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Electricity and Magnetism

5% of exam

Electrical fundamentals and magnetic flux, DC circuit analysis, AC circuits with R, L, and C, and motors and generators.

4 concepts

A. Electrical fundamentals

Electrical Fundamentals: Charge, Fields and Magnetic Flux

Charge, current and voltage from first principles, electric fields and capacitance, magnetic flux and inductance, and the energy stored in each element.

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B. DC circuit analysis

DC Circuit Analysis: Ohm, Kirchhoff and Equivalent Resistance

Ohm's law, KVL and KCL, series-parallel reduction, voltage and current dividers, and power dissipation — the bookkeeping behind every DC problem.

A resistive DC circuit is the most predictable thing on the FE Mechanical exam: no phasors, no time constants, just three ideas — Ohm's law, Kirchhoff's two conservation laws, and the algebra of combining resistors. Points are lost not because the physics is hard but because of bookkeeping: a sign error in a loop, dividing by the wrong resistor in a divider, or confusing the current through a branch with the voltage across it. The NCEES FE Reference Handbook gathers the governing laws — Ohm's law, Kirchhoff's two laws, and the series/parallel resistance rules — across its Electrical and Computer Engineering chapter, and a mechanical candidate who can read those pages fluently can clear these questions in under two minutes each. Worth knowing up front: the voltage- and current-divider shortcuts this concept drills are ·derived· results, not handbook entries — on exam day you reconstruct them from series/parallel reduction plus Ohm's law, which is exactly the fluency this concept builds.

Ohm's law and the sign of a drop

Ohm's law states that the voltage across a resistor is proportional to the current through it,

V = IR

, with the constant

R

in ohms (

\Omega

). The subtlety the exam exploits is direction: current flows from high to low potential through a resistor, so the terminal the current enters is the

+

terminal of the drop. When you write a loop equation you must commit to one current direction per branch and let the algebra return a negative number if you guessed wrong — never flip a sign mid-solution. Conductance

G = 1/R

(siemens) is the same law rearranged,

I = GV

, and is handy when resistors sit in parallel.

V = IR \qquad I = GV,\ \ G = \tfrac{1}{R}

Kirchhoff's two laws

Kirchhoff's current law (KCL) is charge conservation at a node: the currents flowing in equal the currents flowing out,

\sum I_{in} = \sum I_{out}

. Kirchhoff's voltage law (KVL) is energy conservation around any closed loop: the voltage rises equal the voltage drops,

\sum V_{rise} = \sum V_{drop}

. Together they let you solve any circuit, however tangled, by writing one KCL equation per independent node and one KVL equation per independent loop. On the FE you rarely need the full system — most problems collapse to a single equivalent resistance and one application of Ohm's law — but you must recognize when a circuit is irreducible (a bridge, say) and fall back to node or loop equations.

\sum I_{in} = \sum I_{out} \qquad \sum V_{rise} = \sum V_{drop}

Series and parallel resistance

Resistors in series carry the same current and their resistances add:

R_S = R_1 + R_2 + \cdots + R_n

. Resistors in parallel share the same voltage and their conductances add, so the equivalent resistance is the reciprocal of the sum of reciprocals. The two-resistor parallel shortcut, product over sum, is worth memorizing because it appears constantly. A parallel equivalent is always smaller than the smallest branch resistance — a quick sanity check that catches arithmetic slips.

R_P = \left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots + \dfrac{1}{R_n}\right)^{-1}, \qquad R_P = \dfrac{R_1 R_2}{R_1 + R_2}\ (n=2)

Voltage and current dividers

Two patterns recur so often they deserve to be reflexes. In a voltage divider, two resistors in series split the source voltage in proportion to their own resistance: the voltage across

R_2

is

V_2 = V_s\,R_2/(R_1+R_2)

— the resistor you want sits in the numerator. In a current divider, two resistors in parallel split the incoming current in inverse proportion to their resistance: the current through

R_1

carries the ·other· resistor in its numerator,

I_1 = I_s\,R_2/(R_1+R_2)

. The crossed numerator is the single most common error on these problems — more current takes the easier path, the smaller resistor.

V_2 = V_s\,\dfrac{R_2}{R_1+R_2}, \qquad I_1 = I_s\,\dfrac{R_2}{R_1+R_2}

Power dissipation

A resistor converts electrical energy to heat at a rate given by three equivalent forms of one law:

P = VI = I^2R = V^2/R

. Choose the form whose two quantities you already know to avoid an extra step. Power is always positive in a resistor (it absorbs), and the total power delivered by the sources must equal the total dissipated in the resistors — an energy-balance check that should close to the watt. For a mechanical engineer this is the bridge to thermal load: the

I^2R

heat in a motor winding or a heater element is exactly this quantity.

P = VI = I^2 R = \dfrac{V^2}{R}

Thévenin equivalent and maximum power transfer

Any linear network of sources and resistors, viewed from two terminals, behaves like a single voltage source

V_{oc}

(the open-circuit voltage) in series with a single resistance

R_{eq}

(the resistance seen with all independent sources zeroed). This Thévenin equivalent collapses a whole subnetwork into two numbers, which is why it underlies every one-line and load-line analysis downstream. A load drawing power from such a network receives the maximum possible power when the load resistance matches the source,

R_L = R_{eq}

— at which point exactly half the source power reaches the load and the other half is lost internally.

