Analyze series, parallel and mixed DC circuits for current, potential difference, equivalent resistance and power using the junction and loop rules

Learning goal: Explain current, potential difference and resistance in a DC circuit, use the junction rule (conservation of charge) and the loop rule (conservation of energy), apply ΔV = IR for ohmic resistors, find the equivalent resistance of series, parallel and mixed arrangements, calculate the current through, the potential difference across and the power dissipated in each resistor, and plan an investigation of how current depends on resistance at constant voltage.

What moves, and how fast the circuit responds. A direct-current circuit is a closed conducting loop containing a source such as a battery, connecting wires and devices with resistance. When the switch closes, the current and the potential difference appear almost instantaneously in every part of the circuit, even though the electrons themselves drift very slowly (in a typical household wire, a fraction of a millimeter per second; the exact speed depends on the current and the wire). It is the electric field that is established through all parts of the circuit at nearly the speed of light, and that field sets the electrons already present in every part of the wire in motion at once. Current I is the rate at which charge passes a point, in amperes (1 A = 1 C/s). Conventional current is drawn in the direction positive charge would move, from the positive terminal through the circuit to the negative terminal; the electrons drift the other way.

Two conservation rules. Charge is conserved, and in a steady circuit it does not build up anywhere, so at every junction (a branch point) the currents arriving add up to exactly the currents leaving: the junction rule. Because energy is conserved, around any closed loop the energy supplied per unit charge by the source equals the energy transformed per unit charge in the devices: the loop rule. The potential difference, or voltage ΔV, across a source is the potential energy difference it supplies per unit charge, ΔV = ΔE/q, in volts; across a resistor it is the energy per unit charge transformed to thermal energy (or light, or motion) as the charge passes through. For resistors in series around one loop, V_battery = ΔV₁ + ΔV₂ + ΔV₃ + ...; the drops add up to the source voltage.

Ohmic resistors. Resistance R is measured in ohms (Ω). For an ohmic resistor (one whose resistance stays the same as the current changes, which for a metal also means its temperature stays roughly steady), the potential difference across it is the product of the current through it and its resistance: ΔV = IR. Plot ΔV against I and the graph is a straight line through the origin whose slope is R. Only ohmic resistors are studied in this course; a lamp filament, whose resistance rises as it heats, is not one. The equation is understood conceptually, as "energy per charge needed to push a given current through this resistance", and used to find the current or the potential difference anywhere in a circuit; the name attached to it is not the point.

Series, parallel and equivalent resistance. Resistors in series carry the same current, and their resistances add: R_eq = R₁ + R₂ + ...; the source voltage divides among them in proportion to their resistances. Resistors in parallel have the same potential difference across them, the current divides among them (more through the smaller resistance), and 1/R_eq = 1/R₁ + 1/R₂ + ...; the equivalent resistance is always less than the smallest branch, because adding a branch adds a path. A mixed circuit is reduced step by step: replace each parallel group by its equivalent, add the series pieces, find the total current, then work back outward to find the current and voltage for each resistor. Worked model: a 12 V battery drives a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω (the diagram). Parallel pair: 1/R = 1/6 + 1/3 = 1/2, so 2 Ω. Total: 4 + 2 = 6 Ω. Current from the battery: I = 12/6 = 2.0 A. Across the 4 Ω: ΔV₁ = (2.0)(4) = 8 V. Loop rule: the parallel pair must take the remaining 12 − 8 = 4 V. Branch currents: 4/6 = 0.67 A through the 6 Ω and 4/3 = 1.33 A through the 3 Ω; junction rule check: 0.67 + 1.33 = 2.0 A, the current that arrived. Every number has been checked by a conservation rule, which is the habit to build.

Power in a circuit. The rate at which energy is transformed in a resistor is P = ΔV I, in watts, and since ΔV = IR this is also I²R or ΔV²/R. The battery's power output equals the sum of the resistors' powers, each found with that resistor's own current: P_battery = I₁ΔV₁ + I₂ΔV₂ + ... = I_battery V_battery, another statement of energy conservation (only for resistors in series is the same I shared by every term). In the worked model: P₁ = (8)(2.0) = 16 W, P₂ = (4)(0.67) = 2.7 W, P₃ = (4)(1.33) = 5.3 W, total 24 W = (12)(2.0) W. In series the largest resistance dissipates the most power (same current, larger ΔV); in parallel the smallest resistance does (same ΔV, larger current), which is why, across the same household voltage, a heater element of lower resistance delivers more power (P = ΔV²/R).

Misconception: current is used up. The current is the same on both sides of a resistor; what the resistor uses is energy per charge, which shows up as the potential difference across it. Another common error is to treat two lamps drawn side by side as parallel: only the connections decide, and a lamp is in parallel with another only if both ends of each connect to the same two junctions. Strings of lights can be designed in several ways: a plain series string goes dark when one bulb fails because the loop is broken; a bulb could carry a shunt, a path that conducts once its filament burns out, so the rest stay lit; and a string built as several series sections connected in parallel loses only one section when one lamp fails open.

