Electric current, resistance, resistivity, and Ohm's law
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Current, Resistance, and Ohm's Law is part of the AP Physics 2 course on Study Mondo, specifically in the Electric Circuits section. You can explore the full course for more related topics and practice resources.
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I
ΔQ = charge passing point (C)
Δt = time interval (s)
1 Ampere = 1 Coulomb/second
Direction:
Conventional current: Direction positive charges flow (+ to -)
Electron flow: Opposite direction (- to +)
We use conventional current!
Drift Velocity
In conductors, electrons drift slowly:
I=nAvde
where:
n = charge carrier density
A = cross-sectional area
vd = drift velocity (~mm/s, very slow!)
e = elementary charge
💡 Key: Current is fast (~speed of light), but individual electrons drift slowly!
Resistance
Resistance opposes current flow:
R=IV
Units: Ohm (Ω) = V/A
Resistivity
Material property:
R=AρL
where:
ρ = resistivity (Ω·m)
L = length
A = cross-sectional area
Good conductors: Low ρ (copper: 1.7×10−8 Ω·m)
Insulators: High ρ
Temperature Dependence
R=R0(1+αΔT)
where α is temperature coefficient.
Metals: α>0 (R increases with T)
Semiconductors: α<0 (R decreases with T)
Ohm's Law
V=IR
Ohmic materials: Constant R (linear V-I graph)
Non-ohmic: R varies (curved V-I graph, like diodes)
Electric Power
Power dissipated in resistor:
P=IV=I2R=RV2
Units: Watt (W) = J/s
Energy dissipated (heat):
E=Pt=I2Rt=RV2t
Electrical Energy Cost
Power companies charge by kilowatt-hour (kWh):
1 kWh=(1000 W)(3600 s)=3.6×106 J
Cost = (Power in kW) × (time in hours) × (rate per kWh)
AC vs DC
DC (Direct Current): Constant direction (batteries)
AC (Alternating Current): Oscillates (wall outlets, 60 Hz in US)
For AC: Vrms=2V0, Irms=2
Household: 120 V AC is Vrms
Problem-Solving Strategy
Identify knowns: V, I, or R
Choose formula:
Ohm's Law: V=IR
Power: P=IV=I2R=V2/R
Resistance: R=ρL/A
Watch units: A, V, Ω, W
Check reasonableness
Common Mistakes
❌ Confusing current direction (use conventional!)
❌ Using diameter instead of radius in area (A=πr2)
❌ Wrong power formula (choose based on what you know)
❌ Forgetting to convert units (mA → A, kΩ → Ω)
❌ Treating all materials as ohmic
R=6.0
Part (a): Current
Use Ohm's Law:
I=RV=6.012=2.0 A
Part (b): Power dissipated
P=IV=(2.0)(12)=24 W
Or alternatively:
P=RV2=6.0(12)2=6.0144=24 W ✓
Answer:
(a) I = 2.0 A
(b) P = 24 W
2Problem 2easy
❓ Question:
A 12 V battery is connected to a 6.0 Ω resistor. (a) What is the current? (b) What power is dissipated?
💡 Show Solution
Given:
Voltage: V=12 V
Resistance: R=6.0 Ω
Part (a): Current
Use Ohm's Law:
I=RV=6.0
Part (b): Power dissipated
P=IV=(2.0)(12)=24 W
Or alternatively:
P=RV2= ✓
Answer:
(a) I = 2.0 A
(b) P = 24 W
3Problem 3medium
❓ Question:
A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) has length 2.0 m and diameter 1.0 mm. What is its resistance?
💡 Show Solution
Given:
Resistivity: ρ=1.7×10−8 Ω·m
Length: L=2.0 m
Diameter: d=1.0 mm =1.0×10−3 m
Solution:
Step 1: Find cross-sectional area.
r=2d=5.0×10
Step 2: Calculate resistance.
R=AρL=
Answer: R = 0.043 Ω (very low, good conductor!)
4Problem 4medium
❓ Question:
A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) has length 2.0 m and diameter 1.0 mm. What is its resistance?
💡 Show Solution
Given:
Resistivity: ρ=1.7×10−8 Ω·m
Length: L=2.0 m
Diameter: d=1.0 mm =1.0×10−3 m
Solution:
Step 1: Find cross-sectional area.
r=2d=5.0×10
Step 2: Calculate resistance.
R=AρL=
Answer: R = 0.043 Ω (very low, good conductor!)
5Problem 5easy
❓ Question:
A 12 V battery is connected to a 4.0 Ω resistor. (a) What is the current through the resistor? (b) What is the power dissipated? (c) How much energy is dissipated in 5.0 minutes?
💡 Show Solution
Solution:
Given: V = 12 V, R = 4.0 Ω, t = 5.0 min = 300 s
(a) Current (Ohm's Law):
I = V/R = 12/4.0 = 3.0 A
(b) Power:
P = VI = 12 × 3.0 = 36 W
Or: P = V²/R = 144/4.0 = 36 W ✓
Or: P = I²R = (3.0)²(4.0) = 36 W ✓
(c) Energy:
E = Pt = 36 × 300 = 10,800 J or 10.8 kJ
6Problem 6hard
❓ Question:
A 1500 W electric heater operates on 120 V. (a) What is the current? (b) What is the resistance? (c) How much does it cost to run for 8 hours if electricity costs $0.12 per kWh?
💡 Show Solution
Given:
Power: P=1500 W
Voltage: V=120 V
Time: t=8 hours
Rate: $0.12 per kWh
Part (a): Current
P=IVI=VP
Part (b): Resistance
P=RV2
Or using Ohm's Law:
R=IV=12.5 ✓
Part (c): Cost
Energy used:
E=Pt=(1500 W)(8 h)=12,000 Wh=12 kWh
Cost:
Cost=(12 kWh)($0.12/kWh)=$1.44
Answer:
(a) I = 12.5 A
(b) R = 9.6 Ω
(c) Cost = $1.44
7Problem 7medium
❓ Question:
A wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² has resistance 0.50 Ω. (a) What is the resistivity of the material? (b) If the wire is stretched to 3.0 m (volume constant), what is the new resistance?
Alternative: R ∝ L²/V, so R₂ = R₁(L₂/L₁)² = 0.50(3.0/2.0)² = 1.125 Ω ✓
8Problem 8hard
❓ Question:
A 1500 W electric heater operates on 120 V. (a) What is the current? (b) What is the resistance? (c) How much does it cost to run for 8 hours if electricity costs $0.12 per kWh?
💡 Show Solution
Given:
Power: P=1500 W
Voltage: V=120 V
Time: t=8 hours
Rate: $0.12 per kWh
Part (a): Current
P=IVI=VP
Part (b): Resistance
P=RV2
Or using Ohm's Law:
R=IV=12.5 ✓
Part (c): Cost
Energy used:
E=Pt=(1500 W)(8 h)=12,000 Wh=12 kWh
Cost:
Cost=(12 kWh)($0.12/kWh)=$1.44
Answer:
(a) I = 12.5 A
(b) R = 9.6 Ω
(c) Cost = $1.44
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Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.