K323 / 2027
D.C. circuits overview

Topic 3 of 6

Parallel circuits

Parallel branches share the same two endpoints, so they have the same potential difference. Their currents add to the total current entering the group.

Follow the connection points, not the shape of the drawing. Two branches are parallel only when each begins and ends at the same pair of nodes.

V1 = V2 = ... = VgroupEach branch spans the same two potentials. If the group is directly across the supply, Vgroup = Vs.

The entering current divides at a junction and recombines at the other. Charge is conserved, so in steady conditions the total current entering equals the sum of the branch currents.

Itotal = I1 + I2 + ...At equal branch p.d., a smaller resistance carries a larger current. The split need not be equal.

Each branch shares the same two endpoints

The source holds 6.0 V across the whole parallel network. Both finite-resistance branches carry current. Arrow lengths do not compare current magnitudes.

Reconnect the two resistors across the same two nodes

Reconnect the two resistors across the same two nodesAn ideal 6.0 V source connects to nodes A and C. A 2.0 ohm resistor and a 4.0 ohm resistor each connect from A to C, so each has 6.0 V. The upper branch carries 3.0 A and the lower branch 1.5 A, both from A towards C. The source current is 4.5 A. There is no wire directly bridging the two nodes.+-6.0 VAC2.0 ohm4.0 ohm4.5 A3.0 A; 6.0 V1.5 A; 6.0 VThe currents split and then recombine

The equivalent resistance is 4/3 ohm, about 1.33 ohm. This is a different connection from the 6.0 ohm series circuit.

The 2.0 Ω and 4.0 Ω resistors have been reconnected in parallel across 6.0 V. Each branch now has the full 6.0 V: their currents are 3.0 A and 1.5 A, giving 4.5 A in the main lead.

A changed circuit

The same two resistors, connected in parallel

With an ideal fixed 6.0 V source:

  • Current in 2.0 Ω = 6.0/2.0 = 3.0 A.
  • Current in 4.0 Ω = 6.0/4.0 = 1.5 A.
  • Total current = 3.0 + 1.5 = 4.5 A.

Both branches carry current. The equivalent resistance is Vs/Itotal = 6.0/4.5 = 4/3 Ω, approximately 1.33 Ω.

Find the effective resistance

Each branch current is the common p.d. divided by its own resistance. Adding them gives Itotal = V/R1 + V/R2 + ... . Since Itotal = V/Rparallel, dividing by the common nonzero p.d. gives:

1/Rparallel = 1/R1 + 1/R2 + ...Find the sum of reciprocals, then take its reciprocal to obtain resistance.

For exactly two resistors, this can be rearranged to Rparallel = R1R2/(R1 + R2). The general reciprocal relationship also works for three or more branches.

Three parallel branches

6 Ω, 12 Ω and 4 Ω

1/R = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 1/2, so R = 2 Ω.

Across 6 V, the branch currents are 1.0 A, 0.50 A and 1.5 A. Their sum is 3.0 A, which agrees with 6 V / 2 Ω.

For positive finite branch resistances, the parallel equivalent is smaller than the smallest branch resistance. The added path lets more total current flow at the same p.d.; it does not require all the current to choose only one path.

Open one branch

If these branches are connected directly across an ideal fixed-voltage source, opening one branch leaves the same p.d. across the remaining branches. Their currents stay the same, while total source current falls.

This prediction depends on the connection and supply condition. If another resistor is shared in series with the branches, their p.d. can change when a branch opens. The complete-circuit example shows why.

Optional check Fixed 3.0 ohm and 6.0 ohm resistors are connected directly in parallel across an ideal 6.0 V source. What current does the source supply?
Fixed 3.0 ohm and 6.0 ohm resistors are connected directly in parallel across an ideal 6.0 V source. What current does the source supply?