Voltage and Current Division Circuits

You’ll often be asked to calculate the voltage or current across a network element like a resistor, a capacitor, or an inductor, which can be a lengthy task. This is where you can take advantage of the voltage and current division rule to make things easy for yourself.

In this tutorial of the Circuits 101 series, we’ll obtain the mathematical expressions to find out how voltage and current are divided across a network of elements like a resistor. We’ll also solve a couple of problems to get some practice and better clarity regarding how to approach such questions.

Note: For simplicity, we’ll be considering purely resistive networks in this tutorial but the basic mathematical formulas also apply to the impedances of capacitors and inductors.

The Voltage Division Rule (VDR) shows how the voltage gets distributed among different resistors in a series configuration. It is only applicable when there is more than one resistance in a series arrangement. In a parallel combination of resistors, the voltage remains the same.

To understand voltage division, let’s consider the simplest possible network of two resistors in a series configuration connected to a voltage supply.

Our aim here is to obtain an expression that can help us calculate the voltages getting divided across the resistors R1 and R2. For that, we’ll start with the most fundamental equation in electrical circuits, i.e., Ohm’s law.

As we can see, there is a series combination of resistors in the network. We’ll add up the two resistances to get their series equivalent

Great! We know the voltage is Vin, and the resistance is Req, so using Ohm’s law, we can obtain the current in the network as

So, above was how we used Ohm’s law for the overall circuit. In the same way, we can use Ohm’s law for individual resistors as well. Given that the voltage drop across the resistor R1 is V1, and across R2 is V2, we can write

Note that because the two resistors are in series, the current through them stays the same. And the good part is that we’ve already obtained the expression for the current in the circuit. This is the time we substitute it in the above equations

That’s it! This is the voltage division rule. Figuring out the values of voltage drops across resistors in series is so much easier if you know these expressions. If you forget which resistor value should be in the numerator, it may be helpful to remember that the same resistor whose voltage drop you’re trying to find is the resistor in the numerator.

A more generalized version of the voltage division rule is given as follows, which is often used when there are more than two resistors in a series configuration:

Where R_eq= Total equivalent series resistance
V_in = Input voltage
R = Resistor across which divided voltage is being calculated
V_r = Divided voltage across the resistor

Now that we know about the concept, there’s nothing better than solving a few questions that use the principle.

We need to find the voltages V₁ and V₂, which are the voltage drops across the resistor R₁ and R₂ respectively. We can make use of the voltage division expressions to get our answer.

To find V₁, we only need the input voltage V_in and the value of resistors. We don’t need to calculate the current for it.

We already have the value of Vin as 12V, and also the resistances, so let’s plug those into our equation:

Similarly, we can find V₂ as follows:

After obtaining the values, you can verify your answer by making sure that the sum of voltage drops we’ve calculated is equal to the input voltage.

They both come out to be the same so our answer gets verified as true.

This time we’ve got three resistors across which the input voltage is getting divided. This time we’ll make use of the generalized equation to find V₁, V₂, and V₃.

We’ll need R_eq in every expression so let’s calculate it before anything else.

Next, we’ll write the voltage division expressions for each resistor and plug in the values to get our answer:

We have our answers, but just to verify the values we’ll check if their sum equals the input voltage:

And that comes out to be perfect.

Practice problems aside, can you see how we can make use of the divided voltages in real life? Every resistor has a different voltage drop across it, given that they are of different values. This voltage drop can be used as a supply to a part of the network which requires a lesser operating voltage than the main supply in the circuit. This is the reason why such circuits are also known as “Voltage Dividers”, and they are of great use.

In contrast with the voltage division rule, the Current Division Rule (CDR) shows how current is distributed in a parallel circuit. The resistors in parallel configuration have the same voltage across them but the current gets divided.

To obtain the expression for the current division, we'll consider a simple network having two resistors in a parallel configuration.

Once again, we’ll start our derivation with Ohm’s law.

As someone correctly said: “To electronics students, Ohm's Law is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.”

Before we can proceed with the derivation, we’ll have to find the equivalent resistance of this parallel combination of resistors, which is as follows:

We know the input voltage is Vin and the equivalent resistance is Req, so let’s substitute that into Ohm’s law to get the following expression:

Now let’s talk about the current in individual resistors. The voltage across both the resistors is V_in as they are in a parallel configuration. The current through R₁ is I₁, and through R₂ is I₂. Using Ohm’s law, they can be expressed in terms of the individual branch resistance and V_in:

As we’ve obtained Vin already, let’s plug that into the above expressions to get:

The similar resistances in the denominator and numerator cancel out to give us the final expressions:

Awesome! We’ve derived the expression for the current division as well. As you can see, to figure out the branch current we just need the values of the resistances and the input current. Remembering this expression can help you to avoid some not-so-necessary steps and you can get to the answer in fewer steps.

An important thing to note here is that the above equations for each branch current have the opposite resistor in its numerator. That is, to solve for I1 we use R2, and to solve for I2 we use R1 in the expression. The reason is that each branch current is inversely proportional to its resistance resulting in the smaller resistance having the larger current.

The generalized formula for current division when there are more than two resistors in a parallel configuration is:

Where R_eq= Equivalent parallel resistance
I_in = Input current
R = Resistor through which the current is calculated
I_r = Current across the resistor

In case you are wondering how this expression came up, it is the same expression we obtained just before canceling the similar resistances in the numerator and denominator, in the derivation.

We need to find the branch currents I1 and I2. We are already given that the incoming current from the current source is 20 Amps.

For the branch having the resistor R1, the current division expression is:

Substituting the value of I and the resistors, we get:

Similarly, to find I₂, we’ll use the following expression:

Just like voltage division, current division calculation can also be verified by taking the sum of branch currents and comparing it with the incoming current at the node, they should come out as the same. The source current is 20 A and adding I₁ and I₂, we get

That comes out to be the same as the source current hence our answer is verified.

To find the branch currents in this problem, we’ll have to use the general expression of current division for each resistor:

Whenever dealing with more than two resistors, It is always recommended to calculate the equivalent resistance before anything else. If you also feel that finding parallel equivalence is somewhat annoying, this tool can be of great help: Parallel Equivalence - Tools | CircuitBread.

We are already given that the source current is 35 Amps and we’ve just obtained R_eq as 2.72Ω. That’s all we need to find the individual branch currents. The only thing left now is to substitute the values of I_in, R_eq, and the branch resistance in the general expression and we’ll have our answers.

For I₁:

For I₂:

For I₃:

Just to make sure we didn't make any calculation mistakes, let’s add up all the branch currents to check if we get it the same as the source current of 35 Amps.

This verifies our answer as correct.

Voltage and current division are really what make an electrical network function. So understanding this is extremely important and useful if you wish to take a deep dive into the subject of network analysis. Once you develop the intuitive understanding that the greater resistance will have a higher voltage drop and will draw a lesser current, proportional to the value of the resistor, you’ll be able to figure out the divided voltage and current without needing a pen and paper.