Rules:

1. Same current through all components: $I_1 = I_2 = I_3 = ...$
2. Voltages add: $V_{total} = V_1 + V_2 + V_3 + ...$
3. Resistances add: $R_{eq} = R_1 + R_2 + R_3 + ...$

$V = IR_{eq}$

💡 Key: Current is the same everywhere in series!

Voltage Divider:

$V_i = V_{total} \frac{R_i}{R_{eq}}$

Larger resistance gets larger voltage drop.

Parallel Circuits

Components connected across same points (multiple paths).

Rules:

1. Same voltage across all components: $V_1 = V_2 = V_3 = ...$
2. Currents add: $I_{total} = I_1 + I_2 + I_3 + ...$
3. Reciprocal resistances add: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$

💡 Key: Voltage is the same across each branch in parallel!

Special Case (2 resistors):

$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

"Product over sum"

Comparison Table

Current Divider (Parallel)

$I_i = I_{total} \frac{R_{eq}}{R_i}$

Smaller resistance gets larger current! (Inverse relationship)

Combination Circuits

Mix of series and parallel:

Strategy:

1. Identify series/parallel groups
2. Simplify step by step
3. Find equivalent resistance
4. Work backwards to find I and V

Christmas Lights Example

Old (Series): One bulb out → all out! Same current, so if one breaks (open circuit), all go dark.

New (Parallel): One bulb out → others stay on! Each has same voltage, independent paths.

Power in Series vs Parallel

Series: $P_i = I^2 R_i$ (same I, so larger R gets more power)

Parallel: $P_i = \frac{V^2}{R_i}$ (same V, so smaller R gets more power)

Short Circuit

When R ≈ 0 path is created:

• Current → ∞ (theoretically)
• Dangerous! Can cause fire
• Circuit breakers/fuses protect

Open Circuit

When path is broken:

• Current = 0
• Voltmeter measures across (high R, parallel)
• Ammeter measures through (low R, series)

Problem-Solving Strategy

1. Draw circuit diagram (label knowns)
2. Identify series/parallel
3. Find R_eq (simplify step-by-step)
4. Find total current: $I_{total} = V/R_{eq}$
5. Work backwards:
   • Series: Same I, use V = IR for each
   • Parallel: Same V, use I = V/R for each

Common Mistakes

❌ Using wrong formula (series R adds, parallel 1/R adds) ❌ Forgetting current is same in series ❌ Forgetting voltage is same in parallel ❌ Not simplifying combination circuits step-by-step ❌ Connecting ammeter in parallel (should be series!) ❌ Connecting voltmeter in series (should be parallel!)

Part (a): Equivalent resistance

Series: Resistances add $R_{eq} = R_1 + R_2 + R_3 = 2 + 4 + 6 = 12 \text{ Ω}$

Part (b): Total current

$I = \frac{V}{R_{eq}} = \frac{12}{12} = 1.0 \text{ A}$

Part (c): Voltage across each

Series: Same current through each $V_1 = IR_1 = (1.0)(2) = 2.0 \text{ V}$ $V_2 = IR_2 = (1.0)(4) = 4.0 \text{ V}$ $V_3 = IR_3 = (1.0)(6) = 6.0 \text{ V}$

Check: $V_1 + V_2 + V_3 = 2 + 4 + 6 = 12$ V ✓

Answer:

• (a) R_eq = 12 Ω
• (b) I = 1.0 A
• (c) V₁ = 2.0 V, V₂ = 4.0 V, V₃ = 6.0 V

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2}$ $\frac{1}{R_{eq}} = \frac{1 + 2 + 3}{6} = \frac{6}{6} = 1$ $R_{eq} = 1.0 \text{ Ω}$

Part (b): Total current

$I_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}$

Part (c): Current through each

Parallel: Same voltage across each $I_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A}$ $I_2 = \frac{V}{R_2} = \frac{12}{3} = 4.0 \text{ A}$ $I_3 = \frac{V}{R_3} = \frac{12}{2} = 6.0 \text{ A}$

Check: $I_1 + I_2 + I_3 = 2 + 4 + 6 = 12$ A ✓

Answer:

• (a) R_eq = 1.0 Ω (less than any individual!)
• (b) I_total = 12 A
• (c) I₁ = 2.0 A, I₂ = 4.0 A, I₃ = 6.0 A

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ $\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2}$ $\frac{1}{R_{eq}} = \frac{1 + 2 + 3}{6} = \frac{6}{6} = 1$ $R_{eq} = 1.0 \text{ Ω}$

Part (b): Total current

$I_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}$

Part (c): Current through each

Parallel: Same voltage across each $I_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A}$ $I_2 = \frac{V}{R_2} = \frac{12}{3} = 4.0 \text{ A}$ $I_3 = \frac{V}{R_3} = \frac{12}{2} = 6.0 \text{ A}$

Check: $I_1 + I_2 + I_3 = 2 + 4 + 6 = 12$ A ✓

Answer:

• (a) R_eq = 1.0 Ω (less than any individual!)
• (b) I_total = 12 A
• (c) I₁ = 2.0 A, I₂ = 4.0 A, I₃ = 6.0 A

Solution:

Step 1: Find equivalent of series combination. $R_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}$

Step 2: This 12 Ω is in parallel with 6 Ω. $\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3}$ $\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4}$ $R_{eq} = 4.0 \text{ Ω}$

Or using product-over-sum: $R_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}$

Step 3: Total current from battery. $I_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}$

Answer: I_total = 3.0 A

(d) Total power: P_total = VI_total = 12 × 12 = 144 W

Or: P = V²/R_eq = 144/1.0 = 144 W ✓

Solution:

Step 1: Find equivalent of series combination. $R_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}$

Step 2: This 12 Ω is in parallel with 6 Ω. $\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3}$ $\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4}$ $R_{eq} = 4.0 \text{ Ω}$

Or using product-over-sum: $R_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}$

Step 3: Total current from battery. $I_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}$

Answer: I_total = 3.0 A