电流diànliú

Core Concept

Electric current (电流) is the rate at which charge flows past a point in a circuit. Together with resistance it forms 电流和电阻, one of the highest-frequency knowledge points in CSCA physics.

Definition

$I = \frac{Q}{t}$

• $Q$ — charge that passes, in coulombs ($\mathrm{C}$)
• $t$ — time, in seconds
• $I$ — current, in amperes ($\mathrm{A}$); $1\ \mathrm{A} = 1\ \mathrm{C/s}$

Conventional Direction

Current is defined as flowing from the positive terminal to the negative terminal through the external circuit. In a metal the actual carriers are electrons, which drift the opposite way. This is a historical convention, not a mistake — and CSCA questions do ask about it directly.

Ohm's Law

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

Current is proportional to voltage and inversely proportional to resistance. All three DC-circuit knowledge points reduce to this relation plus the series/parallel rules below.

Series and Parallel

	Series 串联	Parallel 并联
Current	same through each element	splits between branches
Voltage	divides across elements	same across each branch
Total resistance	$R = R_1 + R_2$	$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

The line to memorise: series shares current, parallel shares voltage. Nearly every CSCA circuit question is solved by applying that sentence and then Ohm's law.

Kirchhoff's Current Law

Charge is conserved, so at any junction the current in equals the current out:

$\sum I_{\text{in}} = \sum I_{\text{out}}$

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★☆☆☆)

If $12\ \mathrm{C}$ of charge passes through a wire in $4\ \mathrm{s}$, the current is:

Options:

• A. $0.33\ \mathrm{A}$
• B. $3\ \mathrm{A}$
• C. $8\ \mathrm{A}$
• D. $48\ \mathrm{A}$

Detailed Solution:

$I = \frac{Q}{t} = \frac{12}{4} = 3\ \mathrm{A}$

Option A inverts the ratio; option D multiplies instead of dividing. Both come from writing the formula from memory rather than from the meaning ("charge per second").

Answer: B

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

A $12\ \mathrm{V}$ source drives two resistors, $R_1 = 2\ \Omega$ and $R_2 = 4\ \Omega$, connected in series. Find the current and the voltage across $R_2$.

Options:

• A. $2\ \mathrm{A}$, $8\ \mathrm{V}$
• B. $2\ \mathrm{A}$, $4\ \mathrm{V}$
• C. $6\ \mathrm{A}$, $8\ \mathrm{V}$
• D. $3\ \mathrm{A}$, $12\ \mathrm{V}$

Detailed Solution:

1. Series, so resistances add: $R = R_1 + R_2 = 2 + 4 = 6\ \Omega$

2. Ohm's law for the whole loop: $I = \frac{U}{R} = \frac{12}{6} = 2\ \mathrm{A}$

3. Series shares current, so the same $2\ \mathrm{A}$ flows through $R_2$: $U_2 = IR_2 = 2 \times 4 = 8\ \mathrm{V}$

4. Check: $U_1 = 2 \times 2 = 4\ \mathrm{V}$, and $4 + 8 = 12\ \mathrm{V}$ ✓ — the drops must sum to the source.

Answer: A

Common Mistakes

❌ Mistake 1: Thinking current is "used up"

Current is the same at every point of a series circuit. A bulb does not consume current; it converts energy while the same charge flows through.

❌ Mistake 2: Adding parallel resistances directly

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ — the total is always smaller than the smallest branch. If your answer is larger, you added instead.

❌ Mistake 3: Confusing conventional current with electron flow

Conventional current runs positive to negative; electrons drift negative to positive.

❌ Mistake 4: Applying the same voltage to series elements

Series elements share current, not voltage. That swap is the most common circuit error on the exam.

Study Tips

1. ✅ Say "series shares current, parallel shares voltage" out loud before every circuit problem
2. ✅ Find total resistance first, then total current, then work back to individual elements
3. ✅ Verify the voltage drops sum to the source — a free error check
4. ✅ Sanity-check parallel totals: smaller than every branch
5. ✅ Keep $I = Q/t$ and $I = U/R$ distinct — different questions call for different ones

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💡 Exam Tip: 电流和电阻 and 计算电阻串联或并联时的电压 together dominate CSCA DC-circuit questions. Almost all are two steps: reduce the network, then apply Ohm's law.