Electronics Information: mixed review

By gaining or losing electrons. Extra electrons give a net negative charge; missing electrons leave a net positive charge. Protons do not move. Like charges repel, unlike charges attract, and total charge is conserved.

$I=\frac{Q}{t}$: current in amperes equals charge in coulombs divided by time in seconds. 1 A = 1 C/s. Example: 30 C in 6 s is 5 A.

$V=\frac{E}{Q}$: potential difference in volts equals energy in joules per coulomb of charge. 1 V = 1 J/C. A source giving 54 J to 6 C is 9 V.

Conventional current flows from the positive terminal through the load to the negative terminal. Electrons in a metal wire drift the opposite way, negative to positive. Use conventional current unless electron flow is named.

Ammeter: in series, inside the loop, so the current passes through it (nearly zero resistance). Voltmeter: across the component, in parallel (very high resistance). An ammeter across a battery acts as a near short circuit.

Milli (m) divide by 1,000: 750 mA = 0.75 A. Micro ($\mu$) divide by 1,000,000: 40 $\mu$A = 0.00004 A. Kilo (k) multiply by 1,000: 4.7 k$\Omega$ = 4,700 $\Omega$.

No, it is charge: 1 Ah = 1 A for 3,600 s = 3,600 C, so 2 Ah = 7,200 C. Energy also requires the voltage; equal Ah at 12 V holds eight times the energy of 1.5 V.

No. In an unbranched loop the current is the same at every point; the lamp converts energy, not charge. Opening a switch stops current everywhere, but the battery voltage still appears across the gap.

$R=\frac{\rho L}{A}$: resistance ($\Omega$) equals resistivity $\rho$ ($\Omega \cdot \text{m}$) times length L (m) divided by cross-sectional area A (${\text{m}}^2$). R rises with length and falls as area grows.

Resistivity ($\rho$, in $\Omega \cdot \text{m}$) is a property of the material alone. Resistance (R, in ohms) belongs to a specific object and also depends on its length and cross-sectional area.

Area depends on diameter squared ($A=\frac{\pi d^2}{4}$), so area drops to 1/4 and resistance becomes 4 times larger, not 2 times. Always square the diameter ratio.

Metals gain resistance when heated (copper about 0.4 percent per °C) because electrons scatter more. Semiconductors, carbon and NTC thermistors lose resistance as they warm (negative coefficient).

It depends on what is fixed. At the same voltage, $P=\frac{V^2}{R}$, so the smaller R dissipates more. At the same current, $P=I^{2}R$, so the larger R dissipates more.

Only on a de-energized circuit, with the part isolated (at least one lead disconnected), since the meter supplies its own current. Parallel paths, including your fingers on both probes, make the reading too low.

Thicker. A larger gauge number means a thinner wire, so 12 AWG has less resistance per foot and can carry more current than 14 AWG.

Open: a break in the path with effectively infinite resistance, so no current flows (the source voltage can appear across the gap). Short: an unintended very low-resistance path bypassing the load, causing very large current and heat.

$V=IR$, so $I=\frac{V}{R}$ and $R=\frac{V}{I}$. V is volts across the part, I is amperes through it, R is ohms. Example: 36 V across 9 $\Omega$ gives 4 A.

R is V divided by I, not I divided by V (12 V and 3 A gives 4 $\Omega$, not 0.25). Ohm's law is a product, never a sum: 6 V and 2 $\Omega$ never combine to 8.

Volts divided by kilohms gives milliamperes: 12 V / 4 k$\Omega$ = 3 mA. Volts divided by megohms gives microamperes: 5 V / 1 M$\Omega$ = 5 $\mu$A. Otherwise convert to base units first.

Current is proportional to voltage, so it doubles. Power goes with the square ($P=\frac{V^2}{R}$), so it becomes 4 times larger. 25 $\Omega$ at 10 V uses 4 W; at 20 V it uses 16 W.

Series parts share the same current, and their resistances add ($R_{eq}=R_1+R_2$). Parallel branches share the same voltage; branch currents add, and the equivalent resistance is smaller than any branch.

In proportion to resistance, because both carry the same current. 4 $\Omega$ and 8 $\Omega$ across 24 V: I = 2 A, drops of 8 V and 16 V. Equal drops occur only with equal resistors.

