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high school physics Current and Resistance

Electric Current and Resistance

Current and resistance are easy to picture as a water pipe: voltage is the water pressure that pushes, current is the amount of water flowing, and resistance is how narrow the pipe is and how much it holds the flow back. By Ohm's law, V = IR, raising the voltage increases the current, and raising the resistance reduces it. How fast energy is used while the current flows is the power, and the energy actually used over that time is the electrical energy. Move the voltage and resistance sliders and the current around the circuit and the power used change together.

The Idea of Current and Resistance

💧 Comparing with a water pipe
①Voltage V is like water pressure, how hard the charges are pushed
②Current I is like the amount of water flowing: the charge passing a cross-section each second
③Resistance R is like the narrowness of the pipe, how much the flow is held back

Ohm's Law

12 V
4 Ω
💡 What to notice
①Doubling the voltage doubles the current
②Doubling the resistance halves the current
③V = IR is the most basic relation in an electric circuit
④A wire's resistance is R = ρL/A: proportional to its length and inversely proportional to its cross-sectional area
⑤The figure has a single resistor; series and parallel connections are summed up in the exam points

Three Forms of Ohm's Law

Ohm's law
V = IR
Voltage (V) = current (A) × resistance (Ω)
Finding the current
I = VR
Current (A) = voltage (V) ÷ resistance (Ω)
Finding the resistance
R = VI
Resistance (Ω) = voltage (V) ÷ current (A)

Power and Electrical Energy

Power (how fast energy is used)
P = VI = I²R = V²/R
Power (W) = voltage × current; putting in V = IR gives three forms
Electrical energy (energy used)
W = Pt = VIt
Electrical energy (J) = power × time
💰 Electricity bills and energy
①1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
②An electricity bill is the energy used (kWh) times the unit price
③A 100 W bulb left on for 10 hours uses 1 kWh

Worked Examples

Example 1
A bulb with resistance 6 Ω is connected to 12 V. What current flows?
1
Rewrite V = IR as I = V/R.
I = VR
2
Put in V = 12 V and R = 6 Ω.
I = 126 = 2 A
▸ 2 A
The larger the voltage and the smaller the resistance, the larger the current.
Example 2
A 10 Ω resistor carries a current of 2 A. What power does it use?
1
The current and resistance are known, so use P = I²R.
P = I2 R
2
Put in I = 2 A and R = 10 Ω.
P = 22 × 10 = 40 W
▸ 40 W
The three forms P = VI = I²R = V²/R all give the same value; here V = IR = 20 V, so VI = 40 W confirms it.

Summary

Ohm's law
V = IR
Power
P = VI
CSAT-style problem, adapted
For a wire of the same material, if the length is doubled and the cross-sectional area is halved, by what factor does the resistance change?
①2×
②4×
③12×
④14×
⑤No change
▸ ② 4×
1
Resistance is R = ρL/A, proportional to the length and inversely proportional to the cross-sectional area.
R = ρLA
2
With L doubled and A halved, R becomes 2 ÷ (1/2) = 4 times as large.
R ∝ 2LA/2 = 4·LA
🎯 Exam Points
①V = IR: voltage = current × resistance
②From P = VI = I²R = V²/R, pick the form that fits the given quantities
③In series, R = R₁ + R₂, so the resistance grows
④In parallel, 1/R = 1/R₁ + 1/R₂, so the resistance shrinks
⑤Electrical energy is W = Pt, in J or kWh
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Magnetic Field & Induction

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