# Equivalent resistance of a parallel circuit with derivation & numerical problems

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:
VT = V1 = V2 = V3 = ….. (a)
I = I1 + I2 + I3 +…
(b) [ current I gets distributed among parallel paths.]

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

VT/Req = V1/R1 + V2/R2  + V3/R3 ..

=> VT/Req = VT/R1 + VT/R2  + VT/R3 .. [ as we know VT = V1 = V2 = V3 ]
1/Req = 1/R1 + 1/R2  + 1/R3 .. (c)

Req = Equivalent resistance of the resistors in parallel.

When there are just 2 parallel paths then equation (c) looks like this:
1/Req = 1/R1 + 1/R2
=> Req = R1R2 /( R1 + R2 ) (d)

## 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 = Req = (5×20)/(5+20) = 100/25 = 4 ohm
VT = 16 volts
Total circuit current = VT / Req = 16/4 amp = 4 amps
Current through 5 ohm branch = VT /5 amps = 16/5 amps = 3.2 amps
Current through 20 ohm branch = VT /20 amps = 16/20 amps = 4/5 amps=0.8 amps

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 Req = 5 ohm
VT = 20 volts
Total circuit current = VT / Req =20/5 Amps = 4 A
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

Equivalent resistance of a parallel circuit with derivation & numerical problems
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