GEK-2 — DC Circuit Relationships
Key Takeaways
- A series path has one current, additive resistances, and voltage drops that sum to source voltage.
- Parallel branches share the same voltage, branch currents add, and equivalent resistance is less than the smallest branch resistance.
- Kirchhoff's current and voltage checks expose setup errors in series, parallel, and combination circuits.
- Open circuits interrupt a path, shorts bypass impedance, and conductor voltage-drop formulas require a clearly stated length and material assumption.
Start by drawing the current paths
A circuit calculation becomes easier when the diagram is reduced to nodes and paths. Mark the source voltage, each resistance, and the requested quantity. Assume ideal conductors and a steady DC source unless the problem supplies conductor resistance or another condition. Then decide whether components are in series, parallel, or a combination.
Components are in series only when the same current must pass through each component with no branching node between them. Components are in parallel when both ends connect to the same two nodes, so each branch has the same voltage. Physical appearance does not control; electrical connections do.
Series relationships
For series resistors:
- Current is the same through every resistor.
- Equivalent resistance is
R_T = R₁ + R₂ + .... - Individual voltage drops are
V_n = I × R_n. - The sum of all voltage drops equals source voltage.
Consider an ideal 24 V DC source with 4 Ω and 8 Ω resistors in series. Total resistance is R_T = 4 Ω + 8 Ω = 12 Ω. Circuit current is I = V/R_T = 24 V ÷ 12 Ω = 2 A. The drops are V₁ = 2 A × 4 Ω = 8 V and V₂ = 2 A × 8 Ω = 16 V. Check: 8 V + 16 V = 24 V. The larger resistance receives the larger voltage drop because series current is common.
An open anywhere in this single path makes steady current zero throughout the circuit. With zero current, intact resistors have zero IR drop; the source voltage can appear across the open. A short circuit is an unintended path with very low impedance. A short placed across one resistor bypasses it, reducing total resistance and increasing source current. Real source, conductor, and protective-device impedance limits the current; “zero ohms” is an ideal model, not a safe field condition.
Parallel relationships
For parallel resistors:
- Voltage is the same across every branch.
- Branch current is
I_n = V/R_n. - Total current is the sum of branch currents.
- Equivalent resistance follows
1/R_T = 1/R₁ + 1/R₂ + ....
For two resistors, the product-over-sum shortcut is R_T = (R₁R₂)/(R₁ + R₂). For two equal resistors, R_T = R/2. Parallel equivalent resistance must be less than the smallest branch resistance because another current path has been added.
Example: connect 30 Ω and 20 Ω branches across an ideal 120 V DC source. Branch currents are I₁ = 120 V ÷ 30 Ω = 4 A and I₂ = 120 V ÷ 20 Ω = 6 A. Total current is I_T = 4 A + 6 A = 10 A. Equivalent resistance is R_T = 120 V ÷ 10 A = 12 Ω. Check with product over sum: (30 Ω × 20 Ω) ÷ (30 Ω + 20 Ω) = 600 Ω² ÷ 50 Ω = 12 Ω. Total power is P = VI = 120 V × 10 A = 1,200 W; branch powers of 480 W and 720 W also sum to 1,200 W.
An open branch stops current only in that branch if the other parallel paths remain intact. A short across the source is much more severe: it creates a low-impedance source path and can produce high fault current. Never treat a paper short-circuit exercise as permission to test a suspected short by energizing it. Follow applicable OSHA procedures and use properly rated test equipment.
Combination circuits
Reduce one clear group at a time. Suppose a 6 Ω resistor is in series with two 12 Ω resistors in parallel across an ideal 24 V DC source. First reduce the equal parallel pair: 12 Ω ÷ 2 = 6 Ω. Total resistance is then 6 Ω + 6 Ω = 12 Ω, and source current is 24 V ÷ 12 Ω = 2 A.
The series 6 Ω resistor drops 2 A × 6 Ω = 12 V, leaving 12 V across the parallel group. Each 12 Ω branch therefore carries 12 V ÷ 12 Ω = 1 A. The branch currents add to the 2 A entering the node. Do not apply the 2 A total current to each parallel resistor.
These checks are Kirchhoff's laws in practical form. Kirchhoff's current law says current entering a node equals current leaving it. Kirchhoff's voltage law says the algebraic sum of voltage rises and drops around a closed loop is zero.
Conductor voltage drop
A common two-wire DC or single-phase approximation is V_D = 2KID/CM, where K is conductor resistivity in ohm-cmil per foot, I is amperes, D is one-way length in feet, and CM is circular-mil area. The factor 2 accounts for outgoing and returning conductors.
Assume a 120 V, two-wire circuit carries 20 A over a one-way distance of 100 ft using 10 AWG copper with CM = 10,380 and K = 12.9 Ω-cmil/ft. Then V_D = (2 × 12.9 Ω-cmil/ft × 20 A × 100 ft) ÷ 10,380 cmil = 4.97 V. Percentage drop is (4.97 V ÷ 120 V) × 100 = 4.14%. This is an approximation because conductor temperature and connections affect actual resistance. State whether length is one-way; doubling an already round-trip length would double-count the path.
Before accepting any result, ask whether series drops add to the source, parallel branch currents add to total, and equivalent resistance falls on the correct side of the component values.
Diagnostic checkpoint
When a circuit answer seems wrong, return to the drawing before repeating arithmetic. Confirm which nodes are shared, mark the complete current paths, and decide whether the suspected condition is an open path, a bypassing short, or normal load resistance.
Two resistors, 6 Ω and 9 Ω, are in series across an ideal 30 V DC source. What current flows?
Two equal 16 Ω resistors are connected in parallel. What is their equivalent resistance?
A 120 V ideal source supplies parallel branches of 60 Ω and 30 Ω. What is total current?
In a parallel circuit, one branch opens while another intact branch remains connected to an ideal voltage source. What happens?