Series and Parallel Circuits and Power

🔌 Series and Parallel Circuits and Power

Part 1 of 7 — Circuits on the ASVAB

Topic Overview

Why Circuits Matter on EI

Electronics Information gives you 15 questions in 10 minutes on the CAT-ASVAB (about 40 seconds each). Circuit questions are some of the most common: a small diagram of a battery, some resistors or bulbs, maybe a switch, and a question like "what is the current?" or "which bulbs stay lit?" The numbers are always simple enough to do without a calculator.

You will use Ohm's law constantly in this topic:

$V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$

(If that is new, work through the topic Current, Voltage, Resistance and Ohm's Law first.)

The Parts of Every Circuit

Open, Closed and Short

Current flows only around a complete loop from one terminal of the source, through the load, and back to the other terminal.

Key Fact: "Closed" sounds like "shut off," but in electricity a closed switch is ON (the path is complete) and an open switch is OFF (there is a gap).

Reading a Simple Schematic

Electricians draw circuits with symbols, not pictures. You need only a few for now:

Warm-up: read the diagram. 🎯

Name the condition. 🔽

Key Takeaways — Part 1

• Every circuit has a source, a load and conductors; most also have a control (switch) and protection (fuse or breaker)
• Current flows only around a complete loop
• Closed = complete = ON; open = broken = OFF, zero current; short = bypasses the load, very high current
• Ohm's law ($V = IR$) is the main tool in this topic

Up Next: Part 2 — series circuits, where everything sits on one path.

🔌 Series and Parallel Circuits and Power

Part 2 of 7 — Series Circuits

One Path Only

In a series circuit, the parts are connected end to end, so there is only one path for current. Every electron that leaves the battery passes through every part, one after another.

The Three Series Rules

The voltage "drop" across each part follows Ohm's law, $V = I \times R$. Since the current is the same in every part, the biggest resistor gets the biggest share of the voltage.

What Happens When Things Change

• Add another resistor in series: total resistance goes up, so the current goes down (every bulb in the string dims).
• One part burns out or a wire breaks: the only path is broken, so everything goes off. This is the classic behavior of older holiday light strings: one dead bulb and the whole string goes dark.
• A switch or fuse is always placed in series with what it controls or protects, because it has to be able to break the one path.

Worked Example: the Whole Series Routine

A 12-V battery is connected in series with a 2-Ω resistor and a 4-Ω resistor.

Step 1. Total resistance. $R_{total} = 2 + 4 = 6$ Ω.

Step 2. Current (the same everywhere). $I = \frac{V}{R_{total}} = \frac{12}{6} = 2$ A.

Step 3. Each voltage drop. $V_1 = 2 \times 2 = 4$ V and $V_2 = 2 \times 4 = 8$ V.

Step 4. Check. $4 + 8 = 12$ V, the battery voltage. ✓

The 4-Ω resistor has twice the resistance of the 2-Ω resistor, so it takes twice the voltage.

Identical parts

If $n$ identical resistors (or bulbs) are in series across a source, each one gets $\frac{1}{n}$ of the voltage. Four identical bulbs on a 12-V battery: 3 V each.

Run the series routine on Figure 2. ✍️ Numbers only.

1) Total resistance, in ohms:

2) Current through the 3-Ω resistor, in amperes:

3) Voltage drop across the 5-Ω resistor, in volts:

Series reasoning. 🎯

More series practice. ✍️ Numbers only.

1) Resistors of 4 Ω, 6 Ω and 10 Ω are in series across a 40-V source. What is the current, in amperes?

2) Six identical bulbs are wired in series across a 120-V supply. What is the voltage across each bulb, in volts?

Key Takeaways — Part 2

• Series = one path
• Resistances add; current is the same everywhere; voltage drops add up to the source
• The largest resistor takes the largest voltage drop
• Adding resistors in series lowers the current; one break turns everything off
• Switches and fuses go in series with what they control

Up Next: Part 3 — parallel circuits, where current has a choice of paths.

🔌 Series and Parallel Circuits and Power

Part 3 of 7 — Parallel Circuits

Several Paths

In a parallel circuit, each part is connected directly across the source, on its own branch. Current leaving the battery splits at a junction, each share goes through one branch, and the shares rejoin before returning to the battery.

The Three Parallel Rules

Each branch follows Ohm's law on its own: $I_{branch} = \frac{V_{source}}{R_{branch}}$. The smallest resistance carries the largest current.

What Happens When Things Change

• One branch burns out or is switched off: the other branches still have a complete path to the battery, so they stay on at the same brightness.
• Add another branch: total resistance goes down, so the total current from the source goes up (each existing branch keeps its own current).

Key Fact: Your home is wired in parallel. Every outlet and light gets the full household voltage, and turning off the kitchen light does not turn off the TV.

Finding the Equivalent Resistance

You will almost always meet one of three easy cases.

Quick check: the answer must be smaller than the smallest branch. Two resistors of 6 Ω and 3 Ω cannot combine to more than 3 Ω.

Worked Example

A 6-Ω and a 3-Ω resistor are in parallel across a 12-V battery (Figure 3, coming up).

