Get Ready: Module 5 — Charge, Fields & Circuits
Everyone starts somewhere. Before we wire resistor networks and add electric fields, 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.
Reciprocals & Adding Fractions
What it is & why you need it: The reciprocal of a number flips it: the reciprocal of \(4\) is \(\tfrac14\). The parallel-resistance rule is literally a sum of reciprocals: \(\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\). If you can add fractions and flip the result, you can find any parallel resistance.
Worked example: combine \(R_1 = 6\,\Omega\) and \(R_2 = 12\,\Omega\) in parallel
- Write the rule: \(\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{12}\).
- Find a common denominator (\(12\)): \(\dfrac{1}{6} = \dfrac{2}{12}\), so \(\dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12}\).
- Simplify: \(\dfrac{3}{12} = \dfrac{1}{4}\). So \(\dfrac{1}{R_{eq}} = \dfrac{1}{4}\).
- Flip (take the reciprocal) to get \(R_{eq}\): \(R_{eq} = 4\,\Omega\).
Shortcut for exactly two resistors: \(R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{6\cdot 12}{6+12} = \dfrac{72}{18} = 4\,\Omega\). Same answer.
1. What is the reciprocal of \(\tfrac{1}{5}\)? And of \(8\)?
2. Combine \(R_1 = 4\,\Omega\) and \(R_2 = 4\,\Omega\) in parallel using the sum-of-reciprocals rule.
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2. \(\dfrac{1}{R_{eq}} = \dfrac14 + \dfrac14 = \dfrac24 = \dfrac12\), so \(R_{eq} = 2\,\Omega\). Two equal resistors in parallel give half the value.
Inverse-Square Reasoning
What it is & why you need it: Several laws in this module shrink with the square of distance: Coulomb's law \(F = \tfrac{kq_1q_2}{r^2}\) and the point-charge field \(E = \tfrac{kq}{r^2}\). This is the exact same shape as gravitation \(\big(F = \tfrac{Gm_1m_2}{r^2}\big)\), so the intuition you built there transfers directly: double the distance, quarter the strength.
Worked example: a field is \(8000\,\text{N/C}\) at \(2\,\text{m}\) — what is it at \(4\,\text{m}\)?
- Identify the ratio of distances: \(4\,\text{m}\) is twice as far as \(2\,\text{m}\), so \(\dfrac{r_1}{r_2} = \dfrac{2}{4} = \dfrac12\).
- Inverse-square means scale by that ratio squared: \(E_2 = E_1\left(\dfrac{r_1}{r_2}\right)^2\).
- Substitute: \(E_2 = 8000\left(\tfrac12\right)^2 = 8000 \cdot \tfrac14\).
- Compute: \(E_2 = 2000\,\text{N/C}\). Twice the distance → one-quarter the field.
The same logic mirrors gravity: a satellite twice as far from Earth feels one-quarter the gravitational pull. Distance is squared, so its effect grows fast.
1. A point charge gives a field of \(900\,\text{N/C}\) at \(1\,\text{m}\). What is the field at \(3\,\text{m}\)?
2. If you halve the distance to a charge, by what factor does the field change?
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2. Halving \(r\) gives \(\left(\tfrac{1}{1/2}\right)^2 = 2^2 = 4\). The field becomes 4× stronger.
Solving for Any Variable in a Product Relation
What it is & why you need it: Ohm's law \(V = IR\) is a product of three quantities. The module constantly asks you to solve for whichever one is unknown: \(I = \tfrac{V}{R}\) or \(R = \tfrac{V}{I}\). The rule: to isolate a factor in a product, divide both sides by the other factor.
Worked example: a \(24\,\text{V}\) source drives a \(6\,\Omega\) resistor — find the current
- Start with Ohm's law: \(V = IR\), i.e. \(24 = I \cdot 6\).
- The unknown \(I\) is multiplied by \(6\). Undo by dividing both sides by \(6\): \(\dfrac{24}{6} = I\).
- Compute: \(I = 4\,\text{A}\).
- Check: \(V = IR = 4 \cdot 6 = 24\,\text{V}\). True.
The same move works for power \(P = IV\) and \(P = I^2 R\): isolate the one you want by dividing (or taking a square root) to undo what's attached to it.