P_{L,\max} = \dfrac{V_{oc}^{2}}{4R_{eq}} \quad\text{when}\quad R_L = R_{eq}

Exam strategy

Reduce before you solve. Collapse series and parallel groups into one

R_{eq}

, find the total current with one Ohm's-law division, then walk back out using dividers to get individual branch quantities. Sanity-check each step: a parallel equivalent must be smaller than its smallest branch; a divider output must be smaller than its input; the source power must equal the sum of

I^2R

terms. When the network will not reduce (a bridge, multiple sources fighting), switch to KCL at the nodes or KVL around the loops, fix one current direction per branch, and trust the signs. Watch units — milliamps and kilohms multiply to volts, a fact the exam uses to bury a factor of 1000.

Key equations

Ohm's law

V = IR

Voltage across a resistor equals current times resistance. $V$ in volts, $I$ in amperes, $R$ in ohms ( $\Omega$ ).

Kirchhoff's current law (KCL)

\sum I_{in} = \sum I_{out}

Charge conservation at a node: currents entering equal currents leaving. Use one per independent node.

Kirchhoff's voltage law (KVL)

\sum V_{rise} = \sum V_{drop}

Energy conservation around a closed loop: rises equal drops. Use one per independent loop.

Resistors in series

R_S = R_1 + R_2 + \cdots + R_n

Resistors in parallel

R_P = \left(\tfrac{1}{R_1} + \cdots + \tfrac{1}{R_n}\right)^{-1}, \quad R_P = \tfrac{R_1 R_2}{R_1+R_2}

Voltage divider

V_k = V_s\,\dfrac{R_k}{R_1+R_2+\cdots+R_n}

Current divider (two branches)

I_1 = I_s\,\dfrac{R_2}{R_1+R_2}

Power dissipated in a resistor

P = VI = I^2 R = \dfrac{V^2}{R}

Thévenin equivalent

V_{oc} = V_a - V_b, \qquad R_{eq} = R\ \text{seen with sources zeroed}

Maximum power transfer (DC)

P_{L,\max} = \dfrac{V_{oc}^2}{4R_{eq}}\ \ \text{at}\ \ R_L = R_{eq}

Worked examples

Ladder reduction, total current and a branch power

Problem. A $12\,\text{V}$ source drives $R_1 = 4\,\Omega$ in series with the parallel combination of $R_2 = 6\,\Omega$ and $R_3 = 12\,\Omega$ . Find the total current from the source, the voltage across the parallel pair, and the power dissipated in $R_1$ .

Solution. Reduce the parallel pair first:

R_{23} = \dfrac{R_2 R_3}{R_2+R_3} = \dfrac{6\cdot 12}{6+12} = \dfrac{72}{18} = 4\,\Omega

R_{eq} = 4 + \dfrac{6\cdot 12}{6+12} = 8\,\Omega, \quad I = \dfrac{12}{8} = 1.5\,\text{A}

Current divider into parallel branches

Problem. Using the same circuit, a total current of $1.5\,\text{A}$ enters the parallel combination of $R_2 = 6\,\Omega$ and $R_3 = 12\,\Omega$

Thévenin equivalent and maximum power transfer

Problem. A $24\,\text{V}$ source drives $R_1 = 6\,\Omega$ in series; the output is taken across $R_2 = 12\,\Omega$

Common pitfalls

•Crossing up the current divider: the current through $R_1$ carries $R_2$ in the numerator, not $R_1$ . More current takes the lower-resistance path — if your answer sends more current through the larger resistor, you flipped it.
•Putting the wrong resistor on top in a voltage divider: the resistor you want the voltage ACROSS goes in the numerator. (Dividers are opposite each other for this reason.)
•Forgetting that a parallel equivalent is always smaller than its smallest branch — a result larger than any branch resistance signals an arithmetic error.
•Sign errors in KVL: commit to one current direction per branch, write all drops consistently, and let a negative answer mean the real current is reversed. Do not flip a sign mid-loop.
•Mixing prefixes: $\text{mA}\times\text{k}\Omega = \text{V}$ , and $\text{V}/\text{k}\Omega = \text{mA}$ . The exam buries factors of 1000 in milliamps and kilohms.
•Assuming maximum power transfer means maximum efficiency: at $R_L = R_{eq}$ the load gets the most power but only half the total — efficiency there is 50%, not optimal.
•Using a single equivalent resistance on an irreducible network (e.g., a bridge): if no two resistors are cleanly in series or parallel, you must use node (KCL) or mesh (KVL) equations.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering
Nilsson & Riedel, Electric Circuits — ch. 2-4 background on Ohm/Kirchhoff and equivalent circuits

C. AC circuit analysis

AC Circuit Analysis: Phasors, Impedance and Complex Power

RMS values, phasors and impedance of R, L and C, series-parallel AC reduction, resonance, and the real, reactive and complex power triangle with power factor.

