Resistor circuits: follow branches and place meters correctly

MCAT content library / Chemistry & physics / Resistor circuits: follow branches and place meters correctly

The same resistor can carry a different current after another branch changes. Solve the connected circuit before deciding what a meter should read.

About 20 minutes, including the worked examples and checks.

Useful foundations

Electric fields and potential: keep the charge sign

On this page

Read connections before adding resistances
Separate the material from the component’s geometry
Worked application: reduce the inner pair first
Change the conductor and recalculate the connected network
Use measurements to test the assumed resistance
Keep current direction separate from the carrier’s identity
Check your reasoning
Continue learning
Sources

Read connections before adding resistances

A circuit drawing represents connections, not physical distance. Two resistors drawn side by side are parallel only if both ends connect to the same two electrical nodes. Two resistors are in series when the same current must pass through each with no branch between them. AAMC category 4C includes these networks, resistance and conductivity, and electrical measurements.¹

Current is the rate at which charge passes a cross section: I = ΔQ/Δt, measured in amperes, with 1 A =1C/s. Conventional current follows the direction positive charge would move. In a metal wire, electron drift is opposite that direction. This sign convention does not change when the circuit is redrawn or the zero-voltage reference is moved.²

Separate the material from the component’s geometry

For an ohmic component under conditions where its resistance is constant, ΔV = IR. The voltage is measured across that component and the current through it. An ohm is a volt per ampere. A linear current-versus-voltage graph through the origin has slope 1/R; a voltage-versus-current graph has slope R. The axis order determines which quantity the slope represents.

For a uniform conductor, R = ρL/A, where ρ is material resistivity, L is conducting length, and A is cross-sectional area. Resistivity has unitΩ·m and depends on material and conditions such as temperature. Resistance additionally depends on shape. Doubling length doubles R at unchanged ρ and A. Doubling a circular wire’s diameter quadruples A and reduces R to one quarter, provided the other conditions stay fixed.³

As an original geometry example, ρ =2.0×10⁻⁷Ω·m, L =0.50 m, and A =5.0×10⁻⁸m² give R =2.0 Ω. This is the series component used below. Length and area must be expressed in compatible units; a square-millimeter conversion requires squaring the length conversion.

Worked application: reduce the inner pair first

The circuit and measurements are original illustrative models. An ideal 12 V source supplies a2 Ω resistor followed by parallel 6 Ω and 3 Ω resistors. Wires have negligible resistance, and the resistors remain ohmic at fixed temperature.

Ideal source, negligible wire resistance, fixed-temperature ohmic resistors. Node B is the chosen 0 V reference. The 2 Ω component is in series with the entire pair; the 6 Ω and 3 Ω components connect across the same two nodes.

The parallel pair has 1/R_parallel =1/6 +1/3 =1/2, so R_parallel =2 Ω. It must be smaller than either individual branch resistance: a second conducting route increases total conductance. Adding the series 2 Ω gives R_total =4 Ω and source current I_total =12/4 =3 A.⁴

The series component carries all 3 A and drops(3)(2) =6 V. The remaining 6 V is across each parallel branch, giving I_{6 Ω} =6/6 =1 A and I_{3 Ω} =6/3 =2 A. At the junction,1 A +2 A =3 A. Charge does not disappear in the resistors: in steady state, the rates entering and leaving a node balance.

Component	Resistance (Ω)	Voltage across it (V)	Current through it (A)
Series component	2	6	3
Left branch	6	6	1
Right branch	3	6	2
Complete network	4 equivalent	12	3 supplied

Values in the original circuit

Neither branch receives the full 12 V because the series component uses part of the source potential difference. The parallel branches share voltage, not necessarily current. By contrast, the series component and the equivalent parallel pair share total current, not necessarily the same voltage in every possible circuit.

Change the conductor and recalculate the connected network

Now double only the length of the series conductor, preserving its material, temperature, and cross-sectional area. Its resistance becomes 4 Ω. The parallel pair remains 2 Ω, so total resistance becomes 6 Ω and source current falls to 2 A. The pair now receives(2)(2) =4 V: its 6 Ω branch carries 2/3 A and its 3 Ω branch 4/3 A.

The unchanged 6 Ω resistor now carries less current even though its own resistance did not change. Its applied voltage changed because another part of the connected network changed. “Constant resistance” is not “constant current.” Always identify what the source holds fixed and solve the new division of voltage.

