This agrees with the official scheme. Teacher extension: treating 400 ms as 400 s would make the denominator 1000 times too large and current 1000 times too small. Check the time unit before substituting.
At constant temperature, for the same material and cross-sectional area, resistance is directly proportional to wire length. Plot calculated resistance against length; a straight line through the origin supports this relationship. The measured potential difference need not be identical at each length, so calculate each resistance from its own paired readings.
In the second part of this practical, connect two equal resistors in series, then in parallel. With the ammeter measuring total current and the voltmeter across the whole combination, measure total potential difference and current and calculate total resistance. Compare with one resistor: series has greater total resistance; parallel has smaller total resistance. For two identical 10 Ω resistors, ideal totals are 20 Ω in series and 5 Ω in parallel. The parallel result can be explained from the doubled total current at the same potential difference, without needing a reciprocal-resistance formula.
สำรวจ
One operating point: voltage, current and resistance
$I=\frac{V}{R};\quad P=VI$
At the worked operating point, 6 V across 20 ohms produces 0.30 A and 1.8 W.
การปฏิบัติที่จำเป็น 4 วัดกระแสผ่านตัวต้านทาน หลอดไส้ และไดโอดที่แรงดันไฟฟ้าต่าง ๆ ที่วัดได้Across each component. Connect an ammeter in series and a voltmeter in parallel with the component. Vary the pd using a variable dc supply, or a variable resistor in series. Start at zero and stay within component ratings. Record paired readings across a suitable range; repeat and investigate inconsistent readings. Switch off before reversing the supply connections to obtain negative values, using meters that can read the reversed polarity. Plot current vertically against potential difference horizontally.
A และ V วัดกระแสผ่านและแรงดันไฟฟ้า Across the lamp.
สำหรับการตรวจสอบหลอดไส้ใน AQA มิถุนายน 2023 8463/1H Q06.1, Figure 6 ครอบคลุม −6 V ถึง +6 V พร้อมการอ่านค่าที่ 1 V ช่วง Collect positive values, then reverse the supply to obtain the negative values; these settings belong to that lamp investigation, rather than every possible component.
สำหรับไดโอด ให้ใช้ตัวต้านทานป้องกันที่เหมาะสมต่ออนุกรมเพื่อจำกัดกระแส และใช้มิลลิแอมมิเตอร์วัดกระแสขนาดเล็ก The protective resistor, not the milliammeter, protects the diode. Measure pd across the diode alone, excluding the protective resistor. Keep the ohmic resistor near constant temperature; the lamp's changing filament temperature is part of the effect being investigated.
Diode: conducts in the forward direction; reverse current is negligible in this model, so reverse resistance is very high. Do not assume every diode has exactly the same forward voltage.
สำหรับส่วนประกอบแบบขนาน: ระบุว่าแรงดันไฟฟ้าเท่ากัน Across Each ตัว ความรวมของกระแสคือผลบวกของกระแสในแต่ละสาขา และความต้านทานรวมของตัวต้านทานสองตัวมีค่าน้อยกว่าความต้านทานรายตัวที่มีค่าน้อยที่สุด
ส่วนแบ่งเหล่านี้รวมกันได้ 6.0 V ส่วนแบ่งเท่ากันเกิดขึ้นเฉพาะเมื่อความต้านทานเท่ากันที่จุดทำงานเท่านั้น ส่วนประกอบแบบอนุกรมไม่จำเป็นต้องแบ่งแรงดันเท่ากันเสมอไป
Teacher-written comparison with fixed resistors. Two 8.0 Ω resistors are connected to an ideal 12 V supply. In series, $R_{total}=R_1+R_2=16\ \Omega$ and $I=V/R_{total}=0.75\ \text{A}$; each resistor has $V=IR=6.0\ \text{V}$. In parallel, each branch has 12 V, so each branch current is $I=V/R=1.5\ \text{A}$ and $I_{total}=I_1+I_2=3.0\ \text{A}$. Adding another parallel resistor gives another current path and increases total current at the same supply pd. No reciprocal-resistance formula is needed here.
The fixed resistor has the remaining 10 V. This example uses the graph reading supplied above; the complete exam question also requires reading that resistance from the graph. A fixed supply pd does not make the thermistor pd equal to the supply pd.
