Series resistors

R1 and R2 share one intermediate node. Because no third branch is connected to that node, Kirchhoff's current law (KCL) leaves only one possible steady branch current magnitude through the two resistors. KCL is the standard abbreviation for Kirchhoff's Current Law.

This is the useful operational meaning of a true unbranched series connection. If a third branch were attached to the intermediate node, current could split and the two resistor currents would no longer be forced equal.

Let I denote the signed current defined by the arrow in Figure 1.2. Let V₁ and V₂ denote the signed voltages across R1 and R2 shown in Figure 1.3. In the receiver/passive convention, positive current enters the positive voltage terminal of each resistor. Ohm's law therefore gives

Kirchhoff's voltage law (KVL) across the complete two-resistor string says that its terminal voltage is the sum of those two resistor voltages:

Substitute the element laws:

The subscript eq means equivalent. We write R_eq for the single resistance seen from the two external terminals. An equivalent resistor must reproduce the same terminal relation V=R_eqI for every operating point allowed by this linear network. Therefore

The formula is a consequence of three earlier ideas: no current branching at the internal node, KVL for the total voltage, and the resistor constitutive law.

Let N denote the number of resistors in the string. If those N resistors lie on one unbranched path, the same current I passes through all of them:

Using the index k to label one resistor in the list, with V_k=R_kI,

Hence

The symbol Σ means “sum.” The limits k=1 and N mean: add R₁, then R₂, and continue through R_N. This compact notation is only a shorter way to write the expanded sum above.

Because ordinary passive resistances are positive, adding another positive series resistance must increase the equivalent resistance. This gives an immediate sanity check:

The verified circuit in Figure 1.1 contains R1=1 kΩ and R2=2 kΩ in series across an ideal 12 V source. Here kΩ means kilo-ohm: 1 kΩ=1000 Ω. Likewise, mA means milliampere: 1 mA=10⁻³ A.

Step 1 — equivalent resistance.

Step 2 — common current.

Step 3 — individual resistor voltages.

Step 4 — KVL closure.

The larger resistor takes the larger voltage drop because the same current flows through both. Chapter 03 will turn that observation into the voltage-divider formula; for now it is simply Ohm's law applied twice.

Let P₁ and P₂ denote the powers absorbed by R1 and R2. Each passive resistor absorbs power:

Power is written here in milliwatts (mW), where 1 mW=10⁻³ W. Let P_source denote the source power. Under the receiver/sign convention already established, a negative P_source means the source delivers power. The source supplies

Therefore

The equivalent resistor also dissipates I²R_eq=(4 mA)²(3 kΩ)=48 mW, exactly matching the total of the original string. The equivalent preserves terminal V–I behaviour and therefore terminal power, even though it hides how the internal power is distributed.

Replacing R1+R2 by R_eq is valid when the rest of the circuit interacts only through the two external terminals of the string. The replacement reproduces the same total terminal current for the same applied terminal voltage.

It does not preserve internal information such as the midpoint voltage, individual resistor dissipation, component tolerances or maximum voltage across each part. If the midpoint is used elsewhere, collapsing the two resistors into one destroys a port that matters to the larger circuit.

Engineers sometimes place resistors in series to synthesize a nonstandard resistance, distribute dissipation, or share voltage stress. The ideal equivalent remains the sum, but real design must check each component.

Tolerance example. R1=1.0 kΩ ±1% and R2=2.0 kΩ ±1% give nominal R_eq=3.0 kΩ. Worst-case bounds are

In this particular equal-percentage case the series sum is also ±1%, but in general mixed tolerances must be propagated from the actual component limits.

Likewise, the total string may dissipate safely while one resistor exceeds its individual power or voltage rating. The ideal network formula cannot replace component datasheets.

• R1 and R2 share a node, but a load also connects to that node: current can split, so they are not a simple series pair.
• Two resistors happen to carry equal current because of symmetry or a special operating point: equal current alone does not establish a topological series connection.
• Two resistor symbols are drawn in one horizontal line, but a junction branches between them: appearance does not override connectivity.

The durable test is topological: is the shared node unbranched for the elements being reduced?

1. Find the equivalent resistance of 330 Ω, 1.0 kΩ and 4.7 kΩ in series.
2. A 15 V source drives 1 kΩ and 4 kΩ in series. Find the current and both resistor voltages.
3. For problem 2, calculate both resistor powers and verify total power balance.
4. R1 and R2 share a node, but a third resistor also leaves that node. Can you replace R1+R2 by one series equivalent? Explain without using formulas.
5. A 2 kΩ resistor is placed in series with a resistor R. The measured two-terminal equivalent is 7 kΩ. Find R and state the model assumption behind the subtraction.

The next chapter will invert the structure: resistors connected across the same two nodes share a voltage instead of a current. KCL will then produce the parallel-resistance rule.