Circuit Types
Series and parallel circuits behave oppositely, learn the two sets of rules and you can solve any EI circuit question the ASVAB throws at you.
Formula Reference
- Series total resistance: R_total = R1 + R2 + R3 + ...
- Series current: same through every component (I is constant)
- Series voltage: V_total = V1 + V2 + V3 (voltages add up)
- Parallel total resistance: 1/R_total = 1/R1 + 1/R2 + 1/R3 (total R is always less than smallest branch)
- Parallel voltage: same across every branch (V is constant)
- Parallel current: I_total = I1 + I2 + I3 (currents add up)
What the ASVAB is actually testing
Circuit-type questions are some of the most calculation-heavy on EI. They give you a circuit configuration and ask about total resistance, current, or voltage. The key is knowing which of the two rule sets applies, series or parallel, before you start calculating.
Series circuits: everything in one loop
Same current everywhere; resistances add: R = R1 + R2
In a series circuit, components are connected end-to-end in a single loop. There's only one path for current.
Rules for series:
- Current is the same everywhere, one loop means identical flow throughout
- Voltage divides, each component gets a share of the total voltage; they add up to the source
- Resistance adds, R_total = R1 + R2 + R3 + ...
Consequence: if any component fails (opens), the entire circuit goes dark, like old-style Christmas lights.
Parallel circuits: multiple branches
Same voltage across each branch; 1/R = 1/R1 + 1/R2
In a parallel circuit, components are connected side-by-side, sharing the same two connection points. Current has multiple paths to choose from.
Rules for parallel:
- Voltage is the same across every branch, each branch sees the full source voltage
- Current divides, more current flows through lower-resistance branches; branch currents add to the total
- Total resistance decreases, adding more branches always lowers total resistance
This is how household wiring works, each outlet gets full voltage, and appliances operate independently.
Two-resistor parallel shortcut
For exactly two resistors in parallel, skip the reciprocal formula:
R_total = (R1 × R2) / (R1 + R2)
Two 6 Ω resistors in parallel: (6 × 6) / (6 + 6) = 36/12 = 3 Ω
That's always exactly half of either resistor when they're equal.
How to attack circuit questions
R = R1 + R2 · I = V ÷ R
- Identify: series (single loop) or parallel (multiple branches)?
- Write down the appropriate rule set
- Solve for total resistance first, then use V = IR for current or voltage
If the question mentions a short circuit or open circuit: an open (broken wire) stops current; a short (zero resistance path) causes maximum current and often damages the circuit.
Study approach
Draw a simple series and parallel circuit diagram from memory. Then practice each rule set with two and three resistors until you can execute the calculation in under 30 seconds. Speed matters on EI.
Common Pitfalls
- ⚠Thinking total resistance increases when you add a parallel branch, it always decreases (more paths = less total resistance)
- ⚠Applying series voltage rules to a parallel circuit or vice versa, identify the circuit type first
- ⚠Forgetting that in a series circuit, one open component breaks the entire circuit; in parallel, other branches still function
- ⚠Using 1/R_total = 1/R1 + 1/R2 and stopping there, you must take the reciprocal of that sum to get R_total
- ⚠Confusing an open with a short: an open (broken path) stops all current, while a short (near-zero resistance path) causes maximum current and can damage the circuit or blow a fuse
- ⚠Forgetting Kirchhoff's laws as a check: in series the individual voltage drops must add up to the source voltage (KVL), and in parallel the branch currents must add up to the total current (KCL)
Worked Examples
Q1: Two resistors, 6 Ω and 4 Ω, are connected in series to a 20 V battery. What is the total current?
Answer: Series: R_total = 6 + 4 = 10 Ω. I = V/R = 20 V ÷ 10 Ω = 2 A
Q2: The same two resistors (6 Ω and 4 Ω) are now connected in parallel to a 12 V battery. What is the total resistance?
Answer: 1/R_total = 1/6 + 1/4 = 2/12 + 3/12 = 5/12. R_total = 12/5 = 2.4 Ω
Q3: In a parallel circuit with two branches drawing 3 A and 5 A respectively, what is the total current from the battery?
Answer: Parallel: currents add. I_total = 3 + 5 = 8 A
Q4: Three resistors of 5 Ω, 10 Ω, and 15 Ω are in series with a 30 V source. What is the voltage drop across the 10 Ω resistor?
Answer: Series R_total = 5 + 10 + 15 = 30 Ω. I = 30 V ÷ 30 Ω = 1 A. Drop across 10 Ω is V = IR = 1 A × 10 Ω = 10 V. The three drops (5 V, 10 V, 15 V) add to the 30 V source, which is Kirchhoff's voltage law.
Q5: A series-parallel circuit has a 10 Ω resistor in series with two 10 Ω resistors that are in parallel with each other. What is the total resistance?
Answer: Reduce the parallel pair first: two equal 10 Ω in parallel = 10 ÷ 2 = 5 Ω. Then add the series resistor: 5 + 10 = 15 Ω total. Always collapse the parallel section to one value before adding the series part.
Q6: A series circuit of three bulbs is lit. One bulb burns out (opens). What happens, and how would the result differ if the bulbs were wired in parallel?
Answer: In series there is one path, so an open bulb breaks the loop and all three go dark (old-style holiday lights). In parallel each bulb has its own path, so one open bulb goes dark but the other two stay lit because their branches are unaffected.