Electric Circuits
Misconceptions
| Misconception | Correction |
|---|---|
| "Current gets used up in a circuit" | Current is conserved; it flows continuously around a closed loop. The same current enters and exits a resistor |
| "Batteries store charge" | Batteries store chemical energy and provide a voltage (potential difference), not charge |
| "Voltage flows through a wire" | Voltage is a potential difference between two points; current flows, voltage is measured across |
| "All resistors get the same current" | Only true in series; in parallel, current divides inversely proportional to resistance |
| "Adding resistors always increases total resistance" | True for series, but parallel combinations decrease total resistance |
Key Concepts
Electric Current
Current is the rate of charge flow through a conductor.
Definition:
- •I = Q/t (charge per unit time)
- •Unit: Ampere (A) = Coulomb/second (C/s)
- •Conventional current flows from + to - (opposite to electron flow)
Resistance
Resistance opposes the flow of current.
Ohm's Law:
- •V = IR
- •R = V/I
- •Unit: Ohm (Ω) = V/A
Factors affecting resistance:
- •R = ρL/A (resistivity × length / cross-sectional area)
- •Temperature increases resistance in most conductors
Resistors in Series
When resistors are connected end-to-end:
- •Same current through all: I_total = I_1 = I_2 = I_3
- •Voltages add: V_total = V_1 + V_2 + V_3
- •Equivalent resistance: R_eq = R_1 + R_2 + R_3
Resistors in Parallel
When resistors share the same two nodes:
- •Same voltage across all: V_total = V_1 = V_2 = V_3
- •Currents add: I_total = I_1 + I_2 + I_3
- •Equivalent resistance: 1/R_eq = 1/R_1 + 1/R_2 + 1/R_3
Special case for two resistors in parallel:
- •R_eq = (R_1 × R_2) / (R_1 + R_2)
Kirchhoff's Laws
Junction Rule (KCL - Kirchhoff's Current Law):
- •Sum of currents entering a junction = sum of currents leaving
- •ΣI_in = ΣI_out
- •Based on conservation of charge
Loop Rule (KVL - Kirchhoff's Voltage Law):
- •Sum of voltage changes around any closed loop = 0
- •ΣV = 0 (around a loop)
- •Based on conservation of energy
Sign conventions for loop rule:
- •EMF: + if traversing from - to + terminal
- •Resistor: - if traversing in direction of current (voltage drop)
Electrical Power
Power is the rate of energy transfer.
Power formulas:
- •P = IV (general form)
- •P = I²R (useful when current is known)
- •P = V²/R (useful when voltage is known)
- •Unit: Watt (W) = J/s = V·A
Energy dissipated:
- •E = Pt (energy = power × time)
- •Unit: Joule (J) or kilowatt-hour (kWh)
RC Circuits (Basics)
A circuit with a resistor and capacitor in series.
Charging a capacitor:
- •Q(t) = Q_max(1 - e^(-t/τ))
- •τ = RC (time constant)
- •After 5τ, capacitor is ~99% charged
Discharging:
- •Q(t) = Q_0 × e^(-t/τ)
- •Current and voltage decrease exponentially
Equations
[1] I = Q/t (definition of current) [2] V = IR (Ohm's law) [3] R_series = R_1 + R_2 + R_3 + ... (series resistors) [4] 1/R_parallel = 1/R_1 + 1/R_2 + 1/R_3 + ... (parallel resistors) [5] P = IV = I²R = V²/R (electrical power) [6] ΣI_in = ΣI_out (Kirchhoff's current law) [7] ΣV_loop = 0 (Kirchhoff's voltage law) [8] τ = RC (RC time constant)
Worked Examples
Example 1: Ohm's Law
Problem: A 12 V battery is connected to a 4 Ω resistor. What current flows through the resistor?
Solution:
- •Use Ohm's law: V = IR
- •Solve for I: I = V/R
- •I = 12/4 = 3 A
Example 2: Series Circuit
Problem: Three resistors (2 Ω, 3 Ω, and 5 Ω) are connected in series to a 20 V battery. Find the current and voltage across each resistor.
Solution:
- •Find equivalent resistance: R_eq = 2 + 3 + 5 = 10 Ω
- •Find total current: I = V/R_eq = 20/10 = 2 A
- •Voltage drops: V_1 = IR_1 = 2×2 = 4 V, V_2 = 2×3 = 6 V, V_3 = 2×5 = 10 V
- •Check: 4 + 6 + 10 = 20 V ✓
Example 3: Parallel Circuit
Problem: Two resistors (6 Ω and 3 Ω) are connected in parallel to a 12 V source. Find the total current.
Solution:
- •Equivalent resistance: 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2
- •R_eq = 2 Ω
- •Total current: I = V/R_eq = 12/2 = 6 A
Example 4: Power Calculation
Problem: A 60 W light bulb operates at 120 V. What is its resistance and the current through it?
Solution:
- •Find current: P = IV → I = P/V = 60/120 = 0.5 A
- •Find resistance: R = V/I = 120/0.5 = 240 Ω
- •Or directly: P = V²/R → R = V²/P = 14400/60 = 240 Ω
Explanation Patterns
- •Identify circuit type - Is it series, parallel, or combination?
- •Simplify the circuit - Combine series and parallel sections step by step
- •Find total current - Use V_source / R_equivalent
- •Work backward - Find voltage and current in each branch
- •Apply Kirchhoff's laws for complex circuits
- •Check your answer - Verify KCL at junctions and KVL around loops
Common Problem Types
- •Ohm's law calculations: Direct application of V = IR
- •Series circuits: Same current, voltages add
- •Parallel circuits: Same voltage, currents add
- •Combined circuits: Systematic simplification
- •Power calculations: Using P = IV, I²R, or V²/R
- •Kirchhoff's laws: Multi-loop circuit analysis
- •RC circuits: Time constant and exponential behavior