FULL REVIEW
Full Review — Current, Resistance, and Ohm's Law — Algebra-Based
Review the essential ideas, relationships, and problem-solving tools for Current, Resistance, and Ohm's Law.
TIME
45–60 minutes
BEST FOR
A complete unit review
FINISH WITH
A readiness check
After this full review, you'll be able to...
recall the essential ideas, apply them to representative problems, and determine what to study next.
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 45–60 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
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
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.
KEY CONCEPT 3
Resistivity and power
For a uniform conductor R = ρL/A, where ρ is resistivity, ℓ length, and A cross-sectional area. Power relationships include P = I V = I2R = V2/R for ohmic resistors.
Example: Doubling a wire’s length and halving its area increases resistance by a factor of four.
Sensei note: Use P = V2/R only when V is across the resistance 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
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
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.
PRACTICE 3
Geometry change
Work through the question, then check your reasoning.
A wire of resistance 5 Ω has its length doubled and area halved at constant resistivity. Find new resistance.
Reveal Answers
R′/R = (2)/(1/2) = 4, so R′ = 20 Ω.
Why it works: Resistance of a uniform conductor is R = ρL/A. With ρ unchanged, doubling length multiplies R by 2 and halving area multiplies it by another 2, giving four times 5 Ω = 20 Ω.
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.
QUICK CHECK 3
Heating
Work through the question, then check your reasoning.
A 4 Ω resistor carries 3 A. Find the dissipated power.
Reveal Answers
P = I2R = (3 A)²(4 Ω) = 36 W.
Why it works: For an ohmic resistor, substitute V = IR into P = VI to obtain P = I2R. A current of 3 A in 4 Ω dissipates 9 × 4 = 36 W.
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.
KEY TAKEAWAY 3
Energy transfer
P = I V; for ohmic R, also I2R or V2/R.
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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