Voltage and Current Divider Circuits
Introduction: This article focuses on analyzing the voltage divider and current divider circuits
The Essentials
When designing circuits, multiple elements are involved in doing a task. Each of these elements may, at times, require a minimum amount of power, voltage, or current. With this in mind, we are interested in how we can "divide" voltage or current out so each element gets its needed requirements.
Firstly, let's look at voltage division. Figure 1. has a simple example of a voltage divider.
Figure 1: Voltage Divider
When we use nodal analysis, we can find a voltage across any of these resistors in series. Both of these equations describe their respective voltages.
Then for the voltage over \( R_2 \).
This makes sense since both resistors can only take a fraction of the input voltage. If we are to add \( V_1 \) and \( V_2 \), we should get \( V_{in} \) back.
Now for a current divider circuit. Figure 2. has an example of a current divider.
Figure 2: Current Divider
When we use mesh analysis, we can solve for the current through each resistor. Or we can use the following equations.
Both of these equations describe the current through each resistor. Once again, the sum of these currents should equal to \( I_{in} \).
Example
Let's do an example of both voltage and current division. Firstly, voltage division given the schematic in Figure 3.
Figure 3: Voltage Divider Example
Let's say we are curious to know the voltage across each resistor. Now let's implement equation 1 on each resistor.
Note how in each equation the numerator contains the sum of all resistors in series. We need to add each value of the resistors in series, whether it be two or any other number. The resistor fraction term will never be more than 1.
If we want to check our work, we can see if \( V_1 \), \( V_2 \), and \( V_3 \) are correct by adding them and if they equal the input voltage, then these values are correct.
Now for a current divider example in Figure 4.
Figure 4: Current Divider Example
Now let's use Equation 3 and 4 to solve for \( i_1 \) and \( i_2 \).
We can do the same thing with current by adding all current and see if they equal the input current.
Practice
Problem 1.)
Figure 4: Problem 1
Using Figure 5., solve for \( V_1 \), \( V_2 \), \( V_3 \), and \( V_4 \) (voltages across each resistor like \( V_1 \) will be the voltage across the \( R_1 \) resistor, and so forth.).
Problem 2.)
Figure 5: Problem 2
Given in Figure 6., solve for currents \( i_1 \text{(i of 4.5 k$\Omega$)} \), \( i_2\text{(i of 6 k$\Omega$)} \), and \( i_3\text{(i of 2.6 k$\Omega$)} \).
Problem 3.)
Figure 6: Problem 3
Given in Figure 7., solve for the voltage \( V_{3} \) and the current \( i_3 \) and \( i_4 \). (Note that both \( R_3 \) and \( R_4 \) experience the same voltage, so \( V_3 = V_4 \))
Solutions:
Problem 1 Solution.) \( V_1 = 8.4 \, \text{V} \), \( V_2 = 4.76 \, \text{V} \), \( V_3 = 2.63 \, \text{V} \), and \( V_4 = 4.2 \, \text{V} \). Using the previous equations described:
Problem 2 Solution.) \( i_1 = 575 \, \mu \text{A} \), \( i_2 = 431 \, \mu \text{A} \), and \( i_3 = 994 \, \mu \text{A} \). We need to be careful because unlike the logic of the voltage divider equation:
This is due to the nature of the parallel elements Instead, let's use our knowledge of combining parallel resistors to solve for currents.
With this, let's implement the current divider equation
We need to find \( i_2 \) and \( i_3 \). Since we know the current through \( i_1 \), we can solve for the new current going through the 1813.95 resistor.
We can implement the same equation with the new \( i_p \) that we just solved since \( i_p \) represents the sum of the currents \( i_2 \) and \( i_3 \).
Problem 3 Solution.) \( V_3 = 8.79 \, V \), \( i_3 = 836 \, \mu \text{A} \), and \( i_4 = 906 \, \mu \text{A} \). Let's solve for \( V_3 \), a simple way is to combine the parallel resistors and use the voltage divider.
We can do voltage division with the new \( R_p \).
Now, with this and the hint from the problem, this voltage goes across both resistors and we can solve for current using Ohm's Law.