The emf, internal resistance and the closed circuit
Answers are for parents and teachers.
1
A cell of emf 12 V and internal resistance 0.2 Ω delivers 5 A to an outside resistor. Find the outside resistance and the terminal voltage.
The total resistance is 12 ÷ 5 = 2.4 Ω, so R = 2.4 − 0.2 = 2.2 Ω. V = 12 − 5 × 0.2 = 11 V, and IR = 5 × 2.2 = 11 V.
2
A cell reads V = 9.0 V at I = 0.2 A and V = 8.4 V at I = 0.8 A (invented data). Find its internal resistance and its emf.
r = (9.0 − 8.4) ÷ (0.8 − 0.2) = 0.6 ÷ 0.6 = 1.0 Ω. VB = 9.0 + 0.2 × 1.0 = 9.2 V.
3
A 36 V source with no internal resistance feeds a 40 Ω and a 10 Ω resistor in parallel, in series with a 10 Ω resistor. Find the total current and the current in each parallel branch.
The pair gives (40 × 10) ÷ 50 = 8 Ω; total 8 + 10 = 18 Ω; I = 36 ÷ 18 = 2 A. The pair has 2 × 8 = 16 V, so the branches carry 16 ÷ 40 = 0.4 A and 16 ÷ 10 = 1.6 A (sum 2 A).
4
A nearly flat cell reads about 1.5 V on a voltmeter of very high resistance but cannot light a small bulb. Explain why.
The voltmeter draws almost no current, so Ir is tiny and the reading is close to the emf. The bulb draws a larger current, and since the flat cell has a large r, a large part of the emf is lost inside as Ir, leaving too little voltage for the bulb.
5
A cell has emf 6 V and internal resistance 1.2 Ω. For which outside resistance is the terminal voltage 4.8 V?
Ir = 6 − 4.8 = 1.2 V, so I = 1.2 ÷ 1.2 = 1 A. R = V ÷ I = 4.8 ÷ 1 = 4.8 Ω. Check: 6 ÷ (4.8 + 1.2) = 1 A.