Current and Resistance

Electric Current and Resistance

Current and resistance are easy with a water-pipe picture: voltage is pressure, current is flow, resistance is how narrow the pipe is. Ohm's law V = IR says raising voltage increases current and raising resistance decreases it. You can also track power as the rate of energy use and energy as power times time. Slide voltage and resistance here to see how current changes in the circuit.

Intuition for Current and Resistance

💧 Water Pipe Analogy

①Voltage (V) = water pressure — force pushing water

②Current (I) = water flow rate — actual amount flowing

③Resistance (R) = narrowness of the pipe — obstruction to flow

Visualizing Ohm's Law

Voltage V12 V

Resistance R4 Ω

💡 Key Observations

①Increasing voltage increases current proportionally

②Increasing resistance decreases current inversely

③V = IR is the most basic relation in circuits

Ohm's Law

V = IR

Voltage (V) = Current (A) × Resistance (Ω)

Solve for Current

I = VR

Current (A) = Voltage (V) / Resistance (Ω)

Solve for Resistance

R = VI

Resistance (Ω) = Voltage (V) / Current (A)

Power and Energy

Power (rate of consumption)

P = VI = I²R = V²/R

Power (W) = V × I — three convertible forms

Electric Energy

W = Pt = VIt

Energy (J) = power × time — actual energy consumed

💰 Electricity Bill and Energy

①1 kWh = 1000W × 3600s = 3.6 × 10⁶ J

②Electric bill = energy (kWh) × unit price

③A 100W bulb for 10 hours → 1 kWh consumed

Worked Examples

Example 1

A 12 V voltage is applied to a 6 Ω bulb. What current flows?

From Ohm's law V = IR, I = V/R.

I = VR

Substitute V = 12 V, R = 6 Ω.

I = 126 = 2 A

▸ 2 A

Ohm's law V = IR rearranges to I = V/R. Higher voltage and lower resistance give more current.

Example 2

A current of 2 A flows through a 10 Ω resistor. What is the power dissipated?

Use P = I²R (when current and resistance are known).

P = I² R

Substitute I = 2 A, R = 10 Ω.

P = 2² × 10 = 40 W

▸ 40 W

Power has three forms P = VI = I²R = V²/R. Pick the one matching the given quantities.

Summary

Ohm's Law

V = IR

Power

P = VI

CSAT-style

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×

Resistance is R = ρL/A, proportional to length and inversely proportional to area.

R = ρLA

If L doubles and A halves, R becomes 2 ÷ (1/2) = 4×.

R ∝ 2LA/2 = 4·LA

🎯 Exam Points

①V = IR: voltage = current × resistance (most fundamental!)

②P = VI = I²R = V²/R: pick the convenient form

③Series: R_total = R₁ + R₂ (resistance increases)

④Parallel: 1/R_total = 1/R₁ + 1/R₂ (resistance decreases)

⑤Energy W = Pt, units J or kWh

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Electric Field and Potential

Magnetic Field & Induction

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