Branch currents. Each branch has the full 12 V: $I_1 = \frac{12}{6} = 2$ A and $I_2 = \frac{12}{3} = 4$ A.

Total current. $2 + 4 = 6$ A.

Check with $R_{eq}$. $R_{eq} = \frac{6 \times 3}{6 + 3} = 2$ Ω, and $\frac{12}{2} = 6$ A. ✓

Often the fastest route is to find each branch current and add them, without finding $R_{eq}$ at all.

Figure 3 is the circuit from the worked example. ✍️ Now change it. Numbers only.

1) The 3-Ω resistor is replaced by a 12-Ω resistor (the 6-Ω resistor stays). What is the current through the 12-Ω branch, in amperes?

2) For that new circuit (6 Ω and 12 Ω in parallel across 12 V), what is the total current from the battery, in amperes?

3) What is the equivalent resistance of 6 Ω and 12 Ω in parallel, in ohms?

Parallel reasoning. 🎯

Series or parallel? 🔽 Choose the circuit type each statement describes.

Key Takeaways — Part 3

• Parallel = several branches, each connected straight across the source
• Voltage is the same on every branch; branch currents add to the total
• $R_{eq}$ is smaller than the smallest branch: identical → $\frac{R}{n}$; two → product over sum
• The smallest resistance carries the largest current
• One branch out → the others stay on; adding a branch raises the total current
• House wiring is parallel

Up Next: Part 4 — circuits that mix series and parallel, and what happens when something fails.

🔌 Series and Parallel Circuits and Power

Part 4 of 7 — Combination Circuits and Circuit Faults

Series-Parallel Circuits

Many diagrams mix the two: a resistor in series with a parallel group. Solve them by shrinking the circuit one piece at a time.

1. Find a group that is purely parallel (or purely series) and replace it with its equivalent resistance.
2. Repeat until one resistance is left. That is the total.
3. Total current = source voltage ÷ total resistance.
4. Work back out using Ohm's law for each piece.

Worked example (Figure 4, coming up)

A 4-Ω resistor is in series with a parallel pair of 6 Ω and 12 Ω, across 24 V.

• Parallel pair: $\frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4$ Ω.
• Total: $4 + 4 = 8$ Ω (the series resistor plus the pair).
• Total current: $\frac{24}{8} = 3$ A. All 3 A flows through the 4-Ω series resistor.
• Voltage across the series resistor: $3 \times 4 = 12$ V. That leaves $24 - 12 = 12$ V across the parallel pair.
• Branch currents: $\frac{12}{6} = 2$ A and $\frac{12}{12} = 1$ A, which add back to 3 A. ✓

Same circuit, new numbers. ✍️ In Figure 4, the 12-Ω resistor is replaced with a 6-Ω resistor, so the parallel pair is now two 6-Ω resistors, and the battery is replaced with a 21-V battery. The series resistor is still 4 Ω. Numbers only.

1) Equivalent resistance of the parallel pair, in ohms:

2) Total resistance of the circuit, in ohms:

3) Total current from the battery, in amperes:

Circuit Faults: Opens and Shorts

Reading a bulb diagram (Figure 5)

Bulb 1 carries the total current. Bulbs 2 and 3 split it.

• Bulb 1 burns out: it is in series with everything, so all three go out.
• Bulb 2 burns out: Bulb 3's branch is still a complete path through Bulb 1, so Bulbs 1 and 3 stay lit. Now Bulbs 1 and 3 are simply in series, so they share the voltage equally: Bulb 3 gets brighter than before and Bulb 1 gets dimmer.
• A wire shorts across Bulb 1: Bulb 1 goes out, and Bulbs 2 and 3 get the whole battery voltage, so they get brighter.

Key Fact: A short does not "use up" current, it removes resistance. Less resistance means more current, more heat in the wires, and a blown fuse or tripped breaker if the circuit is protected.

Use Figure 5. 🎯 All three bulbs are identical.

Faults in general. 🎯

Key Takeaways — Part 4

• Solve combination circuits by shrinking: replace parallel groups and series groups with their equivalents, find the total current, then work back out
• The resistor in series with everything carries the total current
• Open = a break, zero current in that path; parts in series with it go off
• Short = a bypass with almost no resistance; current rises, bypassed parts go off, other series parts get brighter, fuses blow

Up Next: Part 5 — electrical power: watts, kilowatt-hours and why some bulbs glow brighter.

🔌 Series and Parallel Circuits and Power

Part 5 of 7 — Electrical Power and Energy

What Power Means

Power is how fast electrical energy is used (or delivered). It is measured in watts (W). A 100-W bulb uses energy twice as fast as a 50-W bulb. 1 kilowatt (kW) = 1,000 W.

The Power Formula

$P = I \times V \qquad \text{(watts = amps} \times \text{volts)}$

Combine it with Ohm's law and you get two more forms, useful when you know the resistance:

Rearranged: $I = \frac{P}{V}$ (find the current an appliance draws) and $V = \frac{P}{I}$.

