Series and Parallel Circuits

Series circuits, parallel circuits, combination circuits, equivalent resistance

⚡ Series and Parallel Circuits

Series Circuits

Components connected in a line (same path).

Rules:

Same current through all components: $I_1 = I_2 = I_3 = ...$
Voltages add: $V_{total} = V_1 + V_2 + V_3 + ...$
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:

Same voltage across all components: $V_1 = V_2 = V_3 = ...$
Currents add: $I_{total} = I_1 + I_2 + I_3 + ...$
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

| Property | Series | Parallel | |----------|--------|----------| | Current | Same (I₁ = I₂) | Adds (I_T = I₁ + I₂) | | Voltage | Adds (V_T = V₁ + V₂) | Same (V₁ = V₂) | | Resistance | Adds (R_eq = R₁ + R₂) | Reciprocal (1/R_eq = 1/R₁ + 1/R₂) | | R_eq vs individual | R_eq > any R_i | R_eq < any R_i |

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:

Identify series/parallel groups
Simplify step by step
Find equivalent resistance
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

Draw circuit diagram (label knowns)
Identify series/parallel
Find R_eq (simplify step-by-step)
Find total current: $I_{total} = V/R_{eq}$
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!)

📚 Practice Problems

1Problem 1easy

❓ Question:

Three resistors (2 Ω, 4 Ω, 6 Ω) are connected in series to a 12 V battery. Find (a) equivalent resistance, (b) total current, (c) voltage across each resistor.

💡 Show Solution

Given:

$R_1 = 2$ Ω, $R_2 = 4$ Ω, $R_3 = 6$ Ω (series)
$V = 12$ V

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

2Problem 2easy

❓ Question:

Three resistors (2 Ω, 4 Ω, 6 Ω) are connected in series to a 12 V battery. Find (a) equivalent resistance, (b) total current, (c) voltage across each resistor.

💡 Show Solution

Given:

$R_1 = 2$ Ω, $R_2 = 4$ Ω, $R_3 = 6$ Ω (series)
$V = 12$ V

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

3Problem 3medium

❓ Question:

Three resistors (R₁ = 2.0 Ω, R₂ = 4.0 Ω, R₃ = 6.0 Ω) are connected in series to a 24 V battery. (a) Find the equivalent resistance. (b) Find the current. (c) Find the voltage across each resistor. (d) Find the power dissipated by R₂.

💡 Show Solution

Solution:

Series connection: Same current through all resistors

(a) Equivalent resistance: R_eq = R₁ + R₂ + R₃ = 2.0 + 4.0 + 6.0 = 12 Ω

(b) Current: I = V/R_eq = 24/12 = 2.0 A (same through all)

(c) Voltage across each: V₁ = IR₁ = 2.0 × 2.0 = 4.0 V V₂ = IR₂ = 2.0 × 4.0 = 8.0 V V₃ = IR₃ = 2.0 × 6.0 = 12 V

Check: 4.0 + 8.0 + 12 = 24 V ✓

(d) Power in R₂: P₂ = I²R₂ = (2.0)²(4.0) = 16 W

4Problem 4medium

❓ Question:

💡 Show Solution

Solution:

Series connection: Same current through all resistors

(a) Equivalent resistance: R_eq = R₁ + R₂ + R₃ = 2.0 + 4.0 + 6.0 = 12 Ω

(b) Current: I = V/R_eq = 24/12 = 2.0 A (same through all)

(c) Voltage across each: V₁ = IR₁ = 2.0 × 2.0 = 4.0 V V₂ = IR₂ = 2.0 × 4.0 = 8.0 V V₃ = IR₃ = 2.0 × 6.0 = 12 V

Check: 4.0 + 8.0 + 12 = 24 V ✓

(d) Power in R₂: P₂ = I²R₂ = (2.0)²(4.0) = 16 W

5Problem 5medium

❓ Question:

Three resistors (6 Ω, 3 Ω, 2 Ω) are connected in parallel to a 12 V battery. Find (a) equivalent resistance, (b) total current, (c) current through each resistor.

💡 Show Solution

Given:

$R_1 = 6$ Ω, $R_2 = 3$ Ω, $R_3 = 2$ Ω (parallel)
$V = 12$ 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

6Problem 6medium

❓ Question:

Three resistors (6 Ω, 3 Ω, 2 Ω) are connected in parallel to a 12 V battery. Find (a) equivalent resistance, (b) total current, (c) current through each resistor.

💡 Show Solution

Given:

$R_1 = 6$ Ω, $R_2 = 3$ Ω, $R_3 = 2$ Ω (parallel)
$V = 12$ V

Part (a): Equivalent resistance

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

7Problem 7hard

❓ Question:

Three resistors (R₁ = 6.0 Ω, R₂ = 3.0 Ω, R₃ = 2.0 Ω) are connected in parallel to a 12 V battery. (a) Find the equivalent resistance. (b) Find the current through each resistor. (c) Find the total current from the battery. (d) Find the total power.

💡 Show Solution

Solution:

Parallel connection: Same voltage across all resistors (12 V)

(a) Equivalent resistance: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ 1/R_eq = 1/6.0 + 1/3.0 + 1/2.0 = 1/6 + 2/6 + 3/6 = 6/6 = 1 R_eq = 1.0 Ω

(b) Current through each: I₁ = V/R₁ = 12/6.0 = 2.0 A I₂ = V/R₂ = 12/3.0 = 4.0 A I₃ = V/R₃ = 12/2.0 = 6.0 A

Or: I_total = V/R_eq = 12/1.0 = 12 A ✓

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

Or: P = V²/R_eq = 144/1.0 = 144 W ✓

8Problem 8hard

❓ Question:

Two resistors R₁ = 4 Ω and R₂ = 8 Ω are in series. This combination is in parallel with R₃ = 6 Ω. The whole circuit is connected to a 12 V battery. Find the total current.

💡 Show Solution

Given:

$R_1 = 4$ Ω, $R_2 = 8$ Ω (series)
$R_3 = 6$ Ω (parallel with series combo)
$V = 12$ V

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

9Problem 9hard

❓ Question:

Two resistors R₁ = 4 Ω and R₂ = 8 Ω are in series. This combination is in parallel with R₃ = 6 Ω. The whole circuit is connected to a 12 V battery. Find the total current.

💡 Show Solution

Given:

$R_1 = 4$ Ω, $R_2 = 8$ Ω (series)
$R_3 = 6$ Ω (parallel with series combo)
$V = 12$ V

Solution:

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

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

10Problem 10hard

❓ Question:

💡 Show Solution

Solution:

Parallel connection: Same voltage across all resistors (12 V)

(a) Equivalent resistance: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ 1/R_eq = 1/6.0 + 1/3.0 + 1/2.0 = 1/6 + 2/6 + 3/6 = 6/6 = 1 R_eq = 1.0 Ω

(b) Current through each: I₁ = V/R₁ = 12/6.0 = 2.0 A I₂ = V/R₂ = 12/3.0 = 4.0 A I₃ = V/R₃ = 12/2.0 = 6.0 A

Or: I_total = V/R_eq = 12/1.0 = 12 A ✓

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

Or: P = V²/R_eq = 144/1.0 = 144 W ✓

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