JAMB - Physics (2001 - No. 25)
A cell can supply current of 0.4 A and 0.2 A through a 4.0Ω and 10.0Ω resistors respectively.
This internal resistance of the cell is
This internal resistance of the cell is
2.0Ω
1.0Ω
2.5Ω
1.5Ω
Giải thích
let the internal resistance of the cell = r,
therefore E = I (R+r)
(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)
therefore 0.4 (4+r)= 0.2 (10+r)
1.6 + 0.4r = \(2.0 \times 0.2\text{r}\)
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = \(\frac{0.4}{0.2}\) = 2Ω
therefore r = 2Ω
therefore E = I (R+r)
(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)
therefore 0.4 (4+r)= 0.2 (10+r)
1.6 + 0.4r = \(2.0 \times 0.2\text{r}\)
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = \(\frac{0.4}{0.2}\) = 2Ω
therefore r = 2Ω