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Series and Parallel Circuits

Series circuits, parallel circuits, combination circuits, equivalent resistance

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⚡ Series and Parallel Circuits

Series Circuits

Components connected in a line (same path).

Rules:

  1. Same current through all components: I1=I2=I3=...I_1 = I_2 = I_3 = ...
  2. Voltages add: Vtotal=V1+V2+V3+...V_{total} = V_1 + V_2 + V_3 + ...
  3. Resistances add: Req=R1+R2+R3+...R_{eq} = R_1 + R_2 + R_3 + ...

V=IReqV = IR_{eq}

💡 Key: Current is the same everywhere in series!

Voltage Divider:

Vi=VtotalRiReqV_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: V1=V2=V3=...V_1 = V_2 = V_3 = ...
  2. Currents add: Itotal=I1+I2+I3+...I_{total} = I_1 + I_2 + I_3 + ...
  3. Reciprocal resistances add: 1Req=1R1+1R2+1R3+...\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):

Req=R1R2R1+R2R_{eq} = \frac{R_1 R_2}{R_1 + R_2}

"Product over sum"

Comparison Table

PropertySeriesParallel
CurrentSame (I₁ = I₂)Adds (I_T = I₁ + I₂)
VoltageAdds (V_T = V₁ + V₂)Same (V₁ = V₂)
ResistanceAdds (R_eq = R₁ + R₂)Reciprocal (1/R_eq = 1/R₁ + 1/R₂)
R_eq vs individualR_eq > any R_iR_eq < any R_i

Current Divider (Parallel)

Ii=ItotalReqRiI_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: Pi=I2RiP_i = I^2 R_i (same I, so larger R gets more power)

Parallel: Pi=V2RiP_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: Itotal=V/ReqI_{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!)

📚 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:

  • R1=2R_1 = 2 Ω, R2=4R_2 = 4 Ω, R3=6R_3 = 6 Ω (series)
  • V=12V = 12 V

Part (a): Equivalent resistance

Series: Resistances add Req=R1+R2+R3=2+4+6=12 ΩR_{eq} = R_1 + R_2 + R_3 = 2 + 4 + 6 = 12 \text{ Ω}

Part (b): Total current

I=VReq=1212=1.0 AI = \frac{V}{R_{eq}} = \frac{12}{12} = 1.0 \text{ A}

Part (c): Voltage across each

Series: Same current through each V1=IR1=(1.0)(2)=2.0 VV_1 = IR_1 = (1.0)(2) = 2.0 \text{ V} V2=IR2=(1.0)(4)=4.0 VV_2 = IR_2 = (1.0)(4) = 4.0 \text{ V} V3=IR3=(1.0)(6)=6.0 VV_3 = IR_3 = (1.0)(6) = 6.0 \text{ V}

Check: V1+V2+V3=2+4+6=12V_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:

  • R1=2R_1 = 2 Ω, R2=4R_2 = 4 Ω, R3=6R_3 = 6 Ω (series)
  • V=12V = 12 V

Part (a): Equivalent resistance

Series: Resistances add Req=R1+R2+R3=2+4+6=12 ΩR_{eq} = R_1 + R_2 + R_3 = 2 + 4 + 6 = 12 \text{ Ω}

Part (b): Total current

I=VReq=1212=1.0 AI = \frac{V}{R_{eq}} = \frac{12}{12} = 1.0 \text{ A}

Part (c): Voltage across each

Series: Same current through each V1=IR1=(1.0)(2)=2.0 VV_1 = IR_1 = (1.0)(2) = 2.0 \text{ V} V2=IR2=(1.0)(4)=4.0 VV_2 = IR_2 = (1.0)(4) = 4.0 \text{ V} V3=IR3=(1.0)(6)=6.0 VV_3 = IR_3 = (1.0)(6) = 6.0 \text{ V}

Check: V1+V2+V3=2+4+6=12V_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 (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:

