A First Course in Physics · Ch. XIV, §§ 369–374 · Chapter 14: Electricity in Motion
42. Internal resistance, series and parallel, and shunts
Goal: Your student can use Ohm's law for a whole circuit including the cell's own resistance, find the resistance of wires in series and in parallel, and solve the book's resistance problems.
You'll need
- The book, open to these pages
- Two flashlight bulbs in holders, two AA batteries in a holder, and wires
- A notebook and pencil
- A calculator (optional, for the problems)
What's in the book
The internal resistance of a cell; C = E/(Rₑ + Rᵢ); measuring internal resistance; measuring any resistance with an ammeter and voltmeter or with a high-resistance galvanometer; resistances in series and in parallel; shunts. Sixteen problems follow.
The lesson, step by step
Hook question
SayIf you add a second bulb to a circuit, do both bulbs get brighter, dimmer, or stay the same? It depends on how you connect it.
Read together
Read §§ 369–374. The rule for parallel wires:
When n exactly similar conductors are joined in parallel, i.e. in the manner shown in Fig. 284, the total resistance between a and b is 1/n of the resistance of one of them
Try it: series and parallel
Light one bulb from the batteries and note its brightness. Now connect two bulbs one after the other (series): both are dimmer, because the resistance has doubled. Then connect each bulb across the batteries separately (parallel): both are about as bright as one alone. Unscrew one bulb in each setup and see what happens to the other.
Safety: use only 1.5-volt AA or AAA batteries for every circuit in this course. Never connect anything to a wall outlet, and don't hold a bare wire straight across a battery for more than a few seconds (it gets hot and drains the battery).
Explain the idea
In series, resistances add. In parallel, there are more paths, so the total resistance is less. The cell itself has resistance (the liquid inside), so the full law is current = E.M.F. ÷ (outside resistance + inside resistance) (§ 369). A shunt is a side path: the currents divide inversely as the resistances (§ 374).
Key numbers for the problems
From the book: 9.35 ft of No. 30 copper wire, or 6.2 in. of No. 30 German silver wire, has 1 ohm (§ 367); No. 20 German silver wire has 190.2 ohms per 1000 ft. (problem 11). Resistance goes as length and inversely as the square of the diameter.
Say it back
Your student explains why the lights in a house are wired in parallel rather than in series.
Problems
Work problems 1–16 on book pages 279–280.
Afterwards
Look at a string of holiday lights: when one bulb goes out, does the whole string go dark? That tells you how they are wired.
Answers to the book's problems
1. So that it draws too little current to lower the P.D. it is measuring. 2. About 18 m (German silver resists about 18 times as much as copper). 3. ¼ ohm. 4. No. 20 has about 1/10 the resistance of No. 30 (0.098, from (10.025/31.96)²). 5. About 2 ft. (1.97 ft.). 6. About 56 ft. 7. 0.2 ampere. 8. 30 volts. 9. 220 ohms. 10. 0.54 ampere; 0.27 ampere; about 0.0106 ampere. 11. About 0.95 ohm. 12. About 1.9 ohms. 13. About 1.26 ohms (1.07 + 0.19). 14. 0.5 ampere (the ten wires in parallel make 0.5 ohm; 1 ÷ (1.5 + 0.5)). 15. 1/1001. 16. About 0.47 volt (3 × 3/19).
Worked out for this site from the scan. If a number in your copy differs, trust the book.