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
Explanation
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
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