DC Circuits: Ohm's Law, Series and Parallel Rules

What the rule requires

Current in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance NEETS Module 1 Ch. 3 — Ohm's Law. The flashlight case in the source makes both halves concrete: a 1.5-volt cell across a 5-ohm lamp gives 0.3 ampere. Fit a second cell so 3.0 volts is applied and current doubles to 0.6 ampere. Leave the voltage alone and double the lamp resistance instead, and current halves.

Three quantities, one relation, so any two known values give the third. The circle diagram is the standard aid — E on top, I and R beneath: cover the unknown and the two letters left uncovered show the operation. Cover I and you have E divided by R; cover E and you have I times R; cover R and you have E divided by I NEETS Module 1 Ch. 3 — Application of Ohm's Law. The source cautions against leaning on the diagram alone, and that caution is worth taking on an open-book exam where the distractors are all arithmetically plausible.

One ohm is the resistance through which one volt causes one ampere to flow NEETS Module 1 Ch. 1 — Summary ¶2. One ampere is one coulomb — 6.28 × 10¹⁸ electrons — past a point in one second.

Power

Power is the rate at which work is done, measured in watts, and the basic formula is P = E × I NEETS Module 1 Ch. 3 — Graphical Analysis of the Basic Circuit ¶1. Two derived forms carry most of the exam traffic: P = E²/R and P = I²R. With resistance held constant, power varies as the square of voltage — raise the applied voltage from 1 volt to 3 volts across a fixed 2-ohm load and power goes up ninefold, not threefold. The same square relation holds for current.

The old rhyme is still the quickest way to keep the two derived forms straight: twinkle, twinkle, little star, power equals I squared R; little star up in the sky, power equals E times I.

Series rules

A series circuit has one path for current NEETS Module 1 Ch. 3 — Summary of Characteristics.

The same current flows through every part of the circuit.
Total resistance is the sum of the individual resistances: RT = R1 + R2 + R3 … Rn NEETS Module 1 Ch. 3 — Total Circuit Resistance (Rt) Is Equal to the Sum of the Individual.
Total voltage equals the sum of the individual voltage drops, and that sum must equal the source voltage.
Each drop is proportional to that resistor's ohmic value — same current through all of them, so the largest resistor takes the largest drop.
Total power is the sum of the powers dissipated by each component.

Parallel rules

A parallel circuit has more than one current path connected to a common voltage source NEETS Module 1 Ch. 3 — Parallel Circuit Characteristics ¶1. The same voltage is present in each branch and equals the applied voltage: ES = ER1 = ER2. Source current divides among the paths, so IT = I1 + I2 … In.

Equivalent resistance is always less than the resistance of any branch NEETS Module 1 Ch. 3 — Parallel Circuit Characteristics ¶2. Three ways to get it, and picking the right one saves time:

All branches of equal value — divide one resistor's value by the number of branches. Four 40-ohm resistors in parallel give 10 ohms.
Exactly two unequal resistors — product over the sum. The source says commit this one to memory, and it earns that.
Three or more unequal resistors — the reciprocal method, 1/RT = 1/R1 + 1/R2 + 1/R3, then invert.

Power still adds. Dissipation is heat loss, so it is additive regardless of how the resistors are connected.

Two 90-ohm resistors and one 45-ohm resistor are wired in parallel across a 90-volt source. Before calculating anything, what is the widest range the total resistance can lie in?

Below 45 ohms — the equivalent resistance of a parallel circuit is always less than the smallest branch. Any answer option at or above 45 ohms is wrong on inspection, and on this circuit the reciprocal method gives 1/90 + 1/90 + 1/45 = 4/90, so RT = 22.5 ohms.</details>

Kirchhoff's laws

Kirchhoff's voltage law states that the algebraic sum of the voltage drops in any closed path and the emf's in that path is equal to zero NEETS Module 1 Ch. 3 — Kirchhoff's Voltage Law. It solves circuits that Ohm's law alone cannot — a loop with two sources in it, for instance.

The procedure is fixed: assume a direction of current, assign polarities to every resistor the current passes through, mark the sources, then trace the loop from any point and write each voltage with the polarity it has after the assumed current has passed through the component NEETS Module 1 Ch. 3 — Application of Kirchhoff's Voltage Law. Polarity assignment on a resistor is mechanical: the point where current enters is negative, the point where it leaves is positive.