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D. Motors and generators

Motors, Generators and Transformers

The ideal transformer turns ratio, DC and AC machine basics, synchronous speed and slip, torque-speed behavior, and efficiency.

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Same current through each; resistances add. Equivalent is larger than any single resistor.

Same voltage across each; conductances add. Two-resistor form is product over sum; equivalent is smaller than the smallest branch.

Series string: the target resistor $R_k$ is in the numerator. Voltage splits in direct proportion to resistance.

Parallel pair: the OTHER resistor sits in the numerator. Current splits in inverse proportion to resistance.

Three equivalent forms; always positive (absorbed). $P$ in watts.

Collapses a two-terminal linear network to one source $V_{oc}$ in series with $R_{eq}$ . Zero voltage sources (short) and current sources (open).

Matched load receives maximum power; efficiency is only 50% at the match point.

(smaller than

6\,\Omega

, good). Equivalent resistance:

R_{eq} = R_1 + R_{23} = 4 + 4 = 8\,\Omega

. Total current by Ohm's law:

I = \dfrac{V_s}{R_{eq}} = \dfrac{12}{8} = 1.5\,\text{A}

. Voltage across the parallel pair:

V_{23} = I\,R_{23} = 1.5\times 4 = 6.0\,\text{V}

. Power in

R_1

:

P_1 = I^2 R_1 = (1.5)^2(4) = 9.00\,\text{W}

. Sanity check: source power

P_s = V_s I = 12\times 1.5 = 18\,\text{W}

; the parallel pair dissipates

V_{23}^2/R_{23} = 36/4 = 9\,\text{W}

, and

9 + 9 = 18\,\text{W}

closes the balance. Final:

I = 1.50\,\text{A}

,

V_{23} = 6.00\,\text{V}

,

P_1 = 9.00\,\text{W}

.

. Use the current-divider rule to find the current in each branch, and verify with Ohm's law.

Solution. Current-divider rule (the OTHER resistor in the numerator):

I_2 = I\,\dfrac{R_3}{R_2+R_3} = 1.5\times\dfrac{12}{6+12} = 1.5\times\dfrac{12}{18} = 1.00\,\text{A}

.

I_3 = I\,\dfrac{R_2}{R_2+R_3} = 1.5\times\dfrac{6}{18} = 0.500\,\text{A}

. Verify with the known parallel voltage

V_{23} = 6.0\,\text{V}

:

I_2 = 6/6 = 1.0\,\text{A}

and

I_3 = 6/12 = 0.5\,\text{A}

— agreement. Sanity check: branch currents sum to the input,

1.0 + 0.5 = 1.5\,\text{A}

(KCL satisfied), and the larger current flows through the smaller resistor, as it must. Final:

I_2 = 1.00\,\text{A}

,

I_3 = 0.500\,\text{A}

.

I_2 = I\,\dfrac{R_3}{R_2+R_3}, \qquad I_3 = I\,\dfrac{R_2}{R_2+R_3}

, which connects the output node to the source's return. Find the Thévenin equivalent (

V_{oc}

and

R_{eq}

) seen at the output terminals, the load resistance

R_L

that draws maximum power, and that maximum power.

Solution. Open-circuit voltage (no load,

R_1

and

R_2

form a voltage divider):

V_{oc} = V_s\,\dfrac{R_2}{R_1+R_2} = 24\times\dfrac{12}{6+12} = 24\times\dfrac{12}{18} = 16.0\,\text{V}

. Thévenin resistance (zero the source — replace the ideal voltage source with a short — then look back:

R_1

and

R_2

are now in parallel):

R_{eq} = \dfrac{R_1 R_2}{R_1+R_2} = \dfrac{6\times 12}{18} = 4.00\,\Omega

. Maximum power transfer requires a matched load:

R_L = R_{eq} = 4.00\,\Omega

. Maximum load power:

P_{L,\max} = \dfrac{V_{oc}^2}{4R_{eq}} = \dfrac{(16)^2}{4(4)} = \dfrac{256}{16} = 16.0\,\text{W}

. Sanity check: with

R_L = 4\,\Omega

attached to the real circuit, total resistance is

R_1 + (R_2\Vert R_L) = 6 + 3 = 9\,\Omega

, so

I = 24/9 = 2.667\,\text{A}

; the parallel voltage is

2.667\times 3 = 8.0\,\text{V}

, giving

I_L = 8/4 = 2.0\,\text{A}

and

P_L = (2.0)^2(4) = 16\,\text{W}

— matches, and the load voltage is exactly half

V_{oc}

, the signature of the match point. Final:

V_{oc} = 16.0\,\text{V}

,

R_{eq} = 4.00\,\Omega

,

R_L = 4.00\,\Omega

,

P_{L,\max} = 16.0\,\text{W}

.

V_{oc} = 24\,\dfrac{12}{18} = 16\,\text{V}, \quad R_{eq} = \dfrac{6\cdot12}{18} = 4\,\Omega, \quad P_{L,\max} = \dfrac{16^2}{4(4)} = 16\,\text{W}