In the calibration of an ammeter using faraday ... - JAMB Physics 2003 Question
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
B
0.04A
C
0.05A
D
0.06A
correct option: c
From faraday's law of electrolysis
M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\ \text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
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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