Assume the direction wrong and the magnitude still comes out right — only the sign reverses, and a negative result means nothing more than that the assumed direction was backwards. Carry that negative sign into any further calculation.

Sources of emf driving current the same way are series aiding and their voltages add; sources opposing each other are series opposing and the effective source voltage is the difference, with current in the direction set by the larger source.

Kirchhoff's current law is the parallel-circuit counterpart: the algebraic sum of currents entering and leaving any junction is zero, currents entering counted positive and currents leaving negative.

Source resistance

Every source of emf has internal resistance, and it drops terminal voltage as soon as current flows NEETS Module 1 Ch. 3 — Source Resistance. In the source's example a battery reads 15 volts with the switch open; close it and 2 amperes through 1 ohm of internal resistance drops 2 volts inside the battery, leaving 13 volts at the terminals. Internal resistance cannot be measured directly with a meter — the attempt damages the meter.

Maximum power is transferred to the load when load resistance equals source resistance, and the efficiency of power transfer at that point is 50 percent NEETS Module 1 Ch. 3 — Summary ¶2.

Telling it apart

One criterion decides which rule set applies: how many paths current has from one source terminal to the other.

Series — one path. Current identical at every point; resistances add so RT is larger than any single resistor; source voltage divides. Most often misfiled when a candidate sees a schematic drawn as a neat box and assumes the two resistors along one side must be in parallel.
Parallel — more than one path, all connected to the same pair of points. Voltage identical across every branch; current divides; Req is smaller than the smallest branch. Most often misfiled when the branches are unequal and the candidate reaches for the equal-value shortcut anyway.
Combination — series and parallel elements in the same network, also called a series-parallel circuit NEETS Module 1 Ch. 3 — Series-Parallel Dc Circuits. No new law applies; you reduce it to a series circuit, then to one resistor.

When the drawing is ambiguous, redraw it: trace the current paths, label each junction where current divides, and remember that any unbroken wire is at the same voltage along its whole length until a component interrupts it NEETS Module 1 Ch. 3 — Redrawing Circuits for Clarity. A wire can be stretched or shrunk on paper without changing a single electrical characteristic. A schematic that looks like a simple parallel box frequently turns out to be series-parallel once the junctions are labelled.

Working a question

A 60-volt source feeds R1 = 8 ohms, which is in series with a parallel network of R2 = 20 ohms and R3 = 30 ohms. Find total resistance, total current, every voltage drop, the branch currents and the power NEETS Module 1 Ch. 3 — Solving Combination-Circuit Problems.

Type it. R2 and R3 connect to the same two points, so they are parallel. R1 carries everything that pair carries, so it is in series with the pair. With the values given, the only quantity computable at the outset is the equivalent resistance of R2 and R3 — start there and the rest unlocks in order.
Reduce the parallel pair. Two unequal resistors, so product over the sum: (20 × 30) ÷ (20 + 30) = 600 ÷ 50 = 12 ohms. The circuit is now 8 ohms in series with 12 ohms.
Total resistance. Series now, so add: RT = 8 + 12 = 20 ohms.
Total current. IT = ET ÷ RT = 60 ÷ 20 = 3 amperes. This is the current through R1, because R1 is the series element.
Voltage across R1. Use R1's own current and R1's own resistance: 3 × 8 = 24 volts.
Voltage across the parallel network. 60 − 24 = 36 volts. That 36 volts appears across R2 and across R3, since branch voltages in a parallel network are equal.
Branch currents. IR2 = 36 ÷ 20 = 1.8 amperes. IR3 = 36 ÷ 30 = 1.2 amperes. They sum to 3 amperes, which satisfies Kirchhoff's current law at the junction and confirms steps 2 through 6.
Power. PT = 60 × 3 = 180 watts. R1 dissipates 24 × 3 = 72 watts, R2 dissipates 36 × 1.8 = 64.8 watts, R3 dissipates 36 × 1.2 = 43.2 watts. Those three sum to 180 watts.

