Basically, circuits with parallel components serve as current dividers. Through these circuits, the supply's total current is divided among many parallel pathways. A practical connectivity system is formed by all components sharing the same two connecting points. With this configuration, the current can flow through a variety of branches and routes.
Interestingly, each component's current can have a different value. On the other hand, a distinguishing feature of parallel circuits is that, despite any variations in branch currents, the voltage across every connected branch stays constant. Meaning, let's say VR1 = VR2 = VR3, and so forth.
As a result, it becomes unnecessary to calculate the voltages of individual resistors. This simplifying factor makes it easier to calculate branch currents using Ohm's Law and Kirchhoff's Current Law (KCL).
Current Divider using Resistors
A passive current divider network can be as simple as two resistors linked in parallel, which is also the most straightforward configuration. We may determine the current passing through each parallel resistive branch as a fraction of the total current, using the current divider rule. Take a look at the circuit below.
The two resistors in this simple current divider circuit, R1 and R2, are connected in parallel in order to divide the supply or source current, IS, producing two independent currents, IR1 and IR2 which are then combined and redirected back to the source.
According to Kirchhoffs current law KCL, the total current passing through the circuit is equivalent to the sum of the individual branch currents, or the source current. This equals:
IT = IR1 + IR2
As per Kirchhoff's Current Law (KCL), for R1 and R2's parallel connection to be considered valid, the current passing through resistor R1 must be equal to
IR1 = IT – IR2
and the following will be the amount of current passing via resistor R2:
IR1 – IT = IR2
We can calculate the current (I) flowing through each resistor in accordance with this common voltage using Ohm's Law (V = I * R), which states that the voltage (V) across all of the resistors in a parallel circuit stays constant. The voltage (V) across the whole parallel combination may then be found using the following formula:
IT = IR1 + IR2
IR1 = V / R1 and IR2 = V / R2
IT = V / R1 + V / R2 = V[(1/R1) + (1/R2)]
∴ V = IT[(1/R1) + (1/R2)]-1 = IT[R1*R2 / (R1 + R2)]
For solving IR1 we use the following calculations:
IR1 = V / R1 = IT[(1/R1) / (1/R1 + 1/R2)]
∴ IR1 = IT[R2 / (R1 + R2)]
In the same manner we can further solve for IR2 as shown below:
IR2 = V / R2 = IT[(1/R2) / (1/R1 + 1/R2)]
∴ IR2 = IT[R1 / (R1 + R2)]
It's interesting to note that the resistor value in the numerator of the equations we generated for each branch current (I1 and I2) is the one opposite the branch we're solving for (R2 for I1 and R1 for I2). This makes sense since, to comply with Ohm's Law (I = V/R), current (I) is inversely proportional to resistance (R). Therefore, bigger currents (bigger value in the numerator) will come from a lower resistance (smaller value in the denominator).
Solving a Current Divider Problem
We have two resistors in our parallel circuit: R1 = 10Ω and R2 = 30Ω. A power source of V = 15 volts is connected to them. We wish to determine the total current supplied by the source (it) as well as the current passing through each resistor (I1 and I2).
Step 1: Analyze Parallel Relationships
The resistors' parallel connection is the most important aspect in this case. This indicates that the voltage (V) between their terminals is the same.
Step 2: Use the Kirchhoff Current Law (KCL)
According to KCL, the total current flowing into and out of a junction must be equal. The entire current from the source (IT) in our circuit divides into two branches at the junction point, one of which flows via R1 (I1) and the other through R2 (I2). Thus, in accordance with KCL:
IT = I1 + I2
Step 3: Determine Branch Currents with Ohm's Law
According to Ohm's Law, voltage (V) is equal to current (I) times resistance (R) (V = I * R). For every branch, we are aware of the resistance values (R1 and R2) and the voltage (V = 15V). In order to find the current in each branch (I1 and I2), we may rewrite the equation as follows:
I1 = V / R1 = 15V / 10Ω = 1.5 A
I2 = V / R2 = 15V / 30Ω = 0.5 A
Step 4: Determine the total current
Knowing the individual branch currents now allows us to utilize KCL once again to get the source's total current supply (IT):
IT = I1 + I2 = 1.5 A + 0.5 A = 2.0 A
Final Results:
Current through resistor R1 (I1): 1.5 A
Current through resistor R2 (I2): 0.5 A
Total current supplied by the source (IT): 2.0 A
Current and Resistance in Parallel Circuits:
It can be seen that there is a larger current flowing through the smaller resistor (10Ω). The idea that current takes the route of least resistance is supported by this. Ohm's Law (I = V/R) states that a lower resistance provides a quicker channel for electrons to travel, contributing to a higher current (I).
Short Circuits and Open Circuits:
A path with almost no resistance can be formed by a short circuit. Thus, in accordance with the concept of least resistance, a short circuit would result in an extremely high current flowing, possibly going above safe limits.
