Get Ready: Module 5 — Electricity & Circuits
Everyone starts somewhere. Before we wire batteries and resistors, let's build the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 5 will feel like the next natural step, not a wall.
Adding Fractions and Reciprocals
What it is & why you need it: A reciprocal flips a number: the reciprocal of \(R\) is \(\tfrac{1}{R}\). To combine resistors in parallel, you add their reciprocals and then flip the result back: \(\tfrac{1}{R_{eq}} = \tfrac{1}{R_1} + \tfrac{1}{R_2}\). Adding those fractions over a common denominator is the whole trick — this is the one piece of arithmetic Module 5 leans on most.
Worked example: combine \(6\,\Omega\) and \(12\,\Omega\) in parallel
- Write the reciprocals: \(\tfrac{1}{R_{eq}} = \tfrac{1}{6} + \tfrac{1}{12}\).
- Find a common denominator (12): \(\tfrac{1}{6} = \tfrac{2}{12}\), so \(\tfrac{2}{12} + \tfrac{1}{12} = \tfrac{3}{12}\).
- Simplify the sum: \(\tfrac{3}{12} = \tfrac{1}{4}\). So \(\tfrac{1}{R_{eq}} = \tfrac{1}{4}\).
- Flip back to get \(R_{eq}\): \(R_{eq} = 4\,\Omega\). (Notice it's smaller than either resistor — that always happens in parallel.)
Shortcut for exactly two resistors: \(R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}\) ("product over sum"). Check: \(\dfrac{6 \cdot 12}{6 + 12} = \dfrac{72}{18} = 4\,\Omega\). Same answer.
1. Combine \(4\,\Omega\) and \(4\,\Omega\) in parallel.
2. Combine \(3\,\Omega\) and \(6\,\Omega\) in parallel.
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2. \(\tfrac{1}{R_{eq}} = \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{3}{6} = \tfrac{1}{2}\), so \(R_{eq} = 2\,\Omega\). Or product-over-sum: \(\dfrac{3 \cdot 6}{9} = 2\,\Omega\).
Solving \(V = IR\) for Each Variable
What it is & why you need it: Ohm's law, \(V = IR\), is one equation with three letters. Module 5 asks you to find whichever one is missing, so you need to rearrange it three ways. The golden rule is the same as always: do the same thing to both sides to get the unknown alone.
Worked example: rearrange \(V = IR\) three ways
- Find \(V\) (already solved): \(V = IR\). If \(I = 2\,\text{A}\) and \(R = 5\,\Omega\), then \(V = 2 \cdot 5 = 10\,\text{V}\).
- Find \(I\): divide both sides by \(R\): \(I = \dfrac{V}{R}\). If \(V = 12\,\text{V}\) and \(R = 4\,\Omega\), then \(I = \dfrac{12}{4} = 3\,\text{A}\).
- Find \(R\): divide both sides by \(I\): \(R = \dfrac{V}{I}\). If \(V = 9\,\text{V}\) and \(I = 3\,\text{A}\), then \(R = \dfrac{9}{3} = 3\,\Omega\).
- Check by plugging back: \(R = 3\,\Omega\), \(I = 3\,\text{A}\) gives \(V = IR = 9\,\text{V}\). \(\checkmark\)
A "VIR triangle" helps: cover the letter you want. Cover \(V\) and you see \(I \times R\); cover \(I\) and you see \(\tfrac{V}{R}\); cover \(R\) and you see \(\tfrac{V}{I}\).
1. A \(24\,\text{V}\) source drives a \(6\,\Omega\) resistor. Find the current \(I\).
2. A \(2\,\text{A}\) current and a \(10\,\text{V}\) drop — find the resistance \(R\).
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2. \(R = \dfrac{V}{I} = \dfrac{10}{2} = 5\,\Omega\).
The Electrical Units: V, A, Ω, W
What it is & why you need it: Numbers in physics carry units, and the units tell you which quantity you're holding. In Module 5 you'll juggle four: volts (V) for voltage, amperes (A) for current, ohms (Ω) for resistance, and watts (W) for power. Knowing which unit goes with which formula keeps you from mixing them up.
Worked example: read the units off the formulas
- Voltage is measured in volts (V). From \(V = IR\): amps × ohms gives volts.
- Current is measured in amperes (A): one amp is one coulomb of charge per second.
- Resistance is measured in ohms (Ω): one ohm is one volt per amp \(\big(1\,\Omega = \tfrac{1\,\text{V}}{1\,\text{A}}\big)\).
- Power is measured in watts (W). From \(P = IV\): amps × volts gives watts. One watt is one joule per second.
