4bSeries and Parallel Circuits
Master the analysis of series and parallel resistor networks using Kirchhoff's laws and equivalent resistance techniques.
Kirchhoff's laws are the foundation of all circuit analysis. Every electrical engineer uses these laws daily to design circuits from simple LED flashlights to complex microprocessors. Understanding series and parallel circuits explains why your home is wired in parallel, why batteries can be combined, and how voltage dividers work in electronics.
Why do the lights in your home stay on when you turn off one lamp — and why would they all go out if they were wired in series?
Lesson Overview
In a series circuit, components are connected end-to-end in a single loop. The same current flows through all components; voltages add: V_total = V₁ + V₂ + …; equivalent resistance: R_eq = R₁ + R₂ + …. In a parallel circuit, components share the same two nodes. The same voltage appears across all branches; currents add: I_total = I₁ + I₂ + …; equivalent resistance: 1/R_eq = 1/R₁ + 1/R₂ + …. Kirchhoff's Voltage Law (KVL): the sum of all voltage changes around any closed loop is zero. Kirchhoff's Current Law (KCL): the sum of currents entering a node equals the sum leaving. These laws are consequences of energy conservation and charge conservation, respectively.
Key Equations
Worked Examples
Three resistors R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 10 Ω are connected in series to a 40 V battery. Find the equivalent resistance, current, and voltage across each resistor.
Three resistors R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω are connected in parallel to a 12 V battery. Find the equivalent resistance, total current, and current through each resistor.
A 12 V battery is connected to R₁ = 3 Ω in series with a parallel combination of R₂ = 6 Ω and R₃ = 6 Ω. Find the total current and the voltage across the parallel combination.
Apply KVL to a circuit with a 9 V battery, R₁ = 2 Ω, and R₂ = 7 Ω in series. Find the current.
A node in a circuit has three branches: 3.0 A flowing in, 1.5 A flowing in, and one unknown current flowing out. Find the unknown current (KCL).
Guided Problems
Four resistors of 5 Ω, 10 Ω, 15 Ω, and 20 Ω are connected in series to a 100 V source. Find the equivalent resistance, current, and voltage across the 15 Ω resistor.
Hint: R_eq = sum of all resistors. I = V/R_eq (same for all in series). V₁₅ = I × 15.
Two resistors of 8 Ω and 24 Ω are connected in parallel. Find the equivalent resistance using both the formula 1/R_eq = 1/R₁ + 1/R₂ and the shortcut R₁R₂/(R₁+R₂).
Hint: Both methods should give the same answer. 1/R_eq = 1/8 + 1/24 = 3/24 + 1/24 = 4/24 → R_eq = 6 Ω. Check: (8×24)/(8+24) = 192/32 = 6 Ω ✓
A circuit has a 24 V battery with R₁ = 4 Ω in series, and then R₂ = 6 Ω and R₃ = 12 Ω in parallel. Find the current through R₂.
Hint: Find R_parallel first. Then R_total = R₁ + R_parallel. Find I_total = V/R_total. The voltage across the parallel combination = I_total × R_parallel. Then I₂ = V_parallel/R₂.
Apply KCL at a node where currents of 2.0 A, 3.0 A, and 1.0 A enter, and currents of 4.0 A and I_x leave. Find I_x.
Hint: Sum of currents in = sum of currents out: 2.0 + 3.0 + 1.0 = 4.0 + I_x.
Explain why the equivalent resistance of a parallel combination is always less than the smallest individual resistance.
Hint: Adding a parallel path gives current more routes to flow — it reduces total opposition. Mathematically: 1/R_eq = 1/R₁ + 1/R₂ + … > 1/R_min, so R_eq < R_min.
Key Vocabulary
Series Circuit
A circuit in which components are connected end-to-end in a single path. Same current through all components. Voltages add. R_eq = R₁ + R₂ + …
Example: Christmas lights wired in series: if one bulb burns out, the entire string goes dark (circuit is broken).
