Electronics Information
Electrical Fundamentals & Circuits
Electricity is the movement of electric charge. Circuits provide paths for current to flow. Understanding voltage, current, resistance, power, batteries, conductors, series and parallel paths, and open and short circuits is the foundation of Electronics Information.
Section 1
Voltage, Current & Resistance
Voltage, current, and resistance are core electrical quantities. Voltage is electric potential difference, the electrical push that can cause current to flow. Current is the rate of charge flow. Resistance opposes current.
Voltage is measured in volts (V), current in amperes (A), and resistance in ohms (Ω). A battery provides voltage between terminals. At constant resistance, more voltage generally means more current. At constant voltage, more resistance means less current. Conventional current is shown from positive to negative through an external circuit, even though electrons in metal move oppositely.
Example 1
Which quantity measures the rate at which electric charge flows? A. Voltage | B. Resistance | C. Current | D. Power
Solution
- 1The question asks about charge flow.
- 2Current measures the rate of charge flow.
Why: Voltage is potential difference, resistance opposes flow, and current measures charge moving past a point.
Example 2
A battery voltage stays constant. What happens if circuit resistance increases? A. Current increases. | B. Current decreases. | C. Current remains the same. | D. Current becomes voltage.
Solution
- 1Increasing resistance makes current harder to flow while voltage remains constant.
Why: Current and resistance have an inverse relationship at constant voltage.
Section 2
Ohm's Law
Ohm's law relates voltage, current, and resistance: V = IR. Rearrange it for the missing quantity: I = V/R or R = V/I.
For a 12 V circuit with 4 Ω resistance, I = 12/4 = 3 A. If resistance doubles to 8 Ω at 12 V, current becomes 1.5 A. Identify known values and the missing value before selecting a formula.
Example 1
A 12 V battery is across a 4 Ω resistor. How much current flows? A. 2 A | B. 3 A | C. 8 A | D. 48 A
Solution
- 1Use I = V/R.
- 2I = 12/4 = 3 A.
Why: Current equals voltage divided by resistance.
Example 2
A component has 10 V and 2 A. What is its resistance? A. 20 Ω | B. 12 Ω | C. 8 Ω | D. 5 Ω
Solution
- 1Use R = V/I.
- 2R = 10/2 = 5 Ω.
Why: Resistance is voltage divided by current.
Example 3
A current of 3 A passes through a 6 Ω resistor. What voltage is across it? A. 2 V | B. 9 V | C. 18 V | D. 36 V
Solution
- 1Use V = IR.
- 2V = 3 × 6 = 18 V.
Why: Ohm's law gives voltage as current multiplied by resistance.
Section 3
Electrical Power
Electrical power is the rate electrical energy is transferred or converted. It is measured in watts (W). A lamp converts electrical energy to light and heat; a motor converts it to motion.
The main formula is P = VI. For resistive circuits, P = I²R and P = V²/R also apply. Power is how quickly energy is used; energy is total use. Electrical energy can be measured in watt-hours (Wh) or kilowatt-hours (kWh).
Example 1
A device operates at 120 V and draws 2 A. What power does it use? A. 60 W | B. 118 W | C. 122 W | D. 240 W
Solution
- 1Use P = VI.
- 2P = 120 × 2 = 240 W.
Why: Power equals voltage multiplied by current.
Example 2
Two devices use the same voltage. A draws 2 A and B draws 4 A. Which uses more power? A. A | B. B | C. Same | D. Power does not depend on current
Solution
- 1With the same voltage, B draws twice the current.
- 2P = VI, so B uses twice the power.
Why: At constant voltage, power increases as current increases.
Example 3
A 60 W lamp operates for 3 hours. How much energy does it use? A. 20 Wh | B. 63 Wh | C. 180 Wh | D. 360 Wh
Solution
- 1Energy in watt-hours is power × time.
- 260 × 3 = 180 Wh.
Why: Multiplying power by hours gives total energy use in watt-hours.
Section 4
Batteries, Conductors & Insulators
A circuit needs an energy source and a conducting path. A battery converts stored chemical energy into electrical energy and supplies direct current (DC) to an external circuit.
A cell is one electrochemical unit; a battery may contain one or more cells. Two matching 1.5 V cells in series provide about 3 V. Conductors such as copper and aluminum allow current to flow easily. Insulators such as rubber, plastic, glass, and dry wood strongly resist current. A complete circuit has a source, conducting path, and load, such as a lamp, motor, or resistor.
Example 1
Why is copper commonly used for wiring? A. Good conductor | B. Prevents all current | C. Produces electricity without source | D. Always magnetic
Solution
- 1Copper has relatively low resistance and carries current efficiently.
Why: The metal portion of wiring is meant to carry current.
Example 2
What energy does a battery primarily convert to electrical energy? A. Sound | B. Nuclear | C. Chemical | D. Gravitational
Solution
- 1Chemical reactions inside a battery create electric potential difference.
Why: A battery stores energy chemically and converts it when connected to a circuit.
Example 3
Two 1.5 V cells are connected in series with voltages adding. What is combined nominal voltage? A. 0.75 V | B. 1.5 V | C. 2 V | D. 3 V
Solution
- 1In this series arrangement, voltages add: 1.5 + 1.5 = 3.
Why: Correctly connected series cells add their voltages.