With V vertical and I horizontal, the slope is R (a steeper line means more resistance). With I vertical and V horizontal, the slope is $\frac{1}{R}$ in siemens (a steeper line means less resistance).

Nonohmic parts: a tungsten lamp filament (resistance rises a lot when hot, so switch-on current surges) and a diode (conducts one way after about 0.6 to 0.7 V for silicon). Their V-I graphs curve.

$P=VI$: power in watts equals volts times amperes. Rearranged, $I=\frac{P}{V}$ and $V=\frac{P}{I}$. A 1,200 W microwave on 120 V draws 10 A.

$P=I^{2}R$ (use with current and resistance) and $P=\frac{V^2}{R}$ (use with voltage and resistance). 3 A through 4 $\Omega$ gives 36 W; 6 V across 3 $\Omega$ gives 12 W.

$E=P\times t$. Watts x seconds = joules; watts x hours = watt-hours; kilowatts x hours = kilowatt-hours. A 1.5 kW heater for 2 hours uses 3 kWh. 1 Wh = 3,600 J.

Watts measure a rate (power, joules per second). Watt-hours measure an amount (energy). If time is involved and the question asks how much was used, the answer is in Wh, kWh or J.

Convert to kWh (watts / 1,000 x hours), then multiply by the price per kWh. An 1,800 W hair dryer used 20 minutes a day for 30 days uses 18 kWh, costing $2.70 at 15 cents per kWh.

$\eta =\frac{P_{out}}{P_{in}}\times 100\%$. Lost power = input minus useful output. A charger drawing 50 W and delivering 40 W is 80 percent efficient and loses 10 W as heat. Never above 100 percent.

Energy (Wh) = voltage x capacity (Ah). Runtime (h) = Wh / load watts. A 12 V, 10 Ah battery holds 120 Wh, running a 30 W load about 4 hours. Ah alone is charge, not energy.

It becomes 4 times larger, because $P=I^{2}R$. But at fixed voltage, doubling current only doubles power ($P=VI$). Only current or voltage is ever squared, never resistance.

It is identical everywhere: $I_T=I_1=I_2=I_3$. There is only one path, so resistors lower voltage, never the current passing through them.

Add them: $R_T=R_1+R_2+R_3+\dots$ (ohms). The total is always larger than the largest single resistor, e.g. 10, 22 and 68 $\Omega$ give 100 $\Omega$.

The drops add up to the source voltage: $V_S=V_1+V_2+V_3$, i.e. $V_S-V_1-V_2-V_3=0$. A missing drop is the source minus the known drops.

$V_n=V_S\times \frac{R_n}{R_T}$. The drop is the source voltage times that resistor's share of total resistance; 15 V on 10 $\Omega$ + 20 $\Omega$ puts 10 V on the 20 $\Omega$.

The largest resistor. With a common current, $V_n=I{R}_n$ and $P_n=I^2{R}_n$, so both drop and power are proportional to resistance.

Add resistances to get $R_T$, find $I=V_S/R_T$, find each drop $V_n=I{R}_n$, then check that the drops sum to the source voltage.

An open (broken filament, blown fuse, open switch) stops current everywhere and puts the full source voltage across the gap. A short across one resistor removes it from the total, so current rises.

Amps go in line, volts go across. An ammeter (near-zero resistance) is inserted in series in the current path; a voltmeter (very high resistance) connects in parallel across the part.

Both ends share the same two nodes: one end of each on node A, the other end of each on node B. Drawing position and equal values do not matter.

Every branch gets the same full voltage, and the current divides: $I_T=I_1+I_2+I_3+\dots$. Each branch current is $I_n=V/R_n$.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots$, then flip the sum. Forgetting the final flip leaves an answer in siemens, not ohms.

$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$. Example: 20 $\Omega$ and 30 $\Omega$ give 600/50 = 12 $\Omega$. For n equal resistors use $R/n$.

It is always smaller than the smallest branch. Adding any branch lowers it and raises total current; removing or opening a branch raises it.

Inversely to resistance: $I_1=I_T\times \frac{R_2}{R_1+R_2}$. Example: 2 $\Omega$ and 8 $\Omega$ sharing 10 A carry 8 A and 2 A. The smallest resistor carries the most current.