Use measurements to test the assumed resistance

Voltage across tested component (V)	Current through it (A)
0	0
3	0.50
6	1.00
9	1.50

Original isolated-component data at fixed temperature

For this separate test, ΔV/I =6 Ω at every nonzero point. The current-versus-voltage slope is 1/6 A/V. A slope of 6 would correspond to the reversed axes and have unitsΩ. If measured current ceased increasing proportionally with voltage, the fixed-resistance model would need reconsideration; heating or non-ohmic behavior would be possible explanations, not conclusions established by the curve alone.

To measure branch current, place an ammeter in series within that branch. An ideal ammeter has zero resistance so that inserting it leaves the modeled branch unchanged. An ammeter before the junction measures total current instead. Connecting a low-resistance ammeter directly across two different-potential nodes creates an unintended conducting branch.⁵

To measure the voltage across a resistor, connect a voltmeter across its two endpoints. An ideal voltmeter has infinite resistance and draws no current. A finite-resistance meter draws some current and can alter the network it is measuring. Placing a very high-resistance voltmeter in series would nearly interrupt the branch rather than measure its intended operating voltage.⁵

Keep current direction separate from the carrier’s identity

Electrolyte solutions conduct through mobile ions rather than through electrons traveling freely across the solution as they do in a metal. Positive ions moving with the field and negative ions moving against it both contribute to conventional current in the same field-directed sense. The ability to carry charge depends on mobile charged species; it is not determined merely by whether a substance has dissolved.⁶

For a circuit question, finish with three checks: do branch currents add at the junction, do voltage changes close the loop, and would the proposed meter connection preserve the quantity being measured? Those checks connect a correct numerical answer to the physical circuit.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

An ideal 12 V source is connected to a 2 Ω resistor in series with a parallel pair of two 6 Ω resistors. Wires have negligible resistance and all resistors are ohmic. What current flows through each 6 Ω resistor?

A. 0.80 A

Review this choice. The new parallel pair is 3 Ω, so total resistance is 5 Ω. Dividing12 V by 5 Ω gives 2.4 A total, which splits equally into 1.2 A branches.

B. 1.20 A

Correct. Two6 Ω branches have equivalent resistance 3 Ω. Total current is 12/(2+3)=2.4 A, shared equally; each branch carries 1.2 A.

C. 2.40 A

Review this choice. 2.4 A is the total source current before it splits. Each equal branch carries half.

D. 4.00 A

Review this choice. This ignores the series resistor and misassigns the parallel-pair current. The entire circuit’s current is only 2.4 A.

The principle to carry forward

A branch replacement can reduce total current yet increase voltage across the remaining branches.

Check 2

A uniform circular wire is replaced by one made from the same material at the same temperature. The replacement has twice the length and half the diameter. By what factor does its resistance change?

A. 2

Review this choice. This includes only the doubled length. Halving diameter also reduces cross-sectional area.

B. 4

Review this choice. This accounts for the quartered area but omits the doubled length.

C. 8

Correct. Area scales with diameter squared, so A₂/A₁=1/4. R₂/R₁=(L₂/L₁)/(A₂/A₁)=2/(1/4)=8.

D. 16

Review this choice. The resistance relation contains area, not radius to the fourth power. That fourth-power dependence belongs to a different physical model, such as laminar tube-flow resistance.

The principle to carry forward

Use the geometry required by the stated law; do not transfer a radius exponent from a different system.

Check 3

An ideal 12 V source drives a 2 Ω series resistor followed by parallel 6 Ω and 3 Ω branches. Nodes A and B are the shared endpoints of those two branches. The goal is to measure only the 6 Ω branch current and the voltage across that resistor, with ideal meters. Which arrangement does so?

A. Connect the ammeter across nodes A and B; insert the voltmeter in the 6 Ω branch.

Review this choice. Both placements are reversed. The ammeter creates a shorting path while the series voltmeter interrupts the branch.

B. Insert both meters in series within the 6 Ω branch.

Review this choice. The ammeter placement is suitable, but an ideal series voltmeter has infinite resistance and stops branch current.

C. Insert the ammeter before the junction; connect the voltmeter across the 6 Ω resistor.

Review this choice. The voltage connection is correct. The ammeter before the junction reads total current 3 A, not the 6 Ω branch current 1 A.

D. Insert the ammeter in series within the 6 Ω branch; connect the voltmeter across that resistor.

Correct. The ammeter shares the target branch’s1 A current. The voltmeter spans its endpoints and measures 6 V without drawing current in the ideal model.

The principle to carry forward

“In series” must name the relevant branch, and “across” must name the relevant pair of endpoints.