What is Ohm's Law and how do you use it?

Ohm's Law states V = I × R (voltage = current × resistance). To find any one variable, you need the other two. It's the foundation of all DC circuit analysis.

How do you calculate total resistance in series and parallel circuits?

In series: R_total = R1 + R2 + R3. In parallel: 1/R_total = 1/R1 + 1/R2 + 1/R3. For two parallel resistors: R_total = (R1 × R2)/(R1 + R2).

How do you size an LED resistor?

R = (V_supply - V_LED) / I_LED. For example, a red LED (2V, 20mA) on 5V needs R = (5-2)/0.02 = 150 ohms.

What is impedance in AC circuits?

Impedance Z combines resistance R with reactance from capacitors (Xc) and inductors (XL). Z = sqrt(R² + (XL - XC)²). At resonance, XL = XC and Z = R.

Electronics Calculators - CalcREX

Ohm's Law & DC Circuit Fundamentals

The foundation of all circuit analysis - the relationship between voltage, current, and resistance.

1 Ohm's Law

Georg Ohm discovered that the current through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance. This is the most fundamental law in electronics.

V = I × R

V = voltage (Volts, V)I = current (Amperes, A)R = resistance (Ohms, Ω)

Figure 1: The Ohm's Law Triangle - cover the variable you want to find. What's left tells you the formula.

2 Voltage, Current & Resistance

Voltage (V)

Electrical "pressure" - pushes charges through the circuit

Current (I)

Flow of charge - measured in Amperes (coulombs/second)

Resistance (R)

Opposition to current - measured in Ohms (Ω)

Water analogy: Voltage = water pressure, Current = flow rate, Resistance = pipe narrowing. Higher pressure (V) or wider pipe (lower R) = more water flow (I).

Worked Example 1

Basic Ohm's Law - Finding Current

Problem: A 9V battery is connected to a 470 Ω resistor. What current flows through the circuit?

Step 1: Identify known values

V = 9 V, R = 470 Ω, I = ?

Step 2: Apply Ohm's Law

I = V / R = 9 / 470

I = 0.01914 A = 19.14 mA

Answer: 19.14 mA flows through the resistor. In practice, you'd use the nearest standard value - this is typical for LED current-limiting resistors with a 9V supply.

3 Kirchhoff's Laws

KCL - Current Law (Junction Rule)

The total current entering a junction equals the total current leaving it.

ΣI_in = ΣI_out

KVL - Voltage Law (Loop Rule)

The sum of all voltages around any closed loop equals zero.

ΣV = 0 (around loop)

Worked Example 2

KVL - Voltage Drops in a Loop

Problem: A 12 V battery is connected in series with R₁ = 200 Ω and R₂ = 100 Ω. Find the current, and the voltage across each resistor.

Step 1: Total resistance (series)

R_T = R₁ + R₂ = 200 + 100 = 300 Ω

Step 2: Current

I = V / R_T = 12 / 300

I = 0.04 A = 40 mA

Step 3: Voltage drops (V = IR)

V₁ = 0.04 × 200 = 8 V

V₂ = 0.04 × 100 = 4 V

Check KVL: 12 − 8 − 4 = 0 ✓

Answer: I = 40 mA, V₁ = 8 V, V₂ = 4 V. The larger resistor drops more voltage - voltage divides proportionally to resistance in a series circuit.

Worked Example 3

Resistance from Color Codes

Problem: A resistor has bands: Brown, Black, Red, Gold. What is its resistance and tolerance?

Step 1: Decode bands (4-band)

Band 1 (Brown) = 1

Band 2 (Black) = 0

Band 3 (Red) = ×10² = ×100

Band 4 (Gold) = ±5% tolerance

Step 2: Calculate

R = 10 × 100 = 1000 Ω

R = 1 kΩ ± 5% (950 Ω to 1050 Ω)

Answer: 1 kΩ ± 5%. The color code mnemonic: Black-Brown-Red-Orange-Yellow-Green-Blue-Violet-Gray-White = 0-1-2-3-4-5-6-7-8-9.

