Charge, Current, Voltage, Power & Resistors

TAKEAWAYS:

• Electric charge (Coulomb) is the basic property of electrons and protons; current (Amperes) is the flow of charge per second.
• Voltage (Volts) is the potential difference driving current; power (Watts) is the rate of energy transfer.
• Resistance (Ohms) opposes current flow; Ohm’s Law ($V=IR$) relates voltage, current, and resistance.
• Resistor color bands encode resistance values and tolerances for circuit design.
• Series/parallel circuits combine resistors differently, affecting total resistance and current distribution.

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1. Electric Charge and Current

Electric charge ($Q$) is a fundamental property of matter, measured in Coulombs (C). It exists in two forms:

• Positive charge (protons, +1.6 × 10⁻¹⁹ C)
• Negative charge (electrons, –1.6 × 10⁻¹⁹ C)

Electric current ($I$) is the flow of electric charge through a conductor, measured in Amperes (A): $$I=\frac{Q}{t}$$ where $Q$ = charge (C), $t$ = time (s).

How Current Flows

• In metals, current flows due to free electrons moving from negative to positive terminals (conventional current flows positive to negative).
• In electrolytes, ions carry charge (e.g., batteries, human nerves).

Shows electron flow vs. conventional current direction. (Image: inductiveload, Public domain, via Wikimedia Commons)

Worked Example: Charge Calculation

A current of 5 A flows for 2 minutes. Calculate the total charge transferred. Solution: $$t=2\text{ min}=120\text{ s},I=5\text{ A}$$ $$Q=I\times t=5\times 120=600\text{ C}$$

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2. Voltage (Potential Difference)

Voltage ($V$) is the work done per unit charge to move charge between two points, measured in Volts (V): $$V=\frac{W}{Q}$$ where $W$ = work (Joules), $Q$ = charge (C).

• Battery voltage = electromotive force (emf), the maximum potential difference it can provide.
• Voltage sources (batteries, generators) maintain a fixed potential difference.

Worked Example: Voltage in a Circuit

A 12 V battery moves 3 C of charge. Calculate the work done. Solution: $$W=V\times Q=12\times 3=36\text{ J}$$

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3. Resistance and Ohm’s Law

Resistance ($R$) opposes current flow, measured in Ohms (Ω). It depends on:

• Material (conductivity)
• Length (longer wire = higher resistance)
• Cross-sectional area (thinner wire = higher resistance)
• Temperature (higher temp = higher resistance in metals)

Ohm’s Law relates voltage, current, and resistance: $$V=I\times R$$ Rearranged forms: $$I=\frac{V}{R},R=\frac{V}{I}$$

Resistivity and Conductivity

• Resistivity ($\rho$) is a material property: $$R=\rho \frac{L}{A}$$ where $L$ = length, $A$ = area.
• Conductivity ($\sigma$) is the inverse of resistivity ($\sigma =\frac{1}{\rho }$).

Worked Example: Resistance Calculation

A copper wire has:

• Length ($L$) = 10 m
• Cross-sectional area ($A$) = 1 mm² = 1 × 10⁻⁶ m²
• Resistivity ($\rho$) = 1.68 × 10⁻⁸ Ω·m

Calculate its resistance. Solution: $$R=\rho \frac{L}{A}=1.68\times 10^{-8}\times \frac{10}{1\times 10^{-6}}=0.168\text{ Ω}$$

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4. Power in Electrical Circuits

Electric power ($P$) is the rate of energy transfer, measured in Watts (W): $$P=V\times I$$ Using Ohm’s Law, power can also be expressed as: $$P=I^{2}R\text{or}P=\frac{V^2}{R}$$

Worked Example: Power Dissipation in a Resistor

A 6 Ω resistor carries 2 A of current. Calculate:

1. Power dissipated.
2. Energy consumed in 5 minutes.

Solution:

1. $P=I^{2}R=2^2\times 6=24\text{ W}$
2. Energy ($E$) = Power × Time = $24\times (5\times 60)=7200\text{ J}$

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5. Series and Parallel Resistor Combinations

Resistors can be connected in series or parallel, affecting total resistance ($R_{total}$).