The smallest one. With the same voltage on every branch, $P_n=V^2/R_n$, so power is inversely proportional to resistance. Branch powers add to $P_T=V{I}_T$.

An open branch only loses its own current; with an ideal source the others keep the same voltage and current. A short across the rails shorts every branch and the source, drawing very large current.

AC periodically reverses direction (the signed value crosses zero to the opposite sign). DC flows one way only; it may be steady or vary in size, and still be DC.

Current that rises and falls (often touching zero) but never reverses, such as rectifier output. It changes size but keeps one direction, so it is DC.

$f=\frac{1}{T}$ and $T=\frac{1}{f}$; also $f=N/t$ for N full cycles in t seconds. f is in hertz, T in seconds. A 4 ms period gives 250 Hz.

$V_{rms}=\frac{V_p}{\sqrt{2}}\approx 0.707V_p$ and $V_p\approx 1.414V_{rms}$. Peak-to-peak is $V_{pp}=2V_p$. This 0.707 factor applies only to sine waves.

120 V RMS at 60 Hz (240 V circuits for large appliances). The voltage actually peaks near 170 V, and one cycle lasts about 16.7 ms.

A rectifier (diodes) converts AC to DC. An inverter converts DC to AC. A transformer steps AC voltage up or down and does not work on steady DC.

Use RMS values: $P=V_{rms}{I}_{rms}=\frac{V_{rms}^2}{R}=I_{rms}^{2}R$, in watts. 30 V RMS across 10 $\Omega$ gives 90 W. Peak values overstate power.

Counting every hump as a cycle. One full cycle has one positive and one negative half, so 8 humps (4 up, 4 down) is 4 cycles. Also convert ms to seconds before using f = 1/T.

The longer line is positive; the shorter (often thicker) line is negative. One long-short pair is a single cell; repeated pairs show a multi-cell battery.

Under the usual convention, no: a plain crossing passes over like an overpass. A solid dot marks a real junction where all touching lines form one node. Follow any stated legend.

Open means the contacts are apart and no current passes. Closed means the contacts touch and the switch acts like a wire. Closed completes the circuit (opposite of a water valve).

The triangle points toward the bar; the bar is the cathode (marked by a band on the real part). Conventional current flows easily from anode to cathode when forward biased.

Both are a diode triangle and bar. An LED has small arrows pointing outward (light emitted); a photodiode has arrows pointing inward (light received).

A capacitor is two equal parallel plates (C, farads) and stores energy in an electric field. An inductor is a row of loops (L, henries) and stores energy in a magnetic field.

Both open the circuit on overcurrent and go in series with what they protect. A fuse's link melts and must be replaced with the same rating; a breaker opens a contact and can be reset.

Look at the emitter arrow. Pointing out of the symbol means NPN ('not pointing in'); pointing in means PNP. Terminals are emitter, base and collector, labeled Q.

Dope silicon with a donor that has five valence electrons (phosphorus, arsenic, antimony). The loosely held fifth electron is free, so electrons are the majority carriers.

Dope silicon with an acceptor that has three valence electrons (boron, gallium, aluminum). The missing bond electron leaves a hole, so holes are the majority carriers.

No. Both are electrically neutral overall; each free carrier is balanced by a fixed dopant ion. The letter only names which mobile carrier dominates.

Forward: anode more positive than cathode; the depletion region narrows and a silicon diode conducts above about 0.6 to 0.7 V. Reverse: cathode more positive; the region widens and only tiny leakage flows.

$R=\frac{V_s-V_{LED}}{I}$: supply volts minus LED drop, divided by desired current in amperes. A 6 V supply, 2 V LED and 20 mA give 4/0.02 = 200 $\Omega$.

It is designed to conduct in reverse breakdown at its rated Zener voltage, holding that voltage nearly constant. It serves as a voltage reference or clamp and always needs a series resistor.

$\beta =\frac{I_C}{I_B}$, a unitless ratio; also $I_E=I_C+I_B$. 24 mA collector with 0.2 mA base gives 120. In saturation the outside circuit, not beta, limits $I_C$.

A BJT (emitter, base, collector) is current controlled and needs steady base current. A MOSFET (gate, drain, source) is voltage controlled; its insulated gate draws almost no steady current.

