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Series circuits, parallel circuits, combination circuits, equivalent resistance
Learn step-by-step with practice exercises built right in.
Components connected in a line (same path).
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.
Given:
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💡 Key: Current is the same everywhere in series!
Larger resistance gets larger voltage drop.
Components connected across same points (multiple paths).
💡 Key: Voltage is the same across each branch in parallel!
"Product over sum"
| 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 |
Smaller resistance gets larger current! (Inverse relationship)
Mix of series and parallel:
Strategy:
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.
Series: (same I, so larger R gets more power)
Parallel: (same V, so smaller R gets more power)
When R ≈ 0 path is created:
When path is broken:
❌ 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
Part (b): Total current
Part (c): Voltage across each
Series: Same current through each
Check: V ✓
Answer:
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.
Given:
Part (a): Equivalent resistance
Series: Resistances add
Part (b): Total current
Part (c): Voltage across each
Series: Same current through each
Check: V ✓
Answer:
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.
Given:
Part (a): Equivalent resistance
Parallel: Reciprocal resistances add
Part (b): Total current
Part (c): Current through each
Parallel: Same voltage across each
Check: A ✓
Answer:
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.
Given:
Part (a): Equivalent resistance
Parallel: Reciprocal resistances add
Part (b): Total current
Part (c): Current through each
Parallel: Same voltage across each
Check: A ✓
Answer:
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.
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
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.
Given:
Solution:
Step 1: Find equivalent of series combination.
Step 2: This 12 Ω is in parallel with 6 Ω.
Or using product-over-sum:
Step 3: Total current from battery.
Answer: I_total = 3.0 A
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.
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 ✓
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.
Given:
Solution:
Step 1: Find equivalent of series combination.
Step 2: This 12 Ω is in parallel with 6 Ω.
Or using product-over-sum:
Step 3: Total current from battery.
Answer: I_total = 3.0 A