Physics
JAMB 2003
In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is
-
A.
0.03A
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B.
0.04A
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C.
0.05A
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D.
0.06A
Correct Answer: Option C
Explanation
From faraday's law of electrolysis
M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\
\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
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