1.11.84𝐼=0.60.VΩA

The difference in current between the two circuits is then 𝐼−𝐼=2.12−0.06𝐼−𝐼=1.52.AAA

The difference in total current between the circuit shown in diagram (a) and the circuit shown in diagram (b) is therefore 1.5 A to one decimal place.

Kirchhoff’s laws can also be used to analyze combination circuits. When we are given a combination circuit, we must identify loops and junctions/nodes.

For example, the following circuit diagram contains multiple resistors and batteries in different loops of a circuit.

We can identify two nodes in this circuit and three loops. These are illustrated in the following diagram.

Now, we will work through an example question where we must use Kirchhoff’s laws to analyze a combination circuit.

Example 4: Using Kirchhoff’s Laws to Analyze Combination Circuits

The diagram shows a circuit that contains multiple cells.

  1. What is the current through the 20 Ω resistor?
  2. What is the current at the negative terminal of the 5.0 V battery?
  3. What is the current at the negative terminal of the 10.0 V battery?

Answer

Part 1

Let us begin by labeling the circuit diagram.

The currents in each branch of the circuit are labeled according to the node at the bottom of the circuit. The current from the 10.0 V cell to the node is labeled 𝐼 and is considered the current into the node. The current from the node to the 20 Ω is labeled 𝐼 and is consider as out of the node. The current from the node to the 5.0 V cell is labeled 𝐼 and is considered as out of the node.

We can apply Kirchhoff’s second law to each loop in the circuit to find the potential difference across each resistor.

Starting with the loop containing 𝑉 and 𝑅, we have 𝑉−𝑉=0,V so the potential difference across 𝑅 is 𝑉=10.0.V

Using Ohm’s law, we can calculate the current through 𝑅: 𝐼=𝑉𝑅𝐼=10.020𝐼=0.5.VΩA

This is equal to 𝐼. The current through the 20 Ω resistor is 𝐼 and so is equal to 0.5 A.

Part 2

Next, we can look at the loop containing 𝑉, 𝑉, and 𝑅: 𝑉+𝑉−𝑉=05.0+10.0−𝑉=0.VVVV

So, the potential difference across 𝑅 is 𝑉=15.0.V

From Ohm’s law, we can calculate the current through 𝑅: 𝐼=𝑉𝑅𝐼=15.015𝐼=1.0.VΩA

This is equal to 𝐼. The current at the negative terminal of the 5.0 A battery is 𝐼 and so is equal to 1.0 A.

Part 3

Using Kirchhoff’s first law, we can calculate 𝐼 from the node at the bottom of the circuit: 𝐼=𝐼+𝐼𝐼=1.0+0.5=1.5.A

The current at the negative terminal of the 10.0 V battery is 𝐼 and so is equal to 1.5 A.

Example 5: Using Kirchhoff’s Laws to Analyze Combination Circuits with Unknown Components

In the circuit shown, the resistance of one of the resistors is unknown. The total current in the circuit is 0.25 A.

  1. Find the current 𝐼. Give your answer to two decimal places.
  2. Find the current 𝐼. Give your answer to two decimal places.
  3. Find the potential difference across the unknown resistor. Give your answer to the nearest volt.

Answer

Part 1

We will start by labeling the components of the circuit.

The first part of the question asks us to find 𝐼, the current through 𝑅. To do this, we need to find the potential difference across the parallel section of the circuit.

We can convert the parallel section of the circuit into an equivalent resistor by using the following formula: 𝑅=1𝑅+1𝑅.

Inserting values 𝑅=2.5Ω and 𝑅=3.2Ω gives us 𝑅=12.5+13.2𝑅=1.40.ΩΩΩ

Knowing that the current in the circuit is 0.25 A, the potential difference across the equivalent resistor can be calculated using Ohm’s law: 𝑉=𝐼𝑅𝑉=0.25×1.40𝑉=0.35.AΩV

This potential difference is the same across both branches of the parallel section of the circuit. The current, 𝐼, can then be calculated using Ohm’s law: 𝐼=𝑉𝑅𝐼=0.352.5𝐼=0.14.VΩA

Part 2

The second part of the question asks us to calculate the current 𝐼.

We have already calculated the potential difference across the parallel section of the circuit, so we can apply Ohm’s law to 𝑅: 𝐼=𝑉𝑅𝐼=0.353.2𝐼=0.11.VΩA

Part 3

The third part of this question asks us to calculate the potential difference across the unknown resistor.

We can use Kirchhoff’s second law to solve this problem. Kirchhoff’s second law states that the sum of the potential differences across each individual component in a loop is equal to zero. For this circuit, we will write the potential difference across 𝑅 as 𝑉, across the parallel section as 𝑉, and across 𝑅 as 𝑉: 12−𝑉−𝑉−𝑉=0.V

We have already calculated 𝑉=0.35V. 𝑉 can be calculated using Ohm’s law: 𝑉=𝐼𝑅𝑉=0.25×2.2𝑉=0.55.AΩV

We can substitute these into the equation for Kirchoff’s second law, remembering that these are decreases in potential, so they are negative: 12−𝑉−0.55−0.35=0.VVV

We can then rearrange this to give a value for the potential difference across 𝑅, the unknown resistor: 𝑉=11.1.V

So, the potential difference across the unknown resistor, to the nearest volt, is 11 V.

Let us summarize what we have learned in this explainer in the following key points.

Key Points

  • In combination circuits, we can identify sections of circuits in parallel and circuits in series. We can calculate the equivalent resistance of these sections to analyze the combination circuit.
  • Kirchhoff’s first law states that the sum of the currents into a junction/node in a circuit, 𝐼+𝐼+⋯()()inin, must be the same as the sum of the currents out of the junction/node, 𝐼+𝐼+⋯()()outout: 𝐼+𝐼+⋯=𝐼+𝐼+⋯.()()()()ininoutout
  • Kirchhoff’s second law states that the sum of the potential difference across each component, 𝑉,𝑉,…𝑉, in a loop is equal to zero: 𝑉+𝑉+⋯+𝑉=0.
  • In combination circuits, we can identify loops and nodes that allow us to apply Kirchhoff’s laws to analyze the circuit.

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