Series & Parallel Circuits - Resistor Networks & Voltage Dividers

The two fundamental ways to connect components - and how to analyze any combination.

1 Series vs. Parallel - Key Differences

Property

Series

Parallel

Current

Same through all

Splits between branches

Voltage

Divides across each

Same across all

Total R

R_T = R₁ + R₂ + R₃

1/R_T = 1/R₁ + 1/R₂ + 1/R₃

R_T vs parts

Always larger

Always smaller than smallest

If one opens

Entire circuit breaks

Other branches continue

2 Series Resistors

R_total = R₁ + R₂ + R₃ + ...

Same current flows through every resistorVoltage drops add up to source voltage (KVL)

    +---[R₁ = 100Ω]---[R₂ = 220Ω]---[R₃ = 330Ω]---+
    |                                                 |
   (+) 12V Battery                                   (-)
    |                                                 |
    +-------------------------------------------------+
    R_total = 100 + 220 + 330 = 650 Ω

3 Parallel Resistors

1/R_T = 1/R₁ + 1/R₂ + 1/R₃

For 2 resistors: R_T = (R₁ × R₂)/(R₁ + R₂)Same voltage across every branch

          +---[R₁ = 100Ω]---+
          |                  |
    (+)---+---[R₂ = 200Ω]---+---(-)
          |                  |
          +---[R₃ = 300Ω]---+
    1/R_T = 1/100 + 1/200 + 1/300 = 0.01833
    R_T = 54.55 Ω

Worked Example 1

Series-Parallel Combination Circuit

Problem: Find the total resistance, total current, and voltage across each component. V_S = 24 V, R₁ = 10 Ω (series), R₂ = 30 Ω ∥ R₃ = 60 Ω (parallel block in series with R₁).

Step 1: Solve the parallel block

R₂₃ = (R₂ × R₃)/(R₂ + R₃) = (30 × 60)/(30 + 60)

R₂₃ = 1800/90 = 20 Ω

Step 2: Total resistance

R_T = R₁ + R₂₃ = 10 + 20

R_T = 30 Ω

Step 3: Total current

I_T = V_S/R_T = 24/30

I_T = 0.8 A

Step 4: Voltage drops

V₁ = I_T × R₁ = 0.8 × 10 = 8 V

V₂₃ = I_T × R₂₃ = 0.8 × 20 = 16 V

I₂ = 16/30 = 0.533 A, I₃ = 16/60 = 0.267 A

Check: 8 + 16 = 24 V ✓ | 0.533 + 0.267 = 0.8 A ✓

Answer: R_T = 30 Ω, I = 0.8 A, V₁ = 8 V, V_parallel = 16 V. The parallel block has lower resistance than either individual resistor, and current splits inversely proportional to resistance.

4 Voltage Divider

A voltage divider uses two series resistors to produce a fraction of the input voltage - one of the most common circuits in electronics.

V_out = V_in × R₂ / (R₁ + R₂)

V_out is taken across R₂ (bottom resistor)Works only with no/minimal load on output

Worked Example 2

Voltage Divider Design

Problem: Design a voltage divider to get 3.3 V from a 5 V supply using a 1 kΩ resistor for R₂.

Step 1: Rearrange for R₁

V_out/V_in = R₂/(R₁ + R₂)

3.3/5 = 1000/(R₁ + 1000)

0.66 × (R₁ + 1000) = 1000

R₁ = (1000/0.66) − 1000

R₁ = 515 Ω → use 510 Ω standard value

Answer: R₁ = 510 Ω (nearest standard value), R₂ = 1 kΩ. Actual output: 5 × 1000/1510 = 3.31 V. Voltage dividers are used to level-shift signals (e.g., 5V Arduino to 3.3V sensor).

Electrical Power & Energy - Watts, Kilowatt-Hours & Efficiency

How to calculate power consumption, energy cost, and thermal dissipation in circuits.