  • R1=6R_1 = 6 Ω, R2=3R_2 = 3 Ω, R3=2R_3 = 2 Ω (parallel)
  • V=12V = 12 V

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add 1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} 1Req=16+13+12\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} 1Req=1+2+36=66=1\frac{1}{R_{eq}} = \frac{1 + 2 + 3}{6} = \frac{6}{6} = 1 Req=1.0 ΩR_{eq} = 1.0 \text{ Ω}

Part (b): Total current

Itotal=VReq=121.0=12 AI_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}

Part (c): Current through each

Parallel: Same voltage across each I1=VR1=126=2.0 AI_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A} I2=VR2=123=4.0 AI_2 = \frac{V}{R_2} = \frac{12}{3} = 4.0 \text{ A} I3=VR3=122=6.0 AI_3 = \frac{V}{R_3} = \frac{12}{2} = 6.0 \text{ A}

Check: I1+I2+I3=2+4+6=12I_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

4Problem 4medium

❓ 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:

  • R1=6R_1 = 6 Ω, R2=3R_2 = 3 Ω, R3=2R_3 = 2 Ω (parallel)
  • V=12V = 12 V

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add 1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} 1Req=16+13+12\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} 1Req=1+2+36=66=1\frac{1}{R_{eq}} = \frac{1 + 2 + 3}{6} = \frac{6}{6} = 1 Req=1.0 ΩR_{eq} = 1.0 \text{ Ω}

Part (b): Total current

Itotal=VReq=121.0=12 AI_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}

Part (c): Current through each

Parallel: Same voltage across each I1=VR1=126=2.0 AI_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A} I2=VR2=123=4.0 AI_2 = \frac{V}{R_2} = \frac{12}{3} = 4.0 \text{ A} I3=VR3=122=6.0 AI_3 = \frac{V}{R_3} = \frac{12}{2} = 6.0 \text{ A}

Check: I1+I2+I3=2+4+6=12I_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

5Problem 5medium

❓ 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

6Problem 6hard

❓ 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:

  • R1=4R_1 = 4 Ω, R2=8R_2 = 8 Ω (series)
  • R3=6R_3 = 6 Ω (parallel with series combo)
  • V=12V = 12 V

Solution:

Step 1: Find equivalent of series combination. Rseries=R1+R2=4+8=12 ΩR_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}

Step 2: This 12 Ω is in parallel with 6 Ω. 1Req=1Rseries+1R3\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3} 1Req=112+16=1+212=312=14\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4} Req=4.0 ΩR_{eq} = 4.0 \text{ Ω}

Or using product-over-sum: Req=(12)(6)12+6=7218=4.0 ΩR_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}

Step 3: Total current from battery. Itotal=VReq=124.0=3.0 AI_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}

Answer: I_total = 3.0 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

(c) Total current: I_total = I₁ + I₂ + I₃ = 2.0 + 4.0 + 6.0 = 12 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:

  • R1=4R_1 = 4 Ω, R2=8R_2 = 8 Ω (series)
  • R3=6R_3 = 6 Ω (parallel with series combo)
  • V=12V = 12 V

Solution:

Step 1: Find equivalent of series combination. Rseries=R1+R2=4+8=12 ΩR_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}

Step 2: This 12 Ω is in parallel with 6 Ω. 1Req=1Rseries+1R3\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3} 1Req=112+16=1+212=312=14\frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{6} = \frac{1 + 2}{12} = \frac{3}{12} = \frac{1}{4} Req=4.0 ΩR_{eq} = 4.0 \text{ Ω}

Or using product-over-sum: Req=(12)(6)12+6=7218=4.0 ΩR_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}

Step 3: Total current from battery. Itotal=VReq=124.0=3.0 AI_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}

Answer: I_total = 3.0 A

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📌 Related Topics in Electric Circuits

Current, Resistance, and Ohm's LawKirchhoff's Laws

❓ Frequently Asked Questions

What is Series and Parallel Circuits?▾
Series circuits, parallel circuits, combination circuits, equivalent resistance
How can I study Series and Parallel Circuits effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers Series and Parallel Circuits?▾
Series and Parallel Circuits is part of the AP Physics 2 course on Study Mondo, specifically in the Electric Circuits section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Series and Parallel Circuits?▾
Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.