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Question Number 66106 by Umar last updated on 09/Aug/19

$$\mathrm{In}\mathrm{an}\mathrm{electrolysis}\mathrm{experiment},\mathrm{the}\mathrm{ammeter}$$$$\mathrm{records}\mathrm{a}\mathrm{steady}\mathrm{current}\mathrm{of}1\mathrm{A}.\mathrm{The}\mathrm{mass}$$$$\mathrm{of}\mathrm{copper}\mathrm{deposited}\mathrm{in}30\mathrm{minutes}\mathrm{is}0.66\mathrm{g}.$$$$\mathrm{Calculate}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{ammeter}\mathrm{reading}.$$$$\mathrm{the}\mathrm{electrical}\mathrm{equivalent}\mathrm{of}\mathrm{cu}\mathrm{is}3.3\times {10}^{-4}\mathrm{g}/\mathrm{C}.$$

Answered by MrGaster last updated on 06/Jan/25

$$m=66g,t=30\min =1800s,Z=3.3\times {10}^{-4}(-4)g/c$$$$m=Z\mathrm{lt}\Rightarrow \mathrm{l}\left(\mathrm{m}/Zt\right))\dots \left(\mathrm{i}\right)$$$$\mathrm{l}=\frac{0.66}{3.3+{10}^{-4}\times 1800}(\Rightarrow \left(\mathrm{i}\right))$$$$=1.AA$$$$\mathrm{error}=\mid I_{\mathrm{reality}}-I_{\mathrm{reading}}\mid$$$$=\mid 1.11-1\mid$$$$=0.11A$$$$\mathrm{relative}\mathrm{error}=\frac{\mathrm{error}}{I_{\mathrm{reading}}}\times 100\mathrm{\%}$$$$=\frac{0.11}{1}\times 100\mathrm{\%}$$$$=11\mathrm{\%}$$$$\mathrm{The}\mathrm{reading}\mathrm{error}\mathrm{of}\mathrm{the}\mathrm{electrict}$$$$\mathrm{meer}\mathrm{is}0.11A,\mathrm{relative}\mathrm{error}11\mathrm{\%}$$

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