Steps 7 and 8 are the checks. If branch currents do not sum to total current, or component powers do not sum to total power, you have mixed a total value with a component value somewhere above.

Where candidates lose the point

Feeding a component value into a total-value formula. Asked for the drop across one resistor, the candidate divides source voltage by that resistor's ohms. The distractor built from that error is always on the sheet. Ohm's law demands that all three quantities come from the same part of the circuit NEETS Module 1 Ch. 3 — Series Circuit Analysis — to find a resistor's drop you need that resistor's current and that resistor's resistance.

Adding resistances in parallel. Three resistors of 120, 60 and 40 ohms in parallel invite the answer 220 ohms, and the correct answer is 20 ohms. Screen every parallel answer against the smallest branch before you compute.

Product over the sum applied to three branches. It is a two-resistor formula. Three or more unequal branches need the reciprocal method.

Prefix slips. 100 mA is 0.10 ampere, not 0.01 and not 10. Milli means one-thousandth, micro one-millionth, kilo one thousand, mega one million NEETS Module 1 Ch. 3 — Glossary ¶2. Convert before you calculate, not after.

Current direction inside the source. Electron current flows negative to positive through the external circuit, and inside the voltage source it flows from the positive terminal, through the source, emerging at the negative terminal NEETS Module 1 Ch. 3 — Series Dc Circuits. Questions are written to catch a candidate who applies the external convention throughout.

Choosing 95 percent for efficiency at maximum power transfer. It reads like the better engineering answer, and it is wrong. Matched load means half the power is burned in the source resistance: 50 percent.

Treating low voltage as safe. All live circuits are potential hazards. A fatal shock can occur from 0.1 ampere, and voltages as low as 30 volts have been recorded as producing enough current to be fatal. Electrical fires are extinguished with CO₂.

Check yourself

A series circuit of three resistors — 20, 60 and 80 ohms — carries 3 amperes. What is the total resistance and the source voltage?

160 ohms and 480 volts. Series resistances add: 20 + 60 + 80 = 160. Then ET = IT × RT = 3 × 160 = 480 volts. The same 3 amperes flows through each resistor, so the individual drops are 60, 180 and 240 volts — which also sum to 480.</details>

Three resistors are in parallel: R1 = 30 ohms, R2 = 15 ohms, R3 = 10 ohms. The current through R2 is 4 amperes. What is the source voltage?

60 volts. Work in R2's own branch: ER2 = 4 × 15 = 60 volts. Because every branch of a parallel circuit sees the same voltage as the source, that is the source voltage. Nothing about R1 or R3 is needed.</details>

You assume a direction of current, apply Kirchhoff's voltage law, and the answer comes out as −2 amperes. What does the negative sign tell you, and what do you carry forward?

The magnitude is correct and the assumed direction was wrong. Current is 2 amperes flowing the other way. If further calculations on the circuit use Kirchhoff's law, retain the negative sign in those calculations.</details>

Tracing a loop, you reach a resistor. How are its polarities assigned?

The point where the assumed current enters the resistor is negative; the point where it leaves is positive. Record each voltage with the polarity it has after the assumed current has passed through the component.</details>

A battery reads 15 volts across its terminals with the switch open. With the switch closed and 2 amperes flowing, the terminal voltage reads 13 volts. Why, and what is the internal resistance?

Internal resistance drops part of the emf as soon as current flows. Two volts lost at 2 amperes gives Ri = 2 ÷ 2 = 1 ohm. Do not attempt to measure that 1 ohm directly with a meter — it cannot be measured that way and the attempt damages the meter.</details>

Which of these is not a unit of conductance: siemens, S, G, or ohm?

The ohm. Conductance is the reciprocal of resistance and is expressed in mhos or siemens; the mho has been replaced by the siemens NEETS Module 1 Ch. 3 — Glossary ¶1. Resistance and conductance stand in a reciprocal relationship, not a direct one.</details>

A variable load resistor is adjusted until maximum power is drawn from a dc source with 4 ohms of internal resistance. What is the load resistance, and what is the efficiency of transfer?

4 ohms, and 50 percent. Maximum power transfer occurs when load resistance equals source resistance, and at that point half the total power is dissipated inside the source.</details>