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
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Correct Answer: Option A
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 * 0.2r
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω
therefore r = 2Ω
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 * 0.2r
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω
therefore r = 2Ω