Follow current and voltage through a series path

One path constrains the current

Trace the series diagram from the positive source terminal, through R1 and R2, and back to the negative terminal. There is no branch between the resistors. In steady state, charge does not continually accumulate between them, so the same current passes through both. A resistor transfers electrical energy; it does not consume part of the current.

Set source voltage to 6 V, R1 = 100 Ω and R2 = 200 Ω. Predict which resistor has the larger voltage drop before reading the ledger. The displayed current through each is 20 mA. The drops are 2 V across R1 and 4 V across R2.

Derive the equivalent resistance

The source’s 6 V rise is balanced by resistor drops around the loop. Write V source = I R1 + I R2 = I(R1 + R2). The sum R1 + R2 is the equivalent resistance of this series pair. Here it is 300 Ω, so I = 6/300 = 0.020 A. Multiplying that common current by each resistance gives the two drops.

The intermediate node is 4 V above the negative terminal in this drawing: after traversing R1 from the positive terminal, potential has fallen from 6 V to 4 V; R2 completes the fall to 0 V. The return wire has no voltage drop in the ideal-wire model. These are relative potentials, not absolute charge quantities.

Use a controlled comparison

Hold source voltage at 6 V and compare three settings:

R2	Series current	R1 drop	R2 drop
100 Ω	30 mA	3 V	3 V
200 Ω	20 mA	2 V	4 V
500 Ω	10 mA	1 V	5 V

Capture two cases and explain both changes. Increasing R2 increases the total resistance and reduces the common current. Despite that reduction, R2 receives a larger share of the fixed source voltage. R1 receives a smaller share. The two drops still sum to 6 V.

The divider expression V across R2 = V source × R2/(R1 + R2) makes that share explicit. This expression assumes no additional load connected across R2. Adding another component there changes the network and can change the voltage; an unloaded divider calculation is not automatically a regulated output.

Diagnose a proposed measurement

A report claims a 6 V source drives these series resistors with 20 mA through R1 and 10 mA through R2 in steady state. Explain why the report cannot describe this ideal one-path circuit. Name a possible difference between the diagram and a real setup, rather than silently changing the rule.

Check the series diagnosis

Different steady currents would require charge accumulation or another current path. A misplaced connection, an unreported branch or a measurement error could explain the report. The ideal drawing has none of those. A voltmeter measures between two points, whereas current measurement requires insertion into the relevant path; the browser does not model meter loading or misconnection.

For an unfamiliar 9 V source with 100 Ω and 500 Ω in series, calculate the current and both drops without changing the source beyond the allowed slider steps.

Check the series transfer

Equivalent resistance is 600 Ω; current is 9/600 = 0.015 A = 15 mA. Drops are 1.5 V and 7.5 V, summing to 9 V.

OpenStax: Kirchhoff’s rules connects steady junction balance and loop voltage accounting. The three-case comparison and diagnostic task here are original exercises.