1 The Power Formulas

Electrical power is the rate at which energy is transferred. There are three equivalent formulas - use whichever matches your known values.

P = V × I

When V and I are known

P = I² × R

When I and R are known

P = V² / R

When V and R are known

All three are the same equation! Just substitute Ohm's Law (V = IR) into P = VI. Knowing any two of V, I, R lets you find power.

2 Energy & Cost

E = P × t

E = energy (Joules or Wh)P = power (Watts)t = time (seconds or hours)1 kWh = 3,600,000 J = 3.6 MJ

Your electricity bill is measured in kilowatt-hours (kWh): power in kilowatts multiplied by time in hours.

Worked Example 1

Power Dissipation in a Resistor

Problem: A 220 Ω resistor carries 150 mA. (a) How much power does it dissipate? (b) What minimum wattage rating should the resistor have?

Step 1: Power

P = I² × R = (0.150)² × 220

P = 4.95 W

Step 2: Minimum rating

Use at least 2× the calculated power for safety margin.

Minimum rating: 10 W resistor

Answer: 4.95 W dissipated. Use a 10 W (or higher) resistor. Standard ¼ W resistors would burn out - always check power ratings when designing circuits!

Worked Example 2

Electricity Cost Calculation

Problem: An air conditioner draws 1,500 W and runs 8 hours/day for 30 days. If electricity costs $0.12/kWh, what is the monthly cost?

Step 1: Daily energy

E_daily = 1.5 kW × 8 h = 12 kWh

Step 2: Monthly energy

E_monthly = 12 × 30 = 360 kWh

Step 3: Cost

Cost = 360 × $0.12

Cost = $43.20 per month

Answer: $43.20/month. This is equivalent to 1,296 MJ of energy - enough to lift a 13-ton truck 10 km into the air.

3 Maximum Power Transfer

Maximum power is delivered to a load when the load resistance equals the source's internal resistance.

P_max when R_load = R_source

P_max = V²/(4 × R_source)Efficiency at max power transfer = 50%

Worked Example 3

Maximum Power Transfer

Problem: A battery with EMF = 12 V and internal resistance r = 2 Ω is connected to a load R_L. (a) What R_L gives max power? (b) What is that max power?

Step 1: For max power

R_L = r = 2 Ω

Step 2: Calculate max power

I = V/(r + R_L) = 12/(2 + 2) = 3 A

P_load = I² × R_L = 9 × 2

P_max = 18 W

Answer: R_L = 2 Ω delivers max power of 18 W. But 18 W is also wasted in internal resistance - that's why max power transfer is only 50% efficient. Audio amplifier output impedance matching uses this principle.

Capacitors & Inductors - Energy Storage Components

How capacitors store charge and inductors store magnetic energy - and their behavior in DC and AC circuits.

1 Capacitors - Storing Charge

A capacitor stores energy in an electric field between two conductive plates. Capacitance is measured in Farads (F).

C = Q / V → Q = C × V

C = capacitance (Farads, F)Q = charge (Coulombs, C)V = voltage (Volts)

Energy: E = ½CV²

E = energy stored (Joules)

2 Capacitors in Series & Parallel

Parallel (opposite of resistors!)

C_T = C₁ + C₂ + C₃

Same voltage across each. Total capacitance increases.

Series (opposite of resistors!)

1/C_T = 1/C₁ + 1/C₂ + 1/C₃

Same charge on each. Total capacitance decreases.

Opposite of resistors! Capacitors in parallel add up (like resistors in series). Capacitors in series use the reciprocal formula (like resistors in parallel).

Worked Example 1

Capacitor Energy Storage

Problem: A 470 μF capacitor is charged to 25 V. (a) How much charge does it store? (b) How much energy?

