The **equivalent resistance of a parallel circuit** is found using **Kirchoff’s junction rule**, which states that the sum of the currents entering a branch (or junction) must be equal to the sum of the currents leaving a branch (or junction).

This is illustrated in Figure 1 (the black dot represents the branch point). See how 10 A current is entering the junction and again 6+4 =10 A current is coming out of it through two branches. The concept of a circuit branch is necessary to understand parallel circuits.

## Equivalent resistance of a parallel circuit – derivation

A simple parallel circuit would look like Figure 2. The circuit current **I** splits into two currents, **I1** and **I2** (as measured by ammeters A1 and A2). Those two currents must add up to the total circuit current **I** coming from the battery.

However, experiments show that the potential drops across each resistor in parallel are the same.

Thus, we have for a parallel circuit:**V _{T} = V_{1} = V_{2} = V_{3} = ….. (a)**

I = I_{1} + I_{2} + I_{3 }+…

**(b)**[ current

**I**gets distributed among parallel paths.]

Now, using Ohm’s Law in equation (b) we get,

**V _{T}/R_{eq} = V_{1}/R_{1} + V_{2}/R_{2} + V_{3}/R_{3} ..**

=>

**V**

_{T}/R_{eq}= V_{T}/R_{1}+ V_{T}/R_{2}+ V_{T}/R_{3}..*[ as we know V*

_{T}= V_{1}= V_{2}= V_{3}]

1/R

1/R

_{eq}= 1/R_{1}+ 1/R_{2}+ 1/R_{3}.. (c)_{ }**R**

_{eq}= Equivalent resistance of the resistors in parallel.When there are just 2 parallel paths then equation (c) looks like this:

**1/R**

_{eq}= 1/R_{1}+ 1/R_{2}=>

**R**=

_{eq}**R**/(

_{1}R_{2}**R**) (d)

_{1}+ R_{2}## Numerical Problems with solutions using equivalent resistance formula of parallel circuits

1 ]

Two resistors with resistances of 5 Ω and 20 Ω each are connected in parallel to a 16-V battery. Calculate the equivalent resistance of the circuit, the total circuit current, and the current in each branch of the circuit.

2 ]

Three resistors of 30 Ω, 15 Ω, and 10 Ω are connected in parallel to a 20-V battery. Calculate the equivalent resistance of the circuit, the total current for the circuit, and the current flowing through each resistor.

**Solution **of (1)

Equivalent resistance = **R _{eq}** = (5×20)/(5+20) = 100/25 = 4 ohm

**V**= 16 volts

_{T}Total circuit current =

**V**/

_{T}**R**= 16/4 amp = 4 amps

_{eq}Current through 5 ohm branch =

**V**/5 amps = 16/5 amps = 3.2 amps

_{T}Current through 20 ohm branch =

**V**/20 amps = 16/20 amps = 4/5 amps=0.8 amps

_{T}**Solution **of (2)

Equivalent resistance:

Let’s first find out the eq resistance of 30 ohm and 15 ohm = (30×15)/(30 +15) ohm= (30×15)/(45) ohm = 10 ohm

Then we will find out the final eq resistance by using the above derived resistance 10 ohm and the 3rd resistance (another 10 ohm again) = (10×10)/(10 + 10) ohm =100/20 ohm = 5 ohm

so **R _{eq}** = 5 ohm

**V**= 20 volts

_{T}Total circuit current =

**V**/

_{T}**R**=20/5 Amps = 4 A

_{eq}Current through 30 ohm path = 20/30 A=2/3 A

Current through 15 ohm path = 20/15 A = 4/3 A

Current through 10 ohm path = 20/10 A = 2 A

**Ohm’s Law Numerical problem worksheet** – test your preparation

**Ohm’s Law **– Read the fundamentals