Or equivalently: $\sum I = 0$

(taking current into junction as positive)

Physical Basis:

Conservation of charge - charge cannot accumulate at a point.

💡 Example: If 2 A enters junction, 2 A must leave (maybe 1 A + 1 A in two branches)

Loop Rule (Voltage Law)

Around any closed loop, sum of voltages = 0

$\sum V = 0$

Sign Convention:

• Across resistor (with current): -IR (voltage drop)
• Across resistor (against current): +IR (voltage rise)
• Across battery (+ to -): +ε (voltage rise)
• Across battery (- to +): -ε (voltage drop)

Physical Basis:

Conservation of energy - energy gained = energy lost in complete loop.

Problem-Solving Strategy

Step 1: Label all currents (I₁, I₂, I₃...) with arrows

Step 2: Apply junction rule at each junction

• Write equations: ΣI = 0

Step 3: Apply loop rule for independent loops

• Choose direction (clockwise or counter-clockwise)
• Follow loop, add voltages with signs

Step 4: Solve system of equations

• Typically: n unknowns → n equations needed

Step 5: Check answer

• If current is negative, actual direction is opposite

Example: Multi-Loop Circuit

    +---R₁---I₁--->---+
    |                 |
   ε₁                R₂
    |        I₂       |
    +-------<---------+
            |
           R₃ (I₃ = I₁ - I₂)
            |

Junction: I₁ = I₂ + I₃

Loop 1 (left): ε₁ - I₁R₁ - I₃R₃ = 0

Loop 2 (right): I₃R₃ - I₂R₂ = 0

Solve 3 equations, 3 unknowns!

Internal Resistance

Real batteries have internal resistance r:

$V_{terminal} = ε - Ir$

where:

• ε = EMF (electromotive force, ideal voltage)
• I = current through battery
• r = internal resistance

Load connected: V < ε (voltage drop across r) No load (open circuit): V = ε

RC Circuits (Capacitors)

When capacitor charges through resistor:

$q(t) = Q_{max}(1 - e^{-t/RC})$

$I(t) = I_0 e^{-t/RC}$

Time constant: $\tau = RC$

• After τ: ~63% charged
• After 5τ: ~99% charged

Discharging: $q(t) = Q_0 e^{-t/RC}$

Measuring Instruments

Ammeter:

• Measures current
• Connected in series
• Low internal resistance (ideally 0)

Voltmeter:

• Measures voltage
• Connected in parallel
• High internal resistance (ideally ∞)

Ohmmeter:

• Measures resistance
• Device must be disconnected from circuit

Power Distribution

Why high voltage for transmission?

Power loss in wires: $P_{loss} = I^2 R_{wire}$

For same power transmitted (P = IV):

• Higher V → lower I → much less loss!
• Step-up transformers at power plant
• Step-down transformers at homes

Fuses and Circuit Breakers

Fuse: Thin wire melts if I > rated value Circuit breaker: Switch opens if I > rated value

Both prevent overcurrent → prevent fires!

Household: Typically 15 A or 20 A circuits

Grounding

Ground: Connection to Earth (V = 0 reference)

Safety ground (3rd prong):

• Connects metal case to ground
• If short to case → current flows to ground, not through you!

Common Mistakes

❌ Wrong sign convention in loop rule ❌ Not using all independent loops ❌ Forgetting junction rule ❌ Not accounting for internal resistance ❌ Connecting ammeter in parallel (burns out!) ❌ Connecting voltmeter in series (reads wrong)

Actually, let's define properly:

• I₁: clockwise in left loop
• I₂: clockwise in right loop
• Current through shared 3Ω: I₁ + I₂ (if both clockwise)

Loop 1 (left, clockwise): $12 - I_1(3) = 0$ $I_1 = 4.0 \text{ A}$

Wait, this needs junction. Let me redefine:

Proper setup with junction:

Actually this needs clearer diagram. Let me solve general case:

Loop 1: $12 - 3I_1 = 0$ → $I_1 = 4$ A Loop 2: $6 - 2I_2 = 0$ → $I_2 = 3$ A

(If loops are independent - series configuration)

Answer: Depends on exact configuration. General method:

1. Label currents
2. Junction: ΣI = 0
3. Loops: ΣV = 0
4. Solve system