Step 1: Charge

Q = C × V = 470 × 10⁻⁶ × 25

Q = 0.01175 C = 11.75 mC

Step 2: Energy

E = ½CV² = 0.5 × 470 × 10⁻⁶ × 25²

E = 0.147 J = 147 mJ

Answer: Q = 11.75 mC, E = 147 mJ. Note how energy depends on V² - doubling the voltage quadruples the stored energy. This is why capacitors in camera flashes are charged to high voltages.

3 RC Time Constant

When charging or discharging through a resistor, a capacitor's voltage changes exponentially. The time constant τ defines how fast.

τ = R × C

τ = time constant (seconds)After 1τ: 63.2% charged (or 36.8% remaining)After 5τ: 99.3% charged (effectively full)

Worked Example 2

RC Charging Time

Problem: A 100 μF capacitor charges through a 47 kΩ resistor from a 9 V supply. (a) What is τ? (b) How long to fully charge? (c) What is the voltage at t = τ?

Step 1: Time constant

τ = R × C = 47,000 × 100 × 10⁻⁶

τ = 4.7 seconds

Step 2: Full charge time (5τ)

t_full ≈ 5τ = 5 × 4.7

t_full ≈ 23.5 seconds

Step 3: Voltage at t = τ

V(τ) = V_S × (1 − e⁻¹) = 9 × 0.632

V(τ) = 5.69 V

Answer: τ = 4.7 s, full charge ≈ 23.5 s, V(τ) = 5.69 V. RC circuits are used in timing circuits, debounce filters, and audio tone controls.

4 Inductors - Storing Magnetic Energy

An inductor stores energy in a magnetic field when current flows through a coil. It opposes changes in current.

V_L = L × (dI/dt) | E = ½LI²

L = inductance (Henrys, H)dI/dt = rate of current changeRL time constant: τ = L/R

Worked Example 3

Capacitors in Series and Parallel

Problem: Three capacitors: C₁ = 10 μF, C₂ = 22 μF, C₃ = 47 μF. Find the total capacitance when (a) all in parallel, (b) all in series.

(a) Parallel

C_T = C₁ + C₂ + C₃ = 10 + 22 + 47

C_T = 79 μF

(b) Series

1/C_T = 1/10 + 1/22 + 1/47

= 0.100 + 0.04545 + 0.02128 = 0.16674

C_T = 5.997 μF ≈ 6.0 μF

Answer: Parallel: 79 μF (larger). Series: 6.0 μF (smaller than the smallest capacitor). This is the opposite pattern from resistors - a common exam trap!

AC Circuits - Impedance, Reactance & Resonance

Alternating current analysis - how capacitors and inductors behave differently at different frequencies.

1 AC Fundamentals

Alternating current changes direction periodically. The voltage follows a sine wave described by:

v(t) = V_peak × sin(2πft)

V_RMS = V_peak / √2 = 0.707 × V_peakf = frequency (Hz), T = 1/f (period in seconds)

V_RMS (Root Mean Square) is what your voltmeter reads and what wall outlets are rated at. A "230V" outlet actually swings ±325 V peak.

2 Reactance - AC "Resistance"

Capacitors and inductors oppose AC current, but in a frequency-dependent way called reactance (X), measured in Ohms.

X_C = 1/(2πfC)

Capacitive - decreases with frequency

X_L = 2πfL

Inductive - increases with frequency

Capacitor in AC

Passes high frequencies, blocks low frequencies (DC). Current leads voltage by 90°.

Inductor in AC

Passes low frequencies (DC), blocks high frequencies. Current lags voltage by 90°.

3 Impedance

Impedance (Z) is the total opposition to AC current, combining resistance and reactance. It's a complex quantity.

Z = √(R² + (X_L − X_C)²)

For series RLC circuitV = I × Z (AC Ohm's Law)

Worked Example 1

Capacitive Reactance in an AC Circuit

Problem: A 10 μF capacitor is connected to a 230 V, 50 Hz supply. Find (a) the capacitive reactance and (b) the RMS current.