อธิบายว่าทำไมสายไฟ Live อาจเป็นอันตรายได้แม้ว่าสวิตช์ในวงจร mains จะเปิดอยู่
อธิบายอันตรายของการเชื่อมต่อใดๆ ระหว่างสายไฟ Live กับสายดิน
แหล่งที่มา: หลักสูตร Cambridge International
The UK mains supply is alternating (ac): the potential difference repeatedly changes polarity. Its frequency is 50 Hz, meaning 50 complete cycles per second, and its quoted potential difference is about 230 V. Batteries provide direct (dc) potential difference with one polarity. A dc potential difference need not be perfectly constant in magnitude; its direction does not reverse.
The open switch stops the lamp current in this ideal circuit. Point A remains connected to the live supply, at about 230 V relative to earth. A person making a conducting connection from that live point to earth can receive an electric shock. Do not infer from an unlit appliance that every part of the circuit is isolated. This does not mean that point B after a correctly wired open live switch must also remain live.
The danger depends on the current through the body, its path and duration. A human body is not a zero-resistance wire, but a current much smaller than a typical appliance fuse rating can still cause severe injury. An appliance fuse does not guarantee protection against touching live wiring.
For the classroom fault model, suppose the live wire touches an exposed metal case that has a sound protective-earth connection. The earth conductor supplies a low-resistance fault path; the resulting large current heats and melts a suitably rated fuse in the live wire, breaking the live supply. Without that earth connection, a case can become live without enough current to operate the fuse. A fuse's protection against excessive current is different from a claim that every possible shock current will blow it.
Teacher-written ideal model: a lamp is connected to a single-phase live and neutral supply. The neutral connection breaks between the lamp and the supply. The downstream neutral terminal remains connected to live through the lamp; there is no other return path. No normal load current flows, but that downstream terminal can be at live potential.
Condition
Live-to-earth pd
Load-side neutral-to-earth pd
Pd across lamp
ปกติ
ประมาณ 230 V
ประมาณ 0 V
ประมาณ 230 V
สายกลางขาด
ประมาณ 230 V
ประมาณ 230 V
ประมาณ 0 V
This ideal model explains an unlit lamp with a dangerous downstream terminal. Both lamp terminals are at approximately the same potential, so the lamp pd is near zero; either can still have a large pd relative to earth. It is inconsistent to assign 230 V both across this unlit ideal lamp and from each of its terminals to earth in the stated single-phase model.
Electrical appliances transfer energy from batteries or the mains. A motor transfers energy mechanically to moving objects; a heater transfers energy to the thermal store of its surroundings. Power is the rate of energy transfer: 1 W means 1 J each second. A rating states this rate at the specified working potential difference; it is not the total energy used.
A step-down transformer lowers the transmission pd for consumers; UK domestic appliances use about 230 V. This is a lower and more suitable value than transmission pd; it can still cause a dangerous electric shock. Transformer construction and operation are taught in topic 4.7; this section explains their system-level roles.
คำอธิบายข้อสอบจริง — AQA มิถุนายน 2022 8463/1H Q06.1–06.2 ใบข้อสอบวางหม้อแปลง X ไว้ก่อนสายส่งไฟฟ้าเหนือพื้นดินและ Y ไว้ก่อนผู้ใช้ X เพิ่มแรงต่างศักย์ ลดกระแส ลดการถ่ายโอนความร้อนสู่สิ่งแวดล้อมและเพิ่มประสิทธิภาพการส่งผ่าน Y ลดแรงต่างศักย์ให้เป็นค่าที่ปลอดภัยสำหรับผู้ใช้ อย่าแทนที่คำอธิบาย X ด้วยแค่ “มีประสิทธิภาพมากขึ้น”: ต้องระบุลำดับเหตุการณ์ทางฟิสิกส์
Sheet 2.7 comparison. At 2000 A through 40 Ω, $P_{\text{loss}}=I^2R=(2000\ \text{A})^2\times40\ \Omega=1.6\times10^8\ \text{W}$. At 500 A through the same resistance, $P_{\text{loss}}=I^2R=(500\ \text{A})^2\times40\ \Omega=1.0\times10^7\ \text{W}$. Current is one quarter, so loss is one sixteenth. Without a stated input, do not claim these losses are a small percentage of the total.
Charged objects exert forces without contact: like charges repel; unlike charges attract — a non-contact force. A large potential difference can create a strong electric field across a small air gap. If the field is strong enough, the air becomes conducting (electrical breakdown), and charge flows briefly across the gap as a spark. An earthed conductor can receive a spark; earthing does not remove a nearby high-voltage source.
