Faraday's Laws of Electrolysis

By Faraday's first law of electrolysis: $m = ZIt$
Here $Z$ is the electrochemical equivalent of copper.
$Z = 30 \times 10^{-5}$ g C$^{-1}$, $I = 1.5$ A, $t = 10$ min $= 10 \times 60 = 600$ s
$m = 30 \times 10^{-5} \times 1.5 \times 10 \times 60$
$m = 30 \times 10^{-5} \times 900$
$m = 0.27$ g

By Faraday's first law of electrolysis, the mass liberated is $m = ZIt$, where $Z$ is the electrochemical equivalent.
Power consumed: $P = VI \Rightarrow I = \dfrac{P}{V} = \dfrac{100}{125} = 0.8$ A
Time: $t = 1$ minute $= 60$ s
$m = ZIt = (Z)\left(\dfrac{P}{V}\right)(t) = (0.367 \times 10^{-6})\left(\dfrac{100}{125}\right)(60)$
$m = 0.367 \times 10^{-6} \times 0.8 \times 60 = 0.367 \times 10^{-6} \times 48$
$m = 1.76 \times 10^{-5}$ kg
$m = 17.6$ mg