Step 1: Reactance

X_C = 1/(2πfC) = 1/(2π × 50 × 10 × 10⁻⁶)

X_C = 318.3 Ω

Step 2: Current

I = V/X_C = 230/318.3

I = 0.722 A (RMS)

Answer: X_C = 318.3 Ω, I = 722 mA. At 60 Hz, X_C would be lower (265 Ω) and current would be higher - capacitors pass more current at higher frequencies.

4 Resonance

At the resonant frequency, X_L = X_C and they cancel out. The circuit behaves as purely resistive, and impedance is at its minimum (series) or maximum (parallel).

f₀ = 1 / (2π√(LC))

At resonance: Z = R (minimum for series RLC)Current is maximum at resonance

Worked Example 2

Series RLC Impedance

Problem: R = 100 Ω, L = 0.5 H, C = 10 μF in series. At f = 50 Hz, find (a) X_L, (b) X_C, (c) impedance Z, (d) resonant frequency.

Step 1: Reactances

X_L = 2π × 50 × 0.5 = 157.1 Ω

X_C = 1/(2π × 50 × 10 × 10⁻⁶) = 318.3 Ω

Step 2: Impedance

Z = √(100² + (157.1 − 318.3)²)

= √(10000 + 25985) = √35985

Z = 189.7 Ω

Step 3: Resonant frequency

f₀ = 1/(2π√(0.5 × 10 × 10⁻⁶))

f₀ = 71.2 Hz

Answer: X_L = 157.1 Ω, X_C = 318.3 Ω, Z = 189.7 Ω, f₀ = 71.2 Hz. At 50 Hz the circuit is capacitive (X_C > X_L). At resonance (71.2 Hz), Z would drop to just 100 Ω (= R).

Worked Example 3

Power Factor Correction

Problem: A motor draws 5 A from a 230 V supply at power factor 0.65. Find (a) apparent power, (b) real power, (c) reactive power.

Step 1: Apparent power

S = V × I = 230 × 5

S = 1150 VA

Step 2: Real power

P = S × cos(φ) = 1150 × 0.65

P = 747.5 W

Step 3: Reactive power

Q = S × sin(φ) = 1150 × sin(cos⁻¹(0.65))

Q = 1150 × 0.7599

Q = 873.9 VAR

Answer: S = 1150 VA, P = 747.5 W, Q = 873.9 VAR. Only 65% of the apparent power does useful work. Power factor correction capacitors can bring PF closer to 1.0, reducing utility bills for industrial users.

Semiconductors - Diodes, Transistors & Basic Circuits

The building blocks of modern electronics - how diodes rectify, transistors amplify, and LEDs light up.

1 Diodes - One-Way Valves for Current

A diode allows current to flow in one direction only. In forward bias (anode positive), current flows after overcoming the forward voltage drop. In reverse bias, current is blocked.

Silicon diode V_f

≈ 0.7 V forward drop

Germanium diode V_f

≈ 0.3 V forward drop

LED V_f

1.8–3.3 V (varies by color)

Schottky diode V_f

≈ 0.2–0.3 V (fast switching)

Worked Example 1

LED Current-Limiting Resistor

Problem: You want to power a red LED (V_f = 2.0 V, I_f = 20 mA) from a 5 V supply. What resistor value is needed?

Step 1: Voltage across resistor

V_R = V_supply − V_f = 5.0 − 2.0 = 3.0 V

Step 2: Ohm's Law for resistor

R = V_R/I = 3.0/0.020

R = 150 Ω

Step 3: Power dissipated

P = V_R × I = 3.0 × 0.020 = 0.06 W

¼ W resistor is sufficient

Answer: 150 Ω resistor (standard value). This is the most common beginner electronics calculation - every LED needs a current-limiting resistor to prevent burnout.

2 Transistors - Switches & Amplifiers

A transistor uses a small current or voltage to control a larger current. The two main types are BJT (Bipolar Junction Transistor) and MOSFET.