On the other hand, an open circuit results in a total disruption of the current path, which means that there is no electron path and zero current flow (I = 0).
Equivalent Resistance (Req) in Parallel:
The resistance of the smallest individual resistor is always greater than the total resistance (Req) of resistors connected in parallel. This is so that the total resistance may be reduced if more options for current flow are created by adding parallel channels.
The equivalent resistance (Req), which decreases when more resistors are connected in parallel, makes calculations of current much easier.
Kirchhoff's Current Law (KCL) for Shortcut:
In some cases, calculating all branch currents may not be necessary. When the total current (IT) has been determined, KCL provides a shortcut to calculate the rest of the branch current. To find the total current, just remove the sum of the known branch currents (IT = I1 + I2 +… + In). This method makes use of the condition that the total current flowing into and out of a junction be equal.
Solving a Current Divider Problem for a 110V, 1kW Supply
A circuit for dividing current is created by connecting three resistors with the values 9Ω, 24Ω, and 95Ω. in the event that a 1kW 110 volt supply powers the circuit. Using the current division formula, calculate each branch current separately to determine the equivalent circuit resistance.
Let's calculate the total current IT first:
P = VS * IT
IT = P / V = 1000 / 110 = 9.09 Amps
Next, we calculate the Equivalent Resistance REQ:
REQ = [1 / (1/R1 + 1/R2 + 1/R3)]
= [1 / (1/9 + 1/24 + 1/95)]
= 6.17 Ω
After this, we calculate the branch currents IR1, IR2, IR3.
IR1 = IT(REQ / R1) = 9.09(6.17 / 9) = 6.23 Amps
IR2 = IT(REQ / R2) = 9.09(6.17 / 24) = 2.33 Amps
IR3 = IT(REQ / R3) = 9.09(6.17 / 95) = 0.59 amps
We may verify our results by using KCL, which stipulates that the total current arriving a junction has to equal the total current exiting it. In the above circuit, the total current (IT) should equal the sum of the currents traveling across each branch (IR1, IR2, and IR3).
According to our calculations:
- IT (total current) = IR1 (current through R1) + IR2 (current through R2) + IR3 (current through R3)
- IT ≈ 6.23 A + 2.33 A + 0.59 A ≈ 9.15 A
As predicted, the sum of the individual branch currents (9.15 A) is nearly equal to the total current (IT = 9.09 Amps). The slight variance might be attributable to rounding mistakes during calculation.
Parallel resistors and current distribution:
This demonstrates that the total current (IT) is distributed across the parallel branches according to a simple ratio that is derived from each branch's resistance values. According to Ohm's Law, pathways with lower resistance allow higher current flow.
What Happens When You Add More Parallel Resistors?
It's crucial to remember that the total current (IT) will rise if additional resistors are connected in parallel to the same voltage source (VS). This is to ensure that more current may be pulled from the source and the total resistance can be reduced. Each extra parallel branch creates an additional pathway for current to flow.
Another Quick way to Find Branch Currents using Conductance
Another method to determining branch currents in parallel circuits is to use conductance. The opposite of resistance (R), conductance (G) essentially indicates how readily electricity moves through a component. Consider it as a high-way, where a lane that is larger (has less resistance) has a greater conductance, which permits more traffic (current) to flow.
Conductance is shown by the letter "G," and its value is determined by inverting the resistance (G = 1/R). In the past, conductance was measured in "mho" (℧) which resembles an ohm sign turned backward. These days the unit of conductance used is "Siemens" (S).
For resistors connected in parallel, the overall ease of current flow (total conductance, CT) is just the sum of the resistors' individual conductances. In laymans words if you have several roads side by side (parallel resistors), their combined capacity to handle traffic (current) is equal to the sum of their individual capacities. This differs from resistance in series, in which the total resistance adds up.
Formula for Parallel Conductance
1 / RT = 1 / R1 + 1 / R2 + 1 / R3 + .....etc
GT = G1 + G2 + G3 + ....etc
Understanding Current Flow: Resistance vs Conductance
Assume a water pipe. A bigger pipe (low resistance) causes greater quantities of water to flow smoothly (high conductance). In electronics, conductance (G) is the inverse of resistance (R). It shows how quickly electricity travels through a substance.
Think of resistance (R) as a narrow pipe that restricts water flow. Higher resistance equals less current flow.
Conductance (G) is similar to a larger pipe, allowing for easier water movement. Higher conductance indicates higher current flow.
Resistance and conductance have a simple relationship: G = 1/R. So, for example a 10Ω resistor would have a conductance of 0.1 Siemens (S), the standard unit for conductance.
For extremely small currents, we make use of smaller units such as milli-Siemens (mS), micro-Siemens (uS), and even nano-Siemens (nS). Following our previous analogy, there would be even larger pipelines for maximum water flow.
Current and conductance:
Ohm's Law (I = V/R) describes the relationship between current (I), voltage (V), and resistance. We may use conductance (G) in place of resistance by rearranging the equation as I = V * G. This allows us to determine the current flow depending on voltage and conductance.
