“Adding a parallel resistor increases current” is incomplete. It usually refers to total current drawn from a voltage source. It does not automatically describe the current through every existing branch, especially when the battery’s terminal voltage changes under load.
A stronger solution starts with three questions: which components share two nodes, what does the source hold constant, and does the measuring instrument change the circuit? The AAMC outline includes circuits and their elements. The following examples are original, idealized teaching models.
Read connections rather than the shape of the drawing
Two resistors are in parallel when each connects between the same pair of nodes. They therefore share a potential difference. Components in an unbranched series path carry the same current. Physical proximity on a page does not establish either relationship.
For series resistances, add R values. For parallel resistances, add their reciprocals. These relationships are developed in OpenStax’s series-and-parallel treatment.
For example, two 6 Ω resistors connected between the same nodes have an equivalent resistance of 3 Ω. That equivalent represents the branch pair as seen from those nodes; it does not mean that either individual resistor has changed from 6 Ω to 3 Ω.
Assume throughout the examples that wires have negligible resistance, resistors are ohmic with constant values, and the circuit has reached a steady direct-current state. There are no capacitors or time-dependent charging effects.
Worked example: the battery’s terminal voltage changes
Represent a battery as an ideal 12 V emf source in series with a 2 Ω internal resistance. The emf, ε, is the source’s energy supplied per unit charge in this model. During discharge:
I = ε/(r + Rexternal)
Vterminal = ε − Ir = IRexternal.
This distinguishes emf from the potential difference available across the external load. OpenStax describes this internal-resistance model; real internal resistance can vary, so treating r as constant is an assumption.
Initially the external load is one 6 Ω resistor:
- Total resistance = 2 + 6 = 8 Ω.
- Current = 12/8 = 1.50 A.
- Terminal voltage = 12 − (1.50)(2) = 9.0 V.
Now connect a second 6 Ω resistor in parallel with the first. The external equivalent becomes 3 Ω, while the battery’s internal resistance remains in series with the entire branch pair.
- Total resistance = 2 + 3 = 5 Ω.
- Total current = 12/5 = 2.40 A.
- Terminal voltage = 12 − (2.40)(2) = 7.20 V.
- Current in each 6 Ω branch = 7.20/6 = 1.20 A.
The total current increased from 1.50 A to 2.40 A, but current through the original resistor decreased from 1.50 A to 1.20 A. The additional branch increases the internal voltage drop and lowers the voltage across both external branches.
If the source were ideal with r = 0, both branches would instead remain at 12 V, carrying 2.00 A each. The statement “parallel branches do not affect one another” therefore needs a specified source model.
Check the result through power accounting
With both branches connected, the ideal source supplies power εI = (12)(2.40) = 28.8 W. Internal resistance dissipates I²r = (2.40)²(2) = 11.52 W. The external pair receives VterminalI = (7.20)(2.40) = 17.28 W.
The two dissipation terms add to the source power. Each external resistor dissipates (1.20)²(6) = 8.64 W, half the external total.
This is a useful independent check because an incorrect branch current may still look plausible in isolation. If both 6 Ω branches were mistakenly assigned the full 2.40 A, current conservation would already fail at the junction, and the power accounting would fail as well.
Do not add the two branch currents when calculating a single branch’s power. Conversely, the internal resistor carries the total current, so its power calculation must use 2.40 A.
A meter is another circuit element
An ammeter measures current through a path and belongs in series with that path. Its ideal resistance is zero. A voltmeter measures the potential difference between two points and belongs in parallel between them; its ideal resistance is infinite. Finite meter resistance can alter the measured circuit. OpenStax explains these placements and idealizations.
Return to the single 6 Ω load and the 12 V battery with r = 2 Ω. Suppose a nonideal voltmeter has resistance 12 Ω and is connected across the load.
The load and meter are parallel: their equivalent is 4 Ω. Total current is now 12/(2 + 4) = 2.00 A, and the voltmeter reads 8.00 V. Without the meter, the load voltage was 9.0 V.
The meter did not merely report the previous circuit’s voltage inaccurately; it changed the circuit. The original resistor now carries 8/6 A, while the meter carries 8/12 A. Together they carry 2.00 A. A sufficiently high-resistance voltmeter would make this loading effect small.
Similarly, a real ammeter’s resistance adds to the series path. Inserting it across a component creates a different topology, so the “ammeter resistance is negligible” approximation cannot rescue an incorrect connection.
Transfer exercise: remove a branch
A different battery has ε = 9.0 V and r = 1.0 Ω. It supplies two parallel 4.0 Ω resistors. One branch is disconnected. Predict both total current and current in the remaining resistor.
Before disconnection, the external equivalent is 2.0 Ω. Total current is 9/(1 + 2) = 3.0 A, terminal voltage is 6.0 V, and each branch carries 1.5 A.
After disconnection, the external load is 4.0 Ω. Total current becomes 9/(1 + 4) = 1.8 A, with terminal voltage 7.2 V. The remaining resistor now carries that entire 1.8 A.
Thus total current decreases while current through the surviving branch increases. “Every current decreases because resistance increased” confuses total circuit current with a branch quantity. “The remaining branch takes all 3.0 A” incorrectly treats the battery as a constant-current source.
Use limiting cases to expose a wrong model
For a fixed ε and r, an external resistance approaching infinity gives current approaching zero and terminal voltage approaching ε. An external resistance approaching zero gives terminal voltage approaching zero and current approaching ε/r within this simplified model. It does not give infinite current when r is nonzero.
On your next circuit problem, label nodes, reduce only genuine series or parallel groups, and state whether the source is ideal. Then check one junction and one limiting case. Those checks make the reasoning inspectable before you commit to an answer.