Voltage and Current Divider Rules

Voltage Divider Rule:

The Voltage Divider Rule is a formula we can apply to determine the voltage dropped across a resistor in a series circuit containing multiple resistors.

Note that this rule will apply to any series sections of a circuit or equivalent series circuits which really helps us to efficiently analyze circuits. Consider it as one of the swiss army knives of analog electronics alongside Ohm's Law!

The basic idea behind this rule is that the voltage dropped across each resistor is proportional to the total voltage across the series resistances and the ratio of its resistance to the total resistance. If we refer to the voltage across resistor R as the output voltage Vout, and the voltage applied to the total resistance Rt as the input voltage Vin we then have the Voltage Divider Rule in its general form:

Thinking about this intuitively, we would expect most of the voltage to be dropped across the largest resistor in a series circuit with the amount depending on its resistance compared to the total resistance. As a simple example, if we had a 1kΩ resistor in series with a 2kΩ resistor then we would expect twice the voltage across the 2kΩ and the remainder across the 1kΩ resistor i.e. 1/3Vs and 2/3Vs across them respectively which when summed together gives us Vs as expected for a series circuit.

Shown below is a typical voltage divider circuit where resistance values are chosen such that a particular voltage is output across the second resistor R2 to be used by other circuit elements. We can say that the voltage has been divided to produce an output voltage, hence why this is commonly referred to as a voltage divider circuit. This is a very common technique used in electronic circuit design which you will encounter frequently!

We can determine Vout using the Voltage Divider Rule directly:

In the second half of this section we will learn the current divider rule which looks quite similar. I personally like to remember the voltage divider rule as the resistor whose voltage we want to calculate on top divided by the sum of the resistances multiplied by the voltage across them all.

Derivation:

We can derive the Voltage Divider Rule for the above circuit by using first principles we know for Series Circuits and Ohm's Law.

The output voltage is simply the series current multiplied by R2 and the current is equal to the input voltage divided by the total series resistance:

From here we can subsitute [2] back into [1] to give us the Voltage Divider Rule:

Examples:

1. Determine the voltage across R3 in the circuit below using the voltage divider rule:

2. Determine the voltage across R2 in the circuit below using the voltage divider rule:

This is a Series-Parallel circuit, however, if we combine R2 and R3 in parallel we will have R1 in series with R2||R3 which we can then apply the voltage divider to:

Current Divider Rule:

The Current Divider Rule is a formula we can apply to determine the current through a resistor which is in parallel with other resistors.

This is also considered one of the swiss army knives of analog electronics! It is particularly handy when the circuit contains current sources.

The basic idea behind this rule is that the current through each resistor is proportional to the total current entering the junction and the ratio of the total parallel resistance to its resistance. If we refer to the current through resistor R as the output current Iout, and the voltage applied to the total parallel resistance Rt as the input current Iin we then have the Current Divider Rule in its general form:

As mentioned previously, this formula looks remarkably similar to the Voltage Divider Rule, just with currents, however it is subtly different in that the ratio is inverted and the total resistance is not the summation of the resistances, the case for series resistors, but the combined parallel resistance. It is the ratio of the total parallel resistance to the resistor that we want to calculate the current flowing through.

The similar yet inverted nature of the Current Divider Rule is due to current being inversely proportional to resistance. The majority of the current naturally traverses the path of least resistance (Same voltage across less resistance implies higher current by Ohm's Law). As a simple example of a parallel circuit with two resistors, one large and one small, most of the current will flow through the smaller resistor which is a result of Ohm's Law and is reflected by the Current Divider Rule in that we divide by the resistor we want and a smaller divisor, resistor, results in a larger current.

For the special case of a parallel circuit with two resistors, the Current Divider Rule can be simplified to the following which is even closer to the Voltage Divider Rule:

Where I1 is the current flowing through resistor R1.

This situation arises commonly in circuits so this simplified formula should be memorized well. I remember it as the current we don't want divided by the sum multiplied by the current entering the junction.

Shown below is a typical current divider circuit where a fraction of the input current flows through a parallel resistance as an output current which may be used to drive other electronic components. In a similar manner to voltage division we can say that the input current has been divided to produce a particular output current.

We can determine Iout using the simplified Current Divider Rule for two resistors:

Derivation:

The derivation for the current divider rule is performed in a similar way, if we define the output current in terms of it's resistance, R, and voltage with Ohm's Law we have:

From here we can define the voltage across the resistor in terms of the input current and total circuit resistance:

We can subsitute [2] back into [1] to give us the Current Divider Rule:

For a circuit with 2 parallel resistors R1 and R2 we can replace Rt with the series equivalent resistance given by the product over sum formula:

And if we then replace Rt with this we have:

From which the R1's cancel leaving us with the simplified formula:

Examples:

1. a) Determine the current through R1 in the circuit below using the current divider rule:

For this circuit we need to use the general formula as we have more than 2 resistors:
We need to determine Rt for the above equation for which we need to find the total resistance of the 3 parallel resistors. As there are more than 2 resistors we need to sum the reciprocals and then invert the result to calculate Rt:
The reciprocal is then taken to calculate Rt:
This can then be subsituted into [1] to determine I1:
Note that we could check the above answer by determining the source voltage using Rt and Ohm's Law and then apply Ohm's law again to determine the current through I1 using that voltage and R1. By inspection we would also expect most of the current to flow through R1 as it has a considerably lower resistance than the other two.

b) Determine the current through R2:

As we know now the total parallel resistance we can apply the current divider rule to determine the current flowing through R2 directly:

c) Determine the current through R3:

We could apply the same strategy as b) for this however it is much simply to apply KCL at the junction connecting the resistors to the source and simply subtract I1 and I2 from Is to calculate I3:

Questions:

1. a) Determine the voltage across R2 in the circuit below:

Answer

VR2=1.913V

b) Determine the voltage across R1:

Answer

VR1=4.087V

2. a) Determine the voltage across R1 in the circuit below:

Answer

VR1=1.582V

b) Determine the voltage across R2:

Answer

VR2=3.165V

c) Determine the voltage across R3:

Answer

VR3=256.165mV

3. a) Determine the voltage across R3 in the circuit below:

Answer

VR3=20V

b) Determine the current flowing through R1:

Answer

IR1=2.400mA

c) Determine the current flowing through R2:

Answer

IR2=1.6mA

4. a) Determine the voltage across R4 in the circuit below:

Answer

VR4=8.8V

b) Determine the current flowing through R1:

Answer

IR1=307.693uA

c) Determine the current flowing through R2:

Answer

IR2=615.385uA

d) Determine the current flowing through R3:

Answer

IR3=3.077mA

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