Circuit models: follow nodes, current and voltage

Connections determine the relationship

Two resistors are in series when the same current must pass through them along an unbranched path. Two resistors are in parallel when both ends connect to the same pair of nodes, so they have the same voltage across them. Their appearance on a page does not determine the connection. A text description naming the nodes or branches can be enough. For an ideal ohmic resistor at fixed temperature, voltage, current and resistance satisfy V = IR. Use that relationship on the particular component or equivalent group whose voltage and resistance are known.

Reduce a network, then expand the solution

Series resistances add because the voltage drops add for a common current. Parallel conductances add: 1/R_eq is the sum of the reciprocals of the branch resistances. For positive finite resistors, a parallel equivalent is smaller than the smallest individual resistance. Once a network has been reduced, find the total current from the source voltage and total resistance. Then work backward through the reductions to find group voltages and branch currents. Applying the source voltage to every resistor is valid only when each resistor actually spans the source's two nodes.

Worked example 1: one unbranched path

An ideal 18 V source supplies 3-ohm and 6-ohm resistors in series. Find the current, each voltage drop and the total power. The source and resistors form one complete path, with no additional branches or resistance. Because both resistors carry the same current, the larger resistor must have the larger voltage drop.

• Add the resistances: R_total = 3 + 6 = 9 ohms. The current is I = 18/9 = 2 A.
• Use each component's resistance: the drops are 2 × 3 = 6 V and 2 × 6 = 12 V.
• Check the loop: 6 + 12 = 18 V, matching the ideal source.
• Power is VI: the resistors dissipate 6 × 2 = 12 W and 12 × 2 = 24 W. Their sum is 36 W, matching source power 18 × 2.

Worked example 2: a series resistor feeds two branches

An ideal 24 V source supplies a 4-ohm resistor in series with a parallel pair of 12 ohms and 6 ohms. Find the current in each branch. The 4-ohm resistor carries all the source current before it splits. The two branches share the voltage across their parallel group, which is less than the source voltage because the series resistor has its own drop.

• Reduce the parallel pair: 1/R_parallel = 1/12 + 1/6 = 1/4, so R_parallel = 4 ohms.
• The total resistance is 4 + 4 = 8 ohms, giving source current 24/8 = 3 A.
• The series resistor drops 3 × 4 = 12 V. The parallel pair therefore has the remaining 12 V across it.
• Branch currents are 12/12 = 1 A and 12/6 = 2 A. Their sum is 3 A, matching the current entering the junction.
• The source supplies 72 W. The series resistor uses 36 W and the branches use 12 W and 24 W, again totaling 72 W.

Use a qualitative check before a calculator

Adding a parallel path to an ideal voltage source lowers the equivalent resistance of that group and increases the group's total current. That does not mean every individual branch current must increase in every larger network: a series element can change the voltage available to the group. Keep the boundary clear. Ask which current is total and which is a branch current, and which voltage belongs to the source or a component. These distinctions often expose a wrong answer before you perform a second full calculation.

Misconception: current is used up by a resistor

In the steady-state ideal model, charge does not accumulate in a resistor, so its entering and leaving currents are equal. The resistor transfers electrical energy, producing a voltage drop; it does not consume charge. At a junction, current divides among branches and recombines. Also keep power and energy separate: watts measure the rate of energy transfer, while multiplying constant power by elapsed time gives energy.

Sources & further reading