FOCUSED REVIEW

Focused Review — Current, Resistance, and Ohm's Law — Calculus-Based

Reinforce the highest-leverage ideas and representative problem-solving tools for Current, Resistance, and Ohm's Law.

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TIME

Approximately 15 minutes

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BEST FOR

Targeted reinforcement

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FINISH WITH

A readiness check

After this focused review, you'll be able to...

reinforce the key relationships, apply them to representative problems, and identify what still needs work.

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Course Alignment

This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.

RESOURCE: Physics Sensei Unit Review | UNIT ID: EME-U06 | TOPIC: Current, Resistance, and Ohm's Law | COURSE LEVEL: Calculus-Based

BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam

Your Review Plan

Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.

6 Stages • Approximately 15 minutes.

①

Warm-Up Check

Activate prior knowledge.

②

Core Concepts

Review the essential ideas.

③

Guided Practice

Apply what you learned.

④

Confidence Check

Confirm your understanding.

Warm-Up Check

Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.

ACTIVITY 1

Derivative

Work through the question, then check your reasoning.

If Q(t) = 3t² C (t in seconds), find I(t).

Reveal Answers

I(t) = dQ/dt = 6t A (t measured in seconds).

Why it works: Instantaneous current is the time derivative of charge: I(t) = dQ/dt. Differentiating 3t² gives 6t; with charge in coulombs and time in seconds, the result is in amperes.

ACTIVITY 2

Integral

Work through the question, then check your reasoning.

A wire carries I(t) = 2t A, t in seconds. Calculate charge from 0 to 2 s.

Reveal Answers

ΔQ = ∫₀² 2t dt = [t²]₀² = 4 C.

Why it works: The net charge transported over a time interval is the definite integral of current. Integrating 2t from 0 to 2 s gives the difference t² at the limits, 4 C − 0 C = 4 C.

ACTIVITY 3

Vector flux

Work through the question, then check your reasoning.

For constant uniform J normal to area A, write the surface-integral reduction.

Reveal Answers

I = ∫ J · dA = JA for uniform J perpendicular to the surface.

Why it works: Current through a surface is the flux of current density, I = ∫S J·dA. If J is uniform and normal to a flat cross-section, the dot product is J dA everywhere and integrates to JA.

Ready to strengthen your understanding?

You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?

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Core Concepts

Reinforce the two highest-leverage relationships, then use them in representative situations.

KEY CONCEPT 1

Instantaneous current and charge

Instantaneous current is I(t) = dQ/dt. The charge transported during an interval is ΔQ = ∫ I(t) dt over the chosen time interval. Current is signed relative to the selected surface normal.

$I (t) = \frac{d Q}{d t}; Δ Q = \int_{t_{1}}^{t_{2}} I (t) d t$

Example: For I(t) = (2t) A with t in seconds, charge transported from 0 to 3 s is 9 C.

Sensei note: Average current is not automatically equal to the instantaneous current at an endpoint.

KEY CONCEPT 2

Microscopic current and field

Current density J relates to charge transport: I = ∫ J·dA. In an ohmic isotropic conductor, J = σE and ρ = 1/σ. The lumped relation V = I R follows under steady, uniform geometry assumptions.

$I = \int_{S} \vec{J} \cdot d \vec{A}; \vec{J} = σ \vec{E}$

Example: For uniform J aligned with a wire’s cross section, I = J A.

Sensei note: J is a vector; do not replace the surface integral with J A when field or orientation varies.

Ready to apply these ideas?

You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?

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Guided Practice

Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.

PRACTICE 1

Time-varying current

Work through the question, then check your reasoning.

The net charge transported is Q(t) = (2t² + 1) C, with t in seconds. Determine I(3 s).

Reveal Answers

I(t) = dQ/dt = 4t A, so I(3 s) = 12 A.

Why it works: Current is the instantaneous rate of change of transported charge. Differentiating Q(t) = 2t² + 1 eliminates the constant and gives I(t) = 4t A; evaluating at t = 3 s yields 12 A.

PRACTICE 2

Charge integration

Work through the question, then check your reasoning.

Current is I(t) = (3t²) A with t in seconds. Find the charge moved from 0 to 2 s.

Reveal Answers

ΔQ = ∫₀² (3t²) dt = [t³]₀² = 8 C.

Why it works: Charge transported is the area under the signed current–time curve: ΔQ = ∫₀² 3t² dt. Its antiderivative is t³, so the difference of endpoint values is 8 C.

Ready to check your understanding?

You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?

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Confidence Check

You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.

QUICK CHECK 1

Derivative sign

Work through the question, then check your reasoning.

Q(t) = (5 − t²) C with t in seconds. Find I at t = 2 s.

Reveal Answers

I = dQ/dt = −2t A; at t = 2 s, I = −4 A relative to the chosen orientation.

Why it works: Current is the derivative dQ/dt, not charge divided by time at one instant. Differentiating 5 − t² gives −2t A; the negative sign at 2 s indicates flow opposite the chosen positive orientation.

QUICK CHECK 2

Integral of current

Work through the question, then check your reasoning.

I(t) = 4t A with t in seconds. Find charge moved between 1 and 3 s.

Reveal Answers

ΔQ = ∫₁³4t dt = [2t²]₁³ = 16 C.

Why it works: Net transported charge is the definite time integral of signed current. Integrating 4t yields 2t²; evaluating from 1 to 3 s gives 18 − 2 = 16 C.