Like poles repel and unlike poles attract. Cutting a magnet gives two smaller magnets, each with its own N and S pole; a lone pole cannot be isolated.

Point the right thumb along conventional current and curl the fingers: they show the circular field direction. Current out of the page (dot) gives counterclockwise; into the page (cross) gives clockwise.

Ampere-turns $=N\times I$ (turns times current), the coil's magnetizing strength. Add turns, raise the current, or add a soft iron core. 300 turns at 0.2 A equals 30 turns at 2 A.

$F=BI\ell \sin \theta$: B in teslas, I in amperes, $\ell$ the length in the field in meters, F in newtons. Maximum when the wire crosses the field; zero when parallel. Force is perpendicular to both.

$\Phi =BA\cos \theta$, in webers (1 Wb = 1 T times 1 m${}^2$), where $\theta$ is between the field and the surface normal. 0.5 T straight through 0.04 m${}^2$ gives 0.02 Wb.

$\mathcal{E}=N\frac{\Delta \Phi }{\Delta t}$: N turns, flux change in webers, time in seconds, emf in volts. Faster change or more turns gives more emf; a steady field induces nothing.

The induced current always flows so its own magnetic field opposes the change in flux that caused it. This reflects energy conservation: work must be done to get electrical energy out.

$W=\frac{1}{2}L{I}^2$ joules (L in henries, I in amperes), so doubling current quadruples energy. $\tau =L/R$ seconds; current reaches about 63% after one $\tau$, essentially full after five.

Motor: electrical in, mechanical out. Generator: mechanical in, electrical out. Transformer: AC electrical in, AC electrical out at a different voltage, with no moving parts.

It is a split ring on the shaft fed by carbon brushes. It reverses the coil current every half turn so torque stays in one rotational direction and the motor keeps spinning.

$I=\frac{V-E}{R}$: supply minus back emf, over winding resistance. Back emf grows with speed. At stall E = 0, so $I=V/R$, many times the running current.

$\eta =\frac{P_{out}}{P_{in}}\times 100\%$, with $P_{in}=VI$ and $P_{out}=\tau \omega$. Convert rpm with $\omega =2\pi n/60$ rad/s. Efficiency over 100% means the fraction is upside down.

$f=\frac{pn}{120}$: f in hertz, p the number of poles, n in rpm. Example: 6 poles at 1200 rpm give 60 Hz; a 2-pole machine needs 3600 rpm.

Slip rings are two continuous rings that pass the coil's alternating voltage straight out, giving AC (alternator). A split-ring commutator flips connections each half turn, giving pulsating DC.

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$, $\frac{I_s}{I_p}=\frac{N_p}{N_s}$, and $V_p{I}_p=V_s{I}_s$. The higher-voltage side carries the lower current.

No to both. It needs changing flux, so steady DC gives no ongoing output (and can overheat the primary). Output frequency always equals input frequency; a rectifier is needed for AC to DC.

The current through the body, its path (hand to hand or hand to feet crosses the chest) and its duration. Voltage is the push; a few tens of milliamperes can be dangerous.

It compares current leaving on the hot with current returning on the neutral and trips when they differ by about 4 to 6 mA. It cannot sense a hot-to-neutral path through a person.

Fuse or breaker: overcurrent (overload or short) to protect wiring. GFCI: current leaking to ground, protecting people from shock. AFCI: dangerous arcing, mainly to prevent fires.

No. They respond to amperes of overcurrent, while milliamperes through a body can injure. Shock protection comes from GFCIs, sound insulation and grounding.

Overload: correctly wired loads draw moderately more than the rating, and heat builds gradually. Short: an unintended low-resistance path lets current jump to many times normal almost instantly.

$P=I^{2}R$ (watts, amperes, ohms). Tripling current gives 9 times the heat; 10 A through a 0.2 $\Omega$ joint releases 20 W in one spot, yet the breaker may not trip.

$E=\frac{1}{2}C{V}^2$: E in joules, C in farads, V in volts. Tripling voltage stores 9 times the energy; 2,000 $\mu$F at 100 V holds 10 J even when unplugged.

The neutral carries normal return current as part of the working circuit. The grounding conductor (bare or green) normally carries no current and only provides a fault path that helps the breaker trip.