Investigation: current against resistance at constant voltage. Materials: a 6 V battery pack or a low-voltage regulated supply, resistors of 10 Ω, 22 Ω, 47 Ω and 100 Ω rated at least 2 W (the 10 Ω resistor dissipates 3.6 W at 6 V, so use a 5 W part for it; keep the same supply voltage for every resistor), a multimeter set to measure current, connecting leads, a switch. Steps: draw the circuit diagram first, showing the ammeter in series; connect one resistor, close the switch, read the current, open the switch; repeat for each resistor; measure the actual supply voltage across each resistor while it is running. Plot I against R and I against 1/R. A result that supports the claim: with a regulated supply that holds 6 V, the I vs. R graph curves downward and the I vs. 1/R graph is a straight line through the origin with slope equal to the supply voltage (predicted currents 0.60, 0.27, 0.13 and 0.060 A). Account for unanticipated effects: the terminal voltage of a real battery drops under load because of its internal resistance, so with a battery pack the voltage is not truly constant and the points for the small resistors fall below that line. Measuring the voltage across each resistor does not make it constant, but it lets you allow for the change: plot I ÷ V (the current per volt actually applied) against 1/R, which should be a straight line through the origin with slope 1, or equivalently check that V/I matches each R. The resistance of the leads is neglected because it is far below 10 Ω; a voltmeter is assumed to draw negligible current. Never connect the ammeter directly across the supply, never short the terminals, and stay at or below 9 V.

Worked example 1 — series. A 9.0 V battery with 2.0 Ω and 4.0 Ω in series. R_eq = 6.0 Ω, I = 9.0/6.0 = 1.5 A through both. ΔV₂ = (1.5)(2.0) = 3.0 V, ΔV₄ = (1.5)(4.0) = 6.0 V; loop check 3.0 + 6.0 = 9.0 V. Power: 4.5 W and 9.0 W, total 13.5 W = (9.0)(1.5) W. The larger resistor takes the larger share of the voltage and the power.

Worked example 2 — parallel. 6.0 V across 3.0 Ω and 6.0 Ω in parallel. Each branch sees 6.0 V: I₃ = 2.0 A, I₆ = 1.0 A; total from the source 3.0 A (junction rule). R_eq = 6.0/3.0 = 2.0 Ω, which 1/R = 1/3 + 1/6 = 1/2 confirms. Power: 12 W in the 3 Ω, 6 W in the 6 Ω; the smaller resistance dissipates more because it carries more current at the same voltage.

Worked example 3 — mixed. 24 V battery, 2.0 Ω in series with a parallel pair of 4.0 Ω and 12 Ω. Parallel: 1/R = 1/4 + 1/12 = 1/3, so 3.0 Ω; total 5.0 Ω; I = 24/5.0 = 4.8 A. ΔV across the 2.0 Ω = 9.6 V, leaving 14.4 V across the pair. Branch currents 14.4/4.0 = 3.6 A and 14.4/12 = 1.2 A, summing to 4.8 A. Powers: (4.8)²(2.0) = 46.1 W, (14.4)²/4.0 = 51.8 W, (14.4)²/12 = 17.3 W, total 115.2 W = (24)(4.8) W.

Worked example 4 — reading resistance from a graph. A resistor's ΔV vs. I graph passes through (0.50 A, 6.0 V) and (1.0 A, 12 V). Slope = (12 − 6.0)/(1.0 − 0.50) = 12 Ω, and the line passes through the origin, so the resistor is ohmic with R = 12 Ω. At 0.25 A it would need 3.0 V.

Worked example 5 — a real battery under load. A battery with a 9.0 V open-circuit voltage and 0.50 Ω internal resistance drives a 4.0 Ω resistor. Total resistance 4.5 Ω, I = 2.0 A, and the terminal voltage is only (2.0)(4.0) = 8.0 V: 1.0 V is dropped inside the battery. Assuming 9.0 V across the load would overestimate the current by 12.5 % (9.0/4.0 = 2.25 A instead of 2.0 A), which is why the investigation measures the voltage across each resistor.

Representation task: reduce and label a circuit. Draw a 18 V battery with a 3.0 Ω resistor in series with a parallel pair of 6.0 Ω and 6.0 Ω. Redraw it in two stages of reduction (pair → 3.0 Ω; then 6.0 Ω total), and on the original diagram label the current through and the potential difference across every resistor (3.0 A; 9.0 V across the 3.0 Ω; 9.0 V across the pair with 1.5 A in each branch) and the power in each (27 W, 13.5 W, 13.5 W). Write the junction-rule and loop-rule checks as equations with numbers.

Independent performance: the current-resistance investigation. Plan, draw and carry out the constant-voltage investigation and write it up. Success criteria: a labeled circuit diagram checked for safety before power is applied, with the ammeter in series and the switch shown; a table of resistance, measured voltage across the resistor, measured current and the calculated ratio V/I; graphs of I vs. R and I vs. 1/R with the slope of the second compared with the supply voltage (if the measured voltage changed from resistor to resistor, plot I/V vs. 1/R as well and compare its slope with 1); a claim about how current depends on resistance at constant voltage, supported by the graphs; an explanation of any drop in terminal voltage in terms of the battery's internal resistance; a statement of two assumptions (lead resistance neglected, voltmeter draws negligible current) and why they are reasonable; and a solved mixed-circuit problem of your own design with all currents, voltages and powers checked by the junction and loop rules.