Parallel Circuits

The Main Idea

Parallel Circuit is a circuit that is connected in parallel. All components in parallel circuit are linked to the same set of electric points, and they can create several branches(individual paths) within a circuit. These individual paths provide multiple pathways to the charge, so whenever the charge encounters a branch it would travel to the lower potential. This means adding an additional resistor in a parallel circuit would result in a decreased resistance. In a parallel circuit, the potential difference is identical with each resistor positioned in different branches. If a single circuit is opened(broke), no charge would flow to that path, but other paths will have charges going through them.

In order to understand the parallel circuit, it's critical to know the main difference between series circuit and parallel circuits. In a series circuit the current is identical through out the circuit, whereas parallel circuit has current, depending on the resistor running through the individual branch. On the other hand, in a series circuit the voltage differs,whereas in parallel circuit the voltage is same everywhere.

A Mathematical Model

[math]\displaystyle{ V=IR }[/math]

Using the basic Ohm's law we could determine the voltage, resistance, and current throughout the parallel resistor.

[math]\displaystyle{ \frac{1}{R}_{Total}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...\frac{1}{R_N} }[/math]

When calculating the total Reisistance in a parallel circuit, we need to know the basic principle:

More resistor in parallel circuit, less resistance. So, we need to find the sum of reciprocals of individual resistors to derive the total resistor.

[math]\displaystyle{ V_{Total}=V_1=V_2=V_3...=V_N }[/math]

In a parallel circuit, potential difference through out the circuit is equal everywhere.

[math]\displaystyle{ I_{Total}=I_1+I_2+I_3+...I_N }[/math]

In a parallel circuit, the total amount of current outside the individual branch equals to the sum of individual branches in the circuit. Thus, the individual current in each branch depends on the resistor in the branch.

[math]\displaystyle{ I_n=I_{Total}\frac{R_{Total}}{R_n} }[/math]

Current Divider Law Since the current running through each branch of the circuit is dependent on the impedance of that branch, the current divider law can be used to determine the magnitude the current through each individual branch.

So if we want to find out the current in R1

[math]\displaystyle{ I_1=I_{Total}\frac{R_{Total}}{R_1} }[/math]

A Computational Model

This is a simple way to represent a Parallel Circuit. The schematic version is very useful when we want to compute the values, and understand the concept. R represents a resistor and the battery represents the total emf in the circuit.

This diagram is the actual Parallel Circuit we could see in real life. The actual battery could represent the total emf in the previous schematic sketch, and the light bulb would act as a resistor from the previous parallel circuit.

Examples

These are practical examples we could solve through the formula, and the concept we've learned through this Wiki Page.

Simple

Calculate the total resistance of this parallel circuit.

Solution

[math]\displaystyle{ \frac{1}{R}_{Total}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...\frac{1}{R_N} }[/math]

Using this formula we could derive the total resistance of a parallel circuit. Plugging in 90 for R1, 45 for 42, and 180 for R3, we get

[math]\displaystyle{ {R}_{Total}=25.7143 ohm }[/math]

Middling

Calculate the current running through R1.

Total current running through the circuit is 5 ampere. R1= 4 ohm, and R2= 6 ohm

Solution

In order to find out the current running through R1, we can use the current divider law. First we have to calculate the total resistance.

[math]\displaystyle{ \frac{1}{R}_{Total}=\frac{1}{R_1}+\frac{1}{R_2} }[/math]

Rearranging this formula we get

[math]\displaystyle{ {R}_{Total}=\frac{R_1R_2}{R_1+R_2}=2.4 ohm }[/math]

Now plug in all the value to the current divider law.

[math]\displaystyle{ I_1=I_{Total}\frac{R_{Total}}{R_1} }[/math]

Difficult

Calculate the current running through R2.

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