Section 5
Series Circuits
A series circuit has one current path. The same current passes through every component in that path. Voltage drops across components add to the applied voltage.
Series resistances add: Rtotal = R₁ + R₂ + R₃ + ... Adding resistance in series increases total resistance and reduces total current for the same source voltage. If the one path opens, current stops everywhere in the simple series circuit.
Example 1
Three series resistors are 2 Ω, 3 Ω, and 5 Ω. Total resistance? A. 1 Ω | B. 5 Ω | C. 10 Ω | D. 30 Ω
Solution
- 1Series resistances add.
- 2Rtotal = 2 + 3 + 5 = 10 Ω.
Why: Each series resistance contributes to total opposition along one path.
Example 2
A 12 V battery supplies 2 Ω and 4 Ω series resistors. What drop is across the 4 Ω resistor? A. 2 V | B. 4 V | C. 6 V | D. 8 V
Solution
- 1Total R = 2 + 4 = 6 Ω, so I = 12/6 = 2 A.
- 2The same current passes through 4 Ω: V = IR = 2 × 4 = 8 V.
Why: Current is the same everywhere in series; larger resistance has larger voltage drop.
Example 3
Two identical bulbs are in series. One is removed and the circuit opens. What happens? A. Other brighter | B. Other goes out | C. Other gets all current | D. Battery doubles
Solution
- 1Series has only one path.
- 2Removing a bulb breaks that path, so current cannot flow through either.
Why: An open anywhere in a simple series circuit interrupts the full circuit.
Section 6
Parallel Circuits
A parallel circuit has multiple branches. Each branch provides a separate path for current and has the same voltage when connected across the same two source points.
Total current is the sum of branch currents: Itotal = I₁ + I₂ + I₃ + ... Adding a parallel branch gives current another route, decreasing equivalent resistance. Positive parallel resistance is always less than the smallest individual resistance. An open branch does not automatically stop other branches.
Example 1
Two resistors are parallel across a 9 V battery. What voltage is across each? A. 4.5 V | B. 9 V | C. 18 V | D. 0 V
Solution
- 1Each branch connects across the same two source points.
Why: Voltage is the same across parallel components.
Example 2
A 12 V battery supplies parallel 3 Ω and 6 Ω branches. What total current? A. 2 A | B. 4 A | C. 6 A | D. 12 A
Solution
- 1Branch 1: I₁ = 12/3 = 4 A. Branch 2: I₂ = 12/6 = 2 A.
- 2Add branch currents: Itotal = 4 + 2 = 6 A.
Why: Each branch draws independently; the battery supplies their sum.
Example 3
Two bulbs are parallel. One burns out and its branch opens. What happens to the other? A. It can remain illuminated. | B. It must go out. | C. Resistance becomes zero. | D. It becomes a battery.
Solution
- 1Parallel circuits have separate current paths.
- 2One open branch does not break the remaining branch.
Why: The remaining bulb has its own complete path to the battery.
Example 4
Parallel resistors are 3 Ω and 6 Ω. Equivalent resistance? A. 9 Ω | B. 6 Ω | C. 3 Ω | D. 2 Ω
Solution
- 1Use 1/Rtotal = 1/3 + 1/6 = 2/6 + 1/6 = 1/2.
- 2Take the reciprocal: Rtotal = 2 Ω.
Why: Parallel branches add current paths, making combined resistance less than either individual value.
Section 7
Open Circuits & Short Circuits
An open circuit has an interruption in a conducting path. A short circuit is an unintended low-resistance path that can bypass the intended load and cause excessive current.
An open can be a disconnected wire, open switch, or broken bulb filament. A short can overheat wiring and create fire hazards. Fuses melt under specified excess current; breakers are protective switches that trip under overload or fault conditions. Never attempt to create a short circuit or test live equipment as a study exercise.
Example 1
A battery, bulb, and open switch are in series. What happens? A. Brighter bulb | B. Twice voltage | C. Bulb remains off | D. More energy produced
Solution
- 1An open switch interrupts the conducting path.
- 2Current cannot flow through the bulb.
Why: A complete path is required for a simple battery-powered bulb circuit.
Example 2
Which describes a short circuit? A. Open switch interrupts current. | B. Unintended low-resistance path bypasses load. | C. Resistor works normally. | D. Battery is disconnected.
Solution
- 1A short gives an unintended conducting path with very low resistance.
Why: It can bypass intended resistance and allow excessive current.
Example 3
What is the main purpose of a fuse? A. Increase voltage | B. Store energy | C. Change AC to DC | D. Interrupt excessive current
Solution
- 1A fuse is protective. Its element melts when enough excessive current flows, opening the circuit.
Why: Fuses help protect wiring and equipment from dangerous overcurrent.
Quick Review
Quick Review
- Voltage is electric potential difference in V; current is charge flow in A; resistance opposes current in Ω; power is energy-transfer rate in W.
- Ohm's law: V = IR, I = V/R, R = V/I. Power: P = VI, P = I²R, P = V²/R.
- Series: one path, same current, resistances add. Parallel: multiple paths, same voltage, branch currents add.
- Open circuit: interrupted path. Short circuit: unintended low-resistance path. Fuses and breakers protect against excess current.
The Big Picture
- More voltage → more current at constant resistance. More resistance → less current at constant voltage. More current → more power at constant voltage. Series → one path, same current. Parallel → multiple paths, same voltage.