Worked example 1. A 1,200-W hair dryer runs on 120 V. Current: $I = \frac{1200}{120} = 10$ A.

Worked example 2. A 60-W bulb on a 120-V circuit. Current: $\frac{60}{120} = 0.5$ A. Resistance: $R = \frac{V}{I} = \frac{120}{0.5} = 240$ Ω.

Key Fact: Resistance turns electrical energy into heat. Doubling the current through the same resistor makes four times the heat ($P = I^2R$). That is why a short circuit, with its huge current, heats wires so fast.

Energy and the Electric Bill

Energy = power × time. The power company bills you in kilowatt-hours (kWh): one kilowatt used for one hour.

Cost = kWh × price per kWh. At 15 cents per kWh, the heater above costs 6 × 15 = 90 cents, or $0.90.

Do not mix up kW (how fast energy is used, a power) with kWh (how much energy was used in total).

Which Bulb Is Brighter?

Brightness follows the power each bulb actually uses.

• In parallel (normal house wiring), every bulb gets the full voltage, so the bulb rated at more watts (lower resistance) glows brighter. A 100-W bulb beats a 60-W bulb.
• In series, every bulb carries the same current, so $P = I^2R$ says the bulb with the larger resistance uses more power and glows brighter. (Surprise: wired in series, a 60-W bulb outshines a 100-W bulb, because the 60-W bulb has the higher resistance.)
• In a series circuit, adding bulbs lowers the current, so every bulb gets dimmer. In parallel, adding bulbs leaves each one's brightness unchanged.

Power and energy by hand. ✍️ Numbers only.

1) A car headlight draws 5 A from a 12-V battery. What is its power, in watts?

2) A 900-W microwave runs on 120 V. What current does it draw, in amperes?

3) A current of 2 A flows through a 15-Ω resistor. How much power does the resistor use, in watts?

4) A 2,000-W space heater runs for 3 hours. How many kilowatt-hours does it use?

Reason about power. 🎯

Pick the formula. 🔽

Key Takeaways — Part 5

• Power (watts) = $I \times V$; also $I^2R$ and $\frac{V^2}{R}$; 1 kW = 1,000 W
• Current an appliance draws: $I = \frac{P}{V}$ (1,200 W at 120 V → 10 A)
• Energy = power × time, billed in kilowatt-hours; cost = kWh × price
• Double the current through a resistor → 4× the heat
• In parallel the higher-wattage bulb is brighter; in series the higher-resistance bulb is brighter

Up Next: Part 6 — how to work circuit questions fast and the traps to avoid.

🔌 Series and Parallel Circuits and Power

Part 6 of 7 — Circuit Test Strategy and Common Traps

Timing

On the CAT-ASVAB, EI is 15 questions in 10 minutes: about 40 seconds each. A circuit diagram with numbers usually takes longer than a vocabulary item, so move quickly through recall questions and give diagrams up to about a minute. On the paper test, EI is 20 questions in 9 minutes, so speed matters even more.

You cannot go back on the CAT, and you must answer to move on, so finish each question before you click. Some questions may be unscored tryouts you cannot identify: treat every one as scored. If time is running out, keep working real answers; a string of rushed wrong answers at the end is penalized heavily.

A 20-second first look at any diagram

1. Series or parallel? Trace the path from the + terminal. One path = series. A junction where the wire splits = parallel branches.
2. What does the question ask for? Total resistance, current, a voltage drop, power, or "which bulbs stay lit"?
3. Can you reason instead of calculate? Many questions ask only whether something goes up, down or stays the same.

Fast facts to apply without a formula

The Traps

Elimination tricks

• Parallel total resistance: cross out every choice equal to or larger than the smallest branch.
• Series current: cross out choices that give different currents in different resistors.
• Units: an answer in watts can't be a current; an answer in kWh can't be a power rating.

Spot the trap. 🎯

Up, down or the same? 🔽

Key Takeaways — Part 6

• About 40 s per EI question on the CAT; give diagrams up to a minute by moving fast on recall items
• First look: series or parallel → what is asked → can you reason instead of calculate?
• Parallel totals are smaller than the smallest branch; series current is the same everywhere
• Shorts raise current; kWh is energy, kW is power; double current → 4× heat

Up Next: Part 7 — a mixed mastery set.

🔌 Series and Parallel Circuits and Power

Part 7 of 7 — Mixed Mastery Set

Ten questions mixing everything in this topic, at real EI difficulty. Aim for about 40 seconds each, using scratch paper and no calculator.

Questions 1–3 use Figure 7. 🎯

Questions 4–7. 🎯

Questions 8–10. 🎯

Key Takeaways — Series and Parallel Circuits and Power

• Series: one path; resistances add; same current; voltage drops add to the source; one break → all off
• Parallel: separate branches; same voltage; currents add; total resistance below the smallest branch; one branch out → others stay on
• Combination circuits: shrink parallel groups first, find the total current, then work back out
• Shorts raise current and blow fuses; opens stop current in their path
• Power: $P = IV = I^2R = \frac{V^2}{R}$; energy in kWh = kW × hours

Next topic: Components, Symbols and AC vs DC.