1. A \(3\,\text{A}\) current flows through a resistor with \(18\,\text{V}\) across it. Find \(R\).
2. A device dissipates \(60\,\text{W}\) at \(12\,\text{V}\). Use \(P = IV\) to find the current.
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2. Rearrange \(P = IV\) to \(I = \dfrac{P}{V} = \dfrac{60}{12} = 5\,\text{A}\).
Significant Figures
What it is & why you need it: Circuit answers rarely come out as whole numbers — you'll get things like \(1.4666\dots\,\text{A}\). Significant figures tell you how many digits are meaningful so you can report a clean, honest answer (this module usually asks for 2 decimal places). Round only at the end; keep extra digits while you compute.
Worked example: round \(1.46667\) to two decimal places
- Locate the second decimal place: the digit there is \(6\) (\(1.4\underline{6}667\)).
- Look at the next digit to decide rounding: it's \(6\), which is \(\ge 5\).
- Round the second-decimal digit up: \(6 \to 7\).
- Result: \(1.47\). (Had the next digit been \(< 5\), you would round down / leave it.)
Golden rule: don't round mid-calculation. \(I = \tfrac{V}{R}\) might be \(1.46667\dots\); carry those digits through any later step, then round the final number to 2 places.
1. Round \(3.14159\) to two decimal places.
2. A current works out to \(0.83333\dots\,\text{A}\). Report it to two decimals.
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2. The third decimal is \(3\) (\(< 5\)), so \(0.83\,\text{A}\).
Reading a Circuit Schematic
What it is & why you need it: A schematic is a map. A battery (long line \(+\), short line \(-\)) pushes charge around a loop; a resistor (zig-zag, or a box) resists the flow. The big skill is telling series (one single path) from parallel (the path splits and rejoins). That one judgment decides which formula you use.
Worked example: series or parallel?
- Trace the current from the \(+\) terminal. Put your finger on the wire leaving the long battery plate.
- If there is only one path through every resistor before returning to the \(-\) terminal, the resistors are in series (the current has no choice).
- If the wire splits into separate branches that later rejoin, those branches are in parallel (the current divides).
- A combination circuit does both: a series resistor leading into a parallel block. Reduce the parallel block first, then add it in series.
One single loop through R1 then R2 → series. Add them: \(R_{eq} = R_1 + R_2\).
1. In a series loop the current reaches \(R_1\), then \(R_2\), then \(R_3\). Is the current through \(R_2\) the same as through \(R_1\)? Why?
2. Two resistors are each wired directly across the same battery terminals (the wire splits, each gets its own branch). Series or parallel? What voltage does each one see?
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2. Parallel. Each resistor is connected straight across the source, so each sees 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 \(R_1 = 3\,\Omega\) and \(R_2 = 6\,\Omega\) in parallel.
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\(R_{eq} = \dfrac{3\cdot 6}{3+6} = \dfrac{18}{9} = 2\,\Omega\) (or \(\tfrac13 + \tfrac16 = \tfrac12 \Rightarrow R_{eq} = 2\,\Omega\)). -
A field is \(400\,\text{N/C}\) at \(1\,\text{m}\). What is it at \(2\,\text{m}\)?
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Twice the distance → one-quarter the field: \(400 \cdot \tfrac14 = 100\,\text{N/C}\). -
A \(20\,\text{V}\) source drives a \(5\,\Omega\) resistor. Find the current.
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\(I = \dfrac{V}{R} = \dfrac{20}{5} = 4\,\text{A}\). -
Round \(2.7182\) to two decimal places.
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The third decimal is \(8\) (\(\ge 5\)), so round up: \(2.72\). -
Three resistors are in series. Is the current through each the same or different?
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The same — series circuits have one single path, so one current flows through all of them. -
Add the series resistors \(4\,\Omega\), \(5\,\Omega\), and \(6\,\Omega\). What is \(R_{eq}\)?
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Series resistances add: \(R_{eq} = 4 + 5 + 6 = 15\,\Omega\).
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.)
Reciprocals & fractions
Inverse-square & exponents
Solving literal equations
Significant figures & rounding
Reading circuit schematics
When You're Ready →
No rush. When the skills above feel solid, step into Module 5.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy
AP Physics · enrichment — not a TEKS-graded course. Supplemental, college-level practice in algebra-based
circuits and electrostatics, offered alongside the Mathematical Architects math pathway. "AP" refers to the College Board Advanced Placement framework, not a state standard.