FOCUSED REVIEW
Focused Review — Current, Resistance, and Ohm's Law — Algebra-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Current, Resistance, and Ohm's Law.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
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.
Choose how you want to review
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: Algebra-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.
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
Current expression
Work through the question, then check your reasoning.
Write the definition of average current in terms of charge and elapsed time.
Reveal Answers
I = ΔQ/Δt, where 1 A = 1 C/s.
Why it works: Average current is the net charge crossing a surface divided by elapsed time: Iavg = ΔQ/Δt. A coulomb per second is defined as one ampere.
ACTIVITY 2
Ohm’s law
Work through the question, then check your reasoning.
Find I if 15 V is across a 5 Ω resistor.
Reveal Answers
I = V/R = 15 V / 5 Ω = 3 A.
Why it works: With a constant ohmic resistance, rearranging V = IR gives I = V/R. The units V/Ω reduce to amperes, so 15 V across 5 Ω gives 3 A.
ACTIVITY 3
Resistance scaling
Work through the question, then check your reasoning.
A uniform wire is stretched to twice its length while its volume stays fixed. How does resistance change?
Reveal Answers
Constant volume means A decreases by 2 when L doubles; R′/R = 2/(1/2) = 4.
Why it works: Fixed volume means A L is constant. If length doubles, area must halve; since R = ρL/A at fixed resistivity, the length factor 2 and inverse-area factor 2 multiply to 4.
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?
Core Concepts
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Current, charge, and units
The average current through a surface is I = ΔQ/Δt, with amperes equal to coulombs per second. Conventional current is defined in the positive-charge direction; electrons in metallic conductors drift opposite to I.
Example: If 12 C crosses a cross-section in 4 s, the average current is 3 A.
Sensei note: A negative carrier velocity does not mean a negative current without an explicitly chosen axis.
KEY CONCEPT 2
Ohm’s law and the V–I graph
For an ohmic resistor at fixed temperature, V = I R; resistance equals the slope of a V-versus-I graph. For non-ohmic components that slope can vary.
Example: An 18 V source across a 6 Ω resistor gives I = 3 A.
Sensei note: The slope of I-versus-V is 1/R, not R.
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?
Guided Practice
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Current from charge
Work through the question, then check your reasoning.
A total of 24 C passes through a wire in 6.0 s. Find average current.
Reveal Answers
I = ΔQ/Δt = 24 C/6.0 s = 4.0 A.
Why it works: Average current measures how rapidly charge passes a cross-section. Dividing 24 C by 6.0 s yields 4.0 C/s = 4.0 A.
PRACTICE 2
Resistor power
Work through the question, then check your reasoning.
A 12 Ω resistor is connected across 24 V. Determine current and power.
Reveal Answers
I = V/R = 2.0 A; P = V I = (24 V)(2.0 A) = 48 W.
Why it works: Ohm’s law first gives I = 24 V/12 Ω = 2.0 A. The resistor then dissipates electrical power P = VI = 24 V × 2.0 A = 48 W; equivalently P = V2/R.
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?
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
Graph slope
Work through the question, then check your reasoning.
A resistor has V = 9 V at I = 1.5 A. Determine resistance.
Reveal Answers
R = V/I = 6 Ω.
Why it works: For an ohmic resistor, the ratio of potential difference to current is its resistance. Thus R = 9 V/1.5 A = 6 Ω.
QUICK CHECK 2
Current and time
Work through the question, then check your reasoning.
A steady 0.40 A flows for 30 s. How much charge passes?
Reveal Answers
ΔQ = IΔt = (0.40 A)(30 s) = 12 C.
Why it works: For steady current, I = ΔQ/Δt can be rearranged as ΔQ = IΔt. Multiplying 0.40 C/s by 30 s gives 12 C.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Charge flow
I = ΔQ/Δt; current is measured in amperes.
KEY TAKEAWAY 2
Material and geometry
R = ρL/A; resistance depends on both resistivity and wire dimensions.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
Next Step
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