Tip: a \(60\)-watt bulb on \(120\,\text{V}\) draws \(I = \dfrac{P}{V} = \dfrac{60}{120} = 0.5\,\text{A}\). The units click together: watts ÷ volts = amps.
1. A device runs at \(3\,\text{A}\) and \(12\,\text{V}\). What power does it use, and in what unit?
2. Which unit measures resistance, and which measures current?
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2. Resistance is measured in ohms (Ω); current is measured in amperes (A).
Scientific Notation (with \(k = 8.99\times10^{9}\))
What it is & why you need it: Electric charges are tiny and Coulomb's constant is huge, so we write numbers as a digit times a power of ten: \(8.99\times10^{9}\) means \(8.99\) followed by 9 zeros. In Module 5 you'll multiply and divide these, and the key rule is simple: when you multiply powers of ten, you add the exponents.
Worked example: \((8.99\times10^{9})(2\times10^{-6})\)
- Multiply the front numbers: \(8.99 \times 2 = 17.98\).
- Add the exponents on the tens: \(10^{9} \times 10^{-6} = 10^{\,9 + (-6)} = 10^{3}\).
- Put them together: \(17.98 \times 10^{3}\).
- Tidy into proper form (one digit before the point): \(17.98 \times 10^{3} = 1.798 \times 10^{4} = 17{,}980\).
Remember: \(10^{-6}\) is "micro" (millionths). A charge of \(2\,\mu\text{C}\) is \(2\times10^{-6}\,\text{C}\). To divide powers of ten, subtract the exponents.
1. Compute \((3\times10^{8})(2\times10^{-2})\) in proper scientific notation.
2. Write \(0.000\,005\,\text{C}\) (5 microcoulombs) in scientific notation.
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2. Move the decimal 6 places right to get \(5\): \(0.000005 = 5 \times 10^{-6}\,\text{C}\).
Telling Series from Parallel
What it is & why you need it: Before you can combine resistors, you have to see how they're wired. Series means one single path — the current has nowhere else to go, so it flows through each resistor in turn. Parallel means the path splits into branches that rejoin, so the current divides. This reading skill decides which formula you use.
Worked example: read a circuit's topology
- Trace the current from the battery's \(+\) terminal. If there's only one route back, the parts on it are in series.
- If the wire splits into two or more branches that later rejoin, those branches are in parallel.
- Series clue: remove one resistor and the whole circuit breaks — like old holiday lights where one dead bulb kills the string.
- Parallel clue: each branch sees the same two connection points (nodes), so each feels the same voltage — like the outlets in your home.
Quick test: in series, the current is shared (same \(I\) everywhere). In parallel, the voltage is shared (same \(V\) across each branch).
1. Two resistors lie end-to-end on a single loop with a battery. Series or parallel?
2. Two resistors each connect directly across the same battery terminals. Series or parallel?
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2. Parallel — each spans the same two nodes, so each feels the full battery voltage; the current splits between them.
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Combine \(2\,\Omega\) and \(2\,\Omega\) in parallel.
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\(\tfrac{1}{R_{eq}} = \tfrac{1}{2} + \tfrac{1}{2} = 1\), so \(R_{eq} = 1\,\Omega\). -
A \(20\,\text{V}\) source drives a \(5\,\Omega\) resistor. Find the current.
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\(I = \dfrac{V}{R} = \dfrac{20}{5} = 4\,\text{A}\). -
A \(3\,\text{A}\) current flows through a \(4\,\Omega\) resistor. Find the voltage across it.
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\(V = IR = 3 \cdot 4 = 12\,\text{V}\). -
A device runs at \(2\,\text{A}\) and \(9\,\text{V}\). Find the power, with units.
Show answer
\(P = IV = 2 \cdot 9 = 18\,\text{W}\) (watts). -
Compute \((8.99\times10^{9})(1\times10^{-6})\).
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Front: \(8.99 \times 1 = 8.99\). Exponents: \(10^{9} \times 10^{-6} = 10^{3}\). So \(8.99 \times 10^{3} = 8990\). -
Three resistors share a single loop with a battery. Series or parallel — and what is shared?
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Series. The same current flows through all three; their resistances add.
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Adding fractions & reciprocals
Solving \(V = IR\) (Ohm's law)
Electrical units (V, A, Ω, W)
Scientific notation
Series vs. parallel circuits
When You're Ready →
No rush. When the skills above feel solid, step into Module 5.
Physics · enrichment — not a TEKS-graded course. This Foundations page builds the math prerequisites for the Electricity & Circuits enrichment module: adding fractions and reciprocals, solving \(V = IR\), the electrical units, scientific notation, and reading series vs. parallel topology.