Parallel Circuit
A circuit in which components are connected between the same two nodes, providing multiple current paths. Same voltage across all branches. Currents add. 1/R_eq = 1/R₁ + 1/R₂ + …
Example: Household outlets are wired in parallel: each appliance gets the full 120 V, and turning one off doesn't affect others.
Kirchhoff's Voltage Law (KVL)
The sum of all voltage changes around any closed loop in a circuit equals zero. Based on conservation of energy — a charge returning to its starting point has the same potential energy.
Example: In a loop with a 12 V battery and two resistors: +12 − V₁ − V₂ = 0, so V₁ + V₂ = 12 V.
Kirchhoff's Current Law (KCL)
The sum of currents entering a node (junction) equals the sum of currents leaving. Based on conservation of charge — charge cannot accumulate at a node.
Example: At a junction where 5 A enters from one branch and splits into two branches: I₁ + I₂ = 5 A.
Voltage Divider
A series circuit used to produce a fraction of the supply voltage. V_out = V_in × R₂/(R₁+R₂). Used extensively in electronics to set bias voltages.
Example: Two 10 kΩ resistors in series across 10 V: the midpoint voltage is 5 V (half the supply).
Node (Junction)
A point in a circuit where two or more branches meet. KCL applies at every node. The voltage at a node is well-defined (single value).
Example: In a parallel circuit, the two nodes connecting all the parallel branches are the points where KCL is applied.
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in series. The equivalent resistance is:
Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. The equivalent resistance is:
In a series circuit, which quantity is the same for all resistors?
KVL states that the sum of voltages around any closed loop is:
Two resistors of 4 Ω and 12 Ω are in parallel. Their equivalent resistance is:
Independent Practice
Four resistors of 10 Ω, 20 Ω, 30 Ω, and 40 Ω are connected in series to a 200 V battery. Find R_eq, I, and the voltage across each resistor.
Three resistors of 6 Ω, 12 Ω, and 4 Ω are connected in parallel to an 18 V battery. Find R_eq, total current, and current through each resistor.
A 30 V battery is connected to R₁ = 5 Ω in series with a parallel combination of R₂ = 10 Ω and R₃ = 15 Ω. Find the total current and the current through R₂.
Apply KVL to find the current in a circuit with a 15 V battery, R₁ = 3 Ω, and R₂ = 7 Ω in series. Then verify using KCL at each node.
★ A Wheatstone bridge circuit has R₁ = 100 Ω, R₂ = 200 Ω, R₃ = 150 Ω, and an unknown R₄. The bridge is balanced (no current through the galvanometer). Find R₄ using the balance condition R₁/R₂ = R₃/R₄.
ChallengeCommon Mistakes
Adding resistances in parallel like series resistors (R_eq = R₁ + R₂)
For parallel: 1/R_eq = 1/R₁ + 1/R₂. The equivalent resistance is LESS than the smallest resistor. For two equal resistors R in parallel: R_eq = R/2.
Assuming the same voltage across all resistors in a series circuit
In series, the CURRENT is the same. Voltages are different (V = IR, so larger R gets larger V). In parallel, the VOLTAGE is the same.
Forgetting to account for the direction of EMF when applying KVL
When traversing a battery from − to +, add +EMF. From + to −, add −EMF. For a resistor in the direction of current, add −IR. Against current, add +IR.
Thinking that adding a parallel resistor increases total resistance
Adding any parallel path always decreases total resistance (more paths for current). R_eq < smallest individual R.
Math Tips
Series: R_eq = R₁ + R₂ + … (always larger than any individual R). Parallel: 1/R_eq = 1/R₁ + 1/R₂ + … (always smaller than any individual R).
Two resistors in parallel shortcut: R_eq = R₁R₂/(R₁+R₂). For equal resistors R in parallel: R_eq = R/n (n resistors).
Voltage divider: V₁ = V_total × R₁/(R₁+R₂). The larger resistor gets the larger voltage share.
KVL: assign current directions, then go around loop. Battery + to −: −EMF. Resistor in current direction: −IR. Sum = 0.