Using conductance (G), we can go one step further and determine the total current (IS) passing through our network of parallel resistors. The formula becomes:
IS = GT * VS
IS = (G1 + G2 + G3) * V
Below is an explanation of each term:
IS: The total current passing via every parallel network node.
GT: The total conductance of all the resistors put together (keep in mind that conductance increases parallel).
VS: The circuit's supply voltage.
This equation basically informs us that the voltage pushing the current (voltage source) and the general ease of current flow (conductance) in the parallel network determine the total current.
As we already know, the voltage (V) in a parallel circuit remains identical across all of the components. We also know that Ohm's Law uses the formula V = I * R to calculate the voltage (V), current (I), and resistance (R).
The key idea is that voltage may be expressed in terms of current and conductance by rewriting Ohm's Law since conductance (G) is the reciprocal of resistance (G = 1/R). The result of rearranging the equation is:
I / G = V
The voltage (V) of a component is simply expressed as its conductance (G) divided by the current (I) that flows through it.
We can now describe the current divider rule in terms of conductance (G) rather than resistance (R) by using the relationship between voltage, current, and conductance. In the following section, we'll look into it.
Current Distribution in Parallel Circuits: Using Conductance
IR1 = G1 * V = G1(IT / GT)
∴ IR1 = IT(G1 / GT)
In the same way, currents in parallel resistors R2 and R3 can be calculated as follows:
IR2 = IT(G2 / GT)
IR3 = IT(G3 / GT)
Unlike the previous equations that used resistance, if we use conductance, the numerator includes the value for each branch current (such as I1 or I2) as well as the conductance specific to that branch. This makes sense since conductance (the ease with which current flows) is exactly proportional to resistor value.
Remember that conductance (G) is the opposite of resistance (R). So, implementing conductance simplifies the calculations since each branch current "uses" the conductance that has a direct impact on it.
Solving a Conductance Problem
We have a parallel circuit with three resistors: R1 = 2.2kΩ, R2 = 4.7kΩ, and R3 = 10kΩ connected to a 24V DC supply. We want to find the current flowing through each resistor (I1, I2, and I3).
First we find the total conductance GT
GT = 1 / R1 + 1 / R2 + 1 / R3
= 1 / 2200 + 1 / 4700 + 1 / 10000
= 0.00076731141
= 767.31141 μS
Next, we solve the total supply current IT.
IT = VS * GT = 24 + 0.00076 = 0.018 = 18 mA
Now let's calculate the G1, G2, G3 values.
G1 = 1 / 2200 = 0.00045 = 450 μS
G2 = 1 / 4700 = 0.00021 = 210 μS
G3 = 1 / 10000 = 100 μS
Next, we calculate the current I1, I2 and I3 through individual branches:
IR1 = IT(G1 / GT) = 0.018(0.00045 / 0.00076) = 0.0106 = 10 mA
IR2 = IT(G2 / GT) = 0.018(0.00021 / 0.00076) = 0.0049 = 4.9 mA
IR2 = IT(G3 / GT) = 0.018(0.0001 / 0.00076) = 0.0023 = 2.3 mA
Since conductance is the reciprocal or inverse of resistance, the above circuit's equivalent resistance value is just 1/767.31μS, or 1303 Ω or 1.30 kΩ, which is evidently smaller than R1's lowest resistor value of 2.2kΩ.
Conclusion
Understanding current distribution in parallel circuits (current dividers)
Current dividers help to understand how current flows in parallel circuits, in which every component have an identical voltage. This is where Kirchhoffs Current Law (KCL) comes into play which states that the total current entering a junction and the total current exiting it must be equal…..in other words, what goes in, must come out.
Finding Branch Currents:
There are two primary ways to detect individual branch currents (currents passing through each resistor) in parallel circuits:
1) Using total current and equivalent resistance , (for two branches)
if you know the total current (IT) passing through the circuit you may use the current divider rule for calculating the current in each branch along with the equivalent resistance (Req) of all resistors combined.
In a two branch circuit with equal resistances, the current will be distributed evenly across the branches.
2) Using Conductance (for multiple branches):
Things can become a little more complicated for circuits that include three or more branches. This is where the usefulness of conductance (G) becomes evident.
Conductance, which measures how readily current flows through a component, is the opposite of resistance (G = 1/R).
By applying the current divider rule with conductance, we can more effectively calculate branch currents in a parallel circuit by using the total conductance, which is the sum of the individual conductances.
Why do we Need Conductance?
The arithmetic required to calculate branch currents can be made simpler by using conductance, particularly in circuits with several branches.
The sum of the individual conductances of all the resistors linked in parallel determines the total conductance in a parallel circuit.
Important Points:
When working with current sources, current dividers are particularly helpful.
When you use the principles of current division, you can apply conductance to both DC and AC circuits.
References: Current Dividers