الأسبوع 1 · الوحدة الأولى «الكهربية التيارية والكهرومغناطيسية»، الفصل الأول «التيار الكهربي وقانون أوم» (١): مراجعة التيار وفرق الجهد والقوة الدافعة الكهربية والمقاومة والمقاومة النوعية والتوصيلية وقانون أوم واتجاه التيار الاصطلاحي، ثم توصيل المقاومات على التوالي وعلى التوازي والمقاومة المكافئة · 3 دروس
2
Resistors in series
Physics Sec 3 Term 1, Chapter 1: pp. 8–12
GoalFind the equivalent resistance of resistors in series, then the current and the voltage across each one.
New words
series connection · resistors joined end to end, so that one single path carries the same current through all of themequivalent resistance · the single resistor that would draw the same current from the same sourcevoltage drop · the potential difference across one resistor, V = IR
Explanation
In a series connection the resistors are joined end to end, so the charge has only one path. Therefore the current is the same in every resistor, and the voltages across them add up to the source voltage: V = V₁ + V₂ + V₃. Since V = IR for each one, the equivalent resistance is the sum, R = R₁ + R₂ + R₃, and N equal resistors of r each give R = N·r. The total is larger than every single resistor. As the same current flows through all of them, the larger resistor takes the larger share of the voltage. Until Week 2 the source has no internal resistance. Solve in this order: add the resistances, find I = V ÷ R, then find each drop IR, and check that the drops add up to the source voltage. If one resistor in a series line breaks, the only path is cut and the current stops everywhere.
Worked examples
Resistor
R (Ω)
I (A)
Drop IR (V)
R₁
12
0.2
2.4
R₂
18
0.2
3.6
R₃
30
0.2
6.0
total
60
0.2
12.0
Three resistors of 12 Ω, 18 Ω and 30 Ω are in series on a 12 V source. R = 12 + 18 + 30 = 60 Ω, so I = 12 ÷ 60 = 0.2 A. The drops are 0.2 × 12 = 2.4 V, 0.2 × 18 = 3.6 V and 0.2 × 30 = 6.0 V, and 2.4 + 3.6 + 6.0 = 12 V.
Five equal resistors of 8 Ω each are in series. R = N·r = 5 × 8 = 40 Ω.
A 24 V source drives 0.4 A through a 20 Ω resistor in series with an unknown resistor R. The total resistance is 24 ÷ 0.4 = 60 Ω, so R = 60 − 20 = 40 Ω.
Class activity
Fill this table on paper: a 6 V source with 1, 2 and 3 equal resistors of 10 Ω in series. Find the total resistance and the current each time, and say what happens to the current when you add a resistor. With a grown-up you may also use one 1.5 V cell and a small torch bulb, then add a second identical bulb in series: predict first whether they will be brighter or dimmer, then see what happens. Stop if anything feels warm.
Practice
1
Find the equivalent resistance of 10 Ω, 25 Ω and 5 Ω in series.
R = 10 + 25 + 5 = 40 Ω.
2
Six equal resistors of 7 Ω are in series. Find the equivalent resistance.
R = N·r = 6 × 7 = 42 Ω.
3
A 6 Ω and a 14 Ω resistor are in series on 10 V. Find the current and the drop across each.
R = 6 + 14 = 20 Ω. I = 10 ÷ 20 = 0.5 A. Drops: 0.5 × 6 = 3 V and 0.5 × 14 = 7 V (3 + 7 = 10 V).
4
The total resistance of a series line is 90 Ω. Two of its resistors are 25 Ω and 35 Ω. Find the third.
