Litmux Studio

LessonSeries and parallel circuits · 15 min

Solving combinations and the current divider

Real circuits mix series and parallel. The method is always the same: simplify the network step by step until one resistance is left, find the total current, then work back out to each part with Ohm's law and Kirchhoff's laws.

  1. Find a group that is purely in series (add) or purely in parallel (combine) and replace it by one resistance.
  2. Repeat until one resistance is left. Itotal = V / Rtotal.
  3. Go back through your steps: the current through a series group is the same for each part; the voltage across a parallel group is the same for each branch.

The current divider

When a current splits between two branches, more of it takes the easier path - in inverse proportion to the resistance:

I1 = I × R2 / (R1 + R2) I2 = I × R1 / (R1 + R2)
I
the current arriving at the split
I1, I2
the currents in the branches R1 and R2 - note: each uses the OTHER resistor on top
13.5 mA splits: the 1 kΩ branch takes 13.5 × 2 / 3 = 9 mA, the 2 kΩ branch 13.5 × 1 / 3 = 4.5 mA. Half the resistance, twice the current.
13.5 mA splits: the 1 kΩ branch takes 13.5 × 2 / 3 = 9 mA, the 2 kΩ branch 13.5 × 1 / 3 = 4.5 mA. Half the resistance, twice the current.
Worked example: a ladder

R1 = 1 kΩ in series with [R2 = 2.2 kΩ in parallel with (R3 = 1 kΩ + R4 = 1.2 kΩ)], across 10.5 V. Find the total current and the current in R3.

  1. R3 + R4 in series: 2.2 kΩ.
  2. That in parallel with R2 (2.2 kΩ): 1.1 kΩ.
  3. Plus R1: Rtotal = 2.1 kΩ. I = 10.5 / 2.1 = 5 mA.
  4. The split is between two equal 2.2 kΩ paths: 2.5 mA each, so IR3 = 2.5 mA.

Answer: 5 mA in total, 2.5 mA through R3.

Where you meet it

Ammeter shunts

A sensitive meter movement (say 1 mA full scale, 50 Ω) measures 1 A when a tiny shunt beside it carries the other 999 mA.

Sharing heat

Several resistors in parallel share the current - and the power.

Fault finding

A branch taking far more than its share is shorted; one taking nothing is open.

Practice 130 mA splits between 1 kΩ and 4 kΩ in parallel. How much goes through the 1 kΩ?

Hint

I1 = I × R2 / (R1 + R2).

Show the answer

Answer: 24 mA

30 × 4 / 5 = 24 mA (and 6 mA through the 4 kΩ).

Practice 2What is the total resistance of 470 Ω in series with (1 kΩ ∥ 1 kΩ)?

Show the answer

Answer: 970 Ω

1 k ∥ 1 k = 500 Ω; + 470 = 970 Ω.

Challenge 3A meter movement reads full scale at 1 mA and has 50 Ω resistance. What shunt (in parallel) makes it read 1 A full scale? Answer in mΩ.

Hint

At full scale the movement has 1 mA × 50 Ω = 50 mV across it; the shunt carries the other 999 mA at that same voltage.

Show the answer

Answer: 50.05 mΩ

R = 0.05 V / 0.999 A = 0.05005 Ω = 50.05 mΩ.

Challenge 424 V feeds R1 = 2 kΩ in series with [R2 = 6 kΩ ∥ R3 = 3 kΩ], then R4 = 2 kΩ back to 0 V. What is the voltage across R3?

Hint

R2 ∥ R3 = 2 kΩ, so the loop is 2 + 2 + 2 kΩ.

Show the answer

Answer: 8 V

Rtotal = 6 kΩ, I = 24 / 6 = 4 mA. The parallel pair (2 kΩ) has 4 mA × 2 kΩ = 8 V across it - and so does R3.