电阻diànzǔ

Core Concept

Resistance (电阻) measures how strongly a conductor opposes the flow of current. It is a property of the conductor itself — its material, shape, and temperature — not of the circuit it sits in.

Definition

$R = \frac{U}{I}$

• SI unit: ohm ($\Omega$); $1\ \Omega = 1\ \mathrm{V/A}$

A careful distinction the exam tests: this equation defines how to measure $R$, but it does not mean resistance depends on $U$ and $I$. Double the voltage across an ohmic conductor and the current doubles too — the ratio, and therefore $R$, is unchanged.

What Resistance Actually Depends On

$R = \rho \frac{L}{A}$

• $\rho$ — resistivity (电阻率), a property of the material
• $L$ — length: longer wire, more resistance
• $A$ — cross-sectional area: thicker wire, less resistance

So a wire stretched to twice its length (halving its area, since volume is fixed) has four times the resistance — a favourite exam twist.

Temperature

For metals, resistance rises with temperature. For semiconductors and electrolytes it falls. This reversal is worth remembering.

Combining Resistances

$\text{Series: } R = R_1 + R_2 + \cdots$

$\text{Parallel: } \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

For exactly two in parallel, the shortcut is:

$R = \frac{R_1 R_2}{R_1 + R_2}$

And two equal resistors $R_0$ in parallel give simply $R_0/2$.

Useful check: a series total is larger than the largest element; a parallel total is smaller than the smallest.

Power Dissipated

$P = UI = I^2R = \frac{U^2}{R}$

Which form to pick depends on what is held constant: use $P = I^2R$ in a series circuit (shared current), and $P = U^2/R$ in a parallel circuit (shared voltage).

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★☆☆☆)

Two resistors, $6\ \Omega$ and $3\ \Omega$, are connected in parallel. The total resistance is:

Options:

• A. $2\ \Omega$
• B. $4.5\ \Omega$
• C. $9\ \Omega$
• D. $18\ \Omega$

Detailed Solution:

Use the two-resistor shortcut:

$R = \frac{R_1R_2}{R_1+R_2} = \frac{6 \times 3}{6+3} = \frac{18}{9} = 2\ \Omega$

Check it against the rule: $2\ \Omega$ is smaller than both branches ✓. Option C is the series answer; option B averages them, which is never correct.

Answer: A

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Example 2: Intermediate (Difficulty ★★★☆☆)

A uniform wire of resistance $R$ is cut into three equal pieces, and the three pieces are connected in parallel. The resistance of the combination is:

Options:

• A. $R/9$
• B. $R/3$
• C. $3R$
• D. $9R$

Detailed Solution:

1. Cutting into three equal pieces divides the length by three, and $R \propto L$, so each piece has resistance $R/3$.

2. Three equal resistors in parallel give one third of one of them: $R_{\text{total}} = \frac{R/3}{3} = \frac{R}{9}$

The factor of nine appears because both steps divide by three — once by cutting, once by paralleling.

Answer: A

Common Mistakes

❌ Mistake 1: Believing $R$ changes when $U$ changes

For an ohmic conductor at fixed temperature, $R$ is constant; $U$ and $I$ rise and fall together.

❌ Mistake 2: Averaging parallel resistances

$\frac{R_1+R_2}{2}$ is never the parallel total. Use the reciprocal rule or the two-resistor shortcut.

❌ Mistake 3: Forgetting area changes when a wire is stretched

Stretching conserves volume: doubling the length halves the area, so $R$ quadruples.

❌ Mistake 4: Using the wrong power formula

In series use $P = I^2R$ (bigger $R$ dissipates more); in parallel use $P = U^2/R$ (bigger $R$ dissipates less). Reaching for the wrong one inverts the answer.

Study Tips

1. ✅ Separate the definition $R = U/I$ from the dependence $R = \rho L/A$
2. ✅ Sanity-check totals: series > largest, parallel < smallest
3. ✅ Memorise $R_1R_2/(R_1+R_2)$ for the two-resistor case
4. ✅ On stretching problems, track length and area
5. ✅ Choose the power formula by what is shared — current in series, voltage in parallel

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💡 Exam Tip: 电流和电阻 is a top-three CSCA physics knowledge point, and 计算电阻串联或并联时的电压 builds directly on it. Reduce the network first; everything else follows from Ohm's law.