Parallel Circuits are circuits which satisfy the following rules:
- The same voltage exists across each branch, the current branches off
- Two elements, paths, branches or networks are in parallel if they have two common terminals
- A parallel circuit has two or more branches for current to flow through and the total current entering and exiting the branches is equal to the sum of the currents through each of the branches. (Note that this must be the case for the conservation of charge which states that charge cannot be created or destroyed, it remains the same, to put simply in our context - what goes in must come out)
The key concept is that the voltages across each branch is the same for a parallel circuit.
Two parallel circuits with several resistors are shown below. We can say that the resistors are connected in parallel.
com1 and com2 are 2 common terminals between all three resistors R1, R2 and R2 and R3 hence all 3 resistors are connected in parallel and the same voltage appears across them all.
Resistors in Parallel
When resistors are connected in parallel, the total resistance of the circuit is NOT simply equal to the sum of the resistances. It turns out that the total resistance decreases as we add resistors in parallel.
Thinking about this intuitively each branch has the same voltage and contributes a current I and having more of these branches then increases the overall current which means the equivalent resistance must be lower. The effect is shown in the circuit below:
Note: Most of the current flows through the path of least resistance.
Adding more resistors in parallel decreases the overall circuit resistance.
From this we can derive the mathematical relationship for resistors in parallel:
In other words the total current entering the branches is equal to the current exiting the branches.
We can then replace current with voltage over resistance from Ohm's law:
Each resistor has the same voltage of Vt across it as the voltages are the same for resistances in parallel.
We can then factor out the Vt on the right hand side:
We can then simplify this further by dividing both sides by Vt which gives us the formula for parallel resistance:
Hence we can sum the reciprocal of each resistance together and then take the reciprocal of the result to determine the equivalent resistance.
If we have two resistors in parallel, this formula simplifies further to what is known as the Product over Sum equation for parallel resistance:
Note the R1||R2 in the above which is shorthand for R1 in parallel with R2.
You will make use of the above handy formula many times in your EE journey, remember it well!
Before we dive into some examples, I have an interesting thought experiment for you! If we kept adding more parallel branches to an ideal voltage source wouldn't we be increasing the current without limit? Well, yes, yes we would! The key being that this is true for ideal voltage sources and not the variety you would find in real life.
The reason is that practical voltage sources have a non-zero internal resistance and so what happens is as we add parallel branches we reduce the effective resistance of the resistor in series with the internal resistance which increases the current as we have less resistance overall, however, the upper limit remains bounded by the internal resistance.
1. Calculate the total resistance for the circuit below using the sum of reciprocals formula:Now we take the reciprocal to determine Rt:
2. Calculate the total resistance using the Product over Sum equation:
The above example provides practical insight into the effect of adding resistors in parallel, if we have parallel resistors of the same value then the total resistance is R/N where N is the number of resistors in parallel.
Current Sources in Parallel
Parallel connections of current sources can be represented by a single source with a current equal to the sum of the individual sources.
Shown below is a circuit with two current sources connected in parallel:
Which is equivalent to the circuit below:
1. a) What is the voltage drop across resistors R1 and R2 in the circuit below?We can start solving the above by first simplifying the circuit by combining the two parallel current sources into one to determine the equivalent current source: Redrawing the circuit with the equivalent current source and voltages and currents labelled we now have: Looking simpler already :) Now to calculate the voltage across resistors R1 and R2 we first need to determine if they are in series or parallel. Because the resistors share two common terminals on their left side and right side they are in parallel, another way to check is that the current from Ieq splits into two branches, one branch per resistor with currents I1 and I2 which also means they are in parallel. Note that if the same current went through both resistors then they would be in series. As we know the resistors are in parallel we know they both have the same voltage drop and to determine the voltage drop across a resistor we can use Ohm's Law as long as we know the resistance and current. Thinking this through if we combine R1 and R2 together then we will have the total current, Ieq, flowing through that equivalent resistance. At that point we have a known current and resistance so we can determine the voltage, let's work through that now :) Let's first determine the equivalent resistance by using the handy product over sum rule for resistance as we have the special case of two resistors in parallel: Our equivalent circuit is now as shown below:We now have a simple series circuit with a known resistance and current so we can use Ohm's Law to determine the voltage across Req which will be equal to the voltage across R1 and R2.
b) What is the current flowing through R1?As we now know the voltage across the resistors, we can simply use Ohm's Law to determine the current through R1:
c) What is the current flowing through R2?We can determine the final current by subtracting R1 from the total current as the total current going into the branches is split into I1 and I2:
2. That same poor junior engineering student is once again thrown a bag of parts and asked to design a circuit. This time the lab demonstrator would like to see a circuit sourcing a current of 10mA to a 10k load using a 10k and 2x20k resistors and current sources of 1mA, 9mA and 3mA. How would this circuit look?Once again we can answer the above by recalling the fundamentals, albeit this time for parallel circuits. Let's start with the current, current sources simply add in parallel and so a 1mA current source in parallel with a 9mA source will give us the 10mA. How about the resistance? That's a bit trickier in parallel but fear not as we have a handy rule where two resistances in parallel of the same value make exactly half the value and so 2x20k resistors in parallel give us the 10k! Nice. This is how the circuit would then look to make the demonstrator happy:
Voltage Sources in Parallel
This can only be done if the sources have the SAME VOLTAGE and the SAME POLARITY (otherwise they couldn't be ideal voltage sources).
1. a) Determine the current flowing through resistor R1
b) Determine the current flowing through resistor R2
c) Using the results from a) and b) determine the current It
2. a) Determine the current It in the circuit below:
b) Determine the voltage provided by the current sources:
c) Determine the current flowing through R1:
d) Determine the current flowing through R2:
e) Determine the current flowing through R3:
As per series circuits, let's solidify our knowledge by simulating parallel circuits with the help of LTSpice.
Let's begin with a simple parallel circuit such as below:
We will simulate with a stop time of 1us and measure the voltage at the top of R1 and R2 to confirm we have 3V:
Great. Let's then measure all of the currents by left clicking on the voltage source, R1 and finally R2:
Interesting, we see they overlap. Now why is this the case? The reason is the resistors have the same value and so the current evenly splits. This means we will measure 6mA from the voltage source which we can also quickly check.
-6mA. Why minus? That's odd. The reason is that the simulation software has shown the direction of the current to flow down towards ground which isn't the direction the actual current flows and so the minus sign indicates the current flows the other way.
I would like to intentionally leave further experimentation as an exercise for you to develop intuition in your own way :)