Property

BJT (NPN)

MOSFET (N-channel)

Controlled by

Base current (I_B)

Gate voltage (V_GS)

Key formula

I_C = β × I_B

V_GS > V_th to turn ON

Input impedance

Low (draws current)

Very high (voltage only)

Common use

Audio amplifiers

Digital circuits, power switching

BJT: I_C = β × I_B

I_C = collector current (the load current)I_B = base current (the control signal)β (hFE) = current gain (typically 50–300)

Worked Example 2

BJT as a Switch

Problem: A relay coil requires 100 mA to activate. Using an NPN transistor with β = 100, what base current is needed? If the microcontroller outputs 3.3 V and V_BE = 0.7 V, what base resistor value is needed?

Step 1: Minimum base current

I_B = I_C/β = 100 mA / 100

I_B(min) = 1 mA

Use 2× for reliable saturation:

I_B = 2 mA

Step 2: Base resistor

R_B = (V_out − V_BE)/I_B = (3.3 − 0.7)/0.002

R_B = 1300 Ω → use 1.2 kΩ standard

Answer: R_B = 1.2 kΩ. The microcontroller's 3.3 V GPIO pin drives just 2 mA into the base, which controls 100 mA through the relay. That's 50× current amplification - the essence of transistor switching.

3 Rectification - AC to DC

Diodes convert AC to DC through rectification. A full-bridge rectifier uses 4 diodes to convert both halves of the AC wave.

V_DC ≈ V_peak − 2V_f (full bridge)

V_peak = V_RMS × √22 diodes conduct at any time → 2 × 0.7 V drop

Worked Example 3

Full-Bridge Rectifier Output

Problem: A 12 V_RMS transformer feeds a full-bridge rectifier with silicon diodes. What is the peak DC output voltage (before filtering)?

Step 1: Peak voltage

V_peak = V_RMS × √2 = 12 × 1.414

V_peak = 16.97 V

Step 2: DC output after diode drops

V_DC = V_peak − 2V_f = 16.97 − 2(0.7)

V_DC = 15.57 V (peak)

Answer: 15.57 V peak output. After adding a smoothing capacitor, the DC output would be close to this value with ripple. This is why a "12 V" transformer can feed a 12 V voltage regulator - the peak is higher than RMS.

Comprehensive Electricity & Electronics Calculators

Select a Calculator

How it works

Common Components

Important Values

Electricity & Electronics - In-Depth Guides with Worked Examples

Ohm's Law & DC Circuit Fundamentals

1 Ohm's Law

2 Voltage, Current & Resistance

Basic Ohm's Law - Finding Current

3 Kirchhoff's Laws

KVL - Voltage Drops in a Loop

Resistance from Color Codes

Series & Parallel Circuits - Resistor Networks & Voltage Dividers

1 Series vs. Parallel - Key Differences

2 Series Resistors

3 Parallel Resistors

Series-Parallel Combination Circuit

4 Voltage Divider

Voltage Divider Design

Electrical Power & Energy - Watts, Kilowatt-Hours & Efficiency

1 The Power Formulas

2 Energy & Cost

Power Dissipation in a Resistor

Electricity Cost Calculation

3 Maximum Power Transfer

Maximum Power Transfer

Capacitors & Inductors - Energy Storage Components

1 Capacitors - Storing Charge

2 Capacitors in Series & Parallel

Capacitor Energy Storage

3 RC Time Constant

RC Charging Time

4 Inductors - Storing Magnetic Energy

Capacitors in Series and Parallel

AC Circuits - Impedance, Reactance & Resonance

1 AC Fundamentals

2 Reactance - AC "Resistance"

3 Impedance

Capacitive Reactance in an AC Circuit

4 Resonance

Series RLC Impedance

Power Factor Correction

Semiconductors - Diodes, Transistors & Basic Circuits

1 Diodes - One-Way Valves for Current

LED Current-Limiting Resistor

2 Transistors - Switches & Amplifiers

BJT as a Switch

3 Rectification - AC to DC

Full-Bridge Rectifier Output