Series Connection

• Same current flows through all resistors.
• Total resistance increases: $$R_{total}=R_1+R_2+R_3+\dots$$
• Voltage divides across resistors (voltage divider rule).

Parallel Connection

• Same voltage across all resistors.
• Total resistance decreases: $$\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dots$$
• Current divides (current divider rule).

Shows current division. (Image: Omegatron, CC BY-SA 3.0, via Wikimedia Commons)

Worked Example: Series-Parallel Circuit

Three resistors ($R_1=2\Omega$, $R_2=3\Omega$, $R_3=6\Omega$) are connected:

• $R_1$ and $R_2$ in series.
• Combined with $R_3$ in parallel.
• Supply voltage = 12 V.

Calculate:

1. Total resistance.
2. Current through each resistor.

Solution:

1. $R_{12}=R_1+R_2=2+3=5\Omega$ $$\frac{1}{R_{total}}=\frac{1}{5}+\frac{1}{6}=\frac{11}{30}⟹R_{total}=\frac{30}{11}\approx 2.73\Omega$$
2. Total current ($I_{total}$) = $V/R_{total}=12/2.73\approx 4.4\text{ A}$
   • Current through $R_3$ = $I_{total}\times \frac{R_{12}}{R_{12}+R_3}\approx 4.4\times \frac{5}{11}\approx 2\text{ A}$
   • Current through $R_1$ and $R_2$ = 2 A (same in series).

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6. Resistor Color Coding

Resistors use color bands to indicate:

• Resistance value
• Tolerance (accuracy)
• Temperature coefficient (optional)

Color Code Table:

Example: A resistor with bands Red, Red, Brown, Gold has:

• Resistance = 22 × 10¹ = 220 Ω
• Tolerance = ±5%

Shows band positions and values. (Image: Adim kassn, CC BY-SA 3.0, via Wikimedia Commons)

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7. Real-World Applications

In the Real World

1. Khalti & eSewa (Digital Payments)

   • Uses resistors in circuit boards to limit current to ICs (integrated circuits) and prevent damage.
   • Voltage regulators (using resistors) ensure stable power supply to microcontrollers processing transactions.

2. Ncell & NTC (Telecom & Power Systems)

   • Series resistors in antenna circuits match impedance for efficient signal transmission.
   • Parallel resistors in voltage dividers measure high voltages safely in power grids.

3. Pathao (Ride-Hailing App)

   • Current sensors (using resistors) monitor battery drain in driver apps.
   • Ohm’s Law ensures optimal LED brightness in GPS modules without overheating.

Worked Example: Traffic Light Circuit (Real Scenario)

A traffic light has:

• Red bulb (2 Ω), Yellow bulb (3 Ω), Green bulb (4 Ω) in series.
• Supply voltage = 12 V.

Calculate:

1. Total resistance.
2. Current through each bulb.
3. Power dissipated by the green bulb.

Solution:

1. $R_{total}=2+3+4=9\Omega$
2. $I=V/R=12/9\approx 1.33\text{ A}$ (same for all bulbs in series)
3. $P_{green}=I^{2}R=(1.33{)}^2\times 4\approx 7.11\text{ W}$

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8. Exam Tip

• Memorize Ohm’s Law and power formulas—they appear in every numerical problem.
• Practice resistor color coding—exams often test this quickly (e.g., "A resistor with bands Red, Violet, Brown has resistance ____").
• Understand series vs. parallel:
  • Series: Same current, voltages add.
  • Parallel: Same voltage, currents add.
• Watch units: Always check if answers are in Ω, A, V, or W.
• Real-world links: Relate problems to batteries, household wiring, or mobile phone circuits for better retention.

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Key Formulas Summary: