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In-an-electrolysis-experiment-the-ammeter-records-a-steady-current-of-1A-The-mass-of-copper-deposited-in-30minutes-is-0-66g-Calculate-the-error-in-the-ammeter-reading-the-electrical-equivalent-




Question Number 66106 by Umar last updated on 09/Aug/19
In an electrolysis experiment, the ammeter  records a steady current of 1A. The mass  of copper deposited in 30minutes is 0.66g.  Calculate the error in the ammeter reading.     the electrical equivalent of cu is 3.3×10^(−4) g/C.
$$\mathrm{In}\:\mathrm{an}\:\mathrm{electrolysis}\:\mathrm{experiment},\:\mathrm{the}\:\mathrm{ammeter} \\ $$$$\mathrm{records}\:\mathrm{a}\:\mathrm{steady}\:\mathrm{current}\:\mathrm{of}\:\mathrm{1A}.\:\mathrm{The}\:\mathrm{mass} \\ $$$$\mathrm{of}\:\mathrm{copper}\:\mathrm{deposited}\:\mathrm{in}\:\mathrm{30minutes}\:\mathrm{is}\:\mathrm{0}.\mathrm{66g}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ammeter}\:\mathrm{reading}. \\ $$$$\:\:\:\mathrm{the}\:\mathrm{electrical}\:\mathrm{equivalent}\:\mathrm{of}\:\mathrm{cu}\:\mathrm{is}\:\mathrm{3}.\mathrm{3}×\mathrm{10}^{−\mathrm{4}} \mathrm{g}/\mathrm{C}. \\ $$
Answered by MrGaster last updated on 06/Jan/25
m=66g,t=30min=1800s,Z=3.3×10^(−4) (−4)g/c  m=Zlt⇒l(m/Zt))…(i)  l=((0.66)/(3.3+10^(−4) ×1800))(⇒(i))  =1.AA  error=∣I_(reality) −I_(reading) ∣  =∣1.11−1∣  =0.11A  relative error=((error)/I_(reading) )×100%  =((0.11)/1)×100%  =11%  The reading error of the electrict  meer is 0.11A,relative error 11%
$${m}=\mathrm{66}{g},{t}=\mathrm{30min}=\mathrm{1800}{s},{Z}=\mathrm{3}.\mathrm{3}×\mathrm{10}^{−\mathrm{4}} \left(−\mathrm{4}\right){g}/{c} \\ $$$$\left.{m}={Z}\mathrm{lt}\Rightarrow\mathrm{l}\left(\mathrm{m}/{Zt}\right)\right)\ldots\left(\mathrm{i}\right) \\ $$$$\mathrm{l}=\frac{\mathrm{0}.\mathrm{66}}{\mathrm{3}.\mathrm{3}+\mathrm{10}^{−\mathrm{4}} ×\mathrm{1800}}\left(\Rightarrow\left(\mathrm{i}\right)\right) \\ $$$$=\mathrm{1}.{AA} \\ $$$$\mathrm{error}=\mid{I}_{\mathrm{reality}} −{I}_{\mathrm{reading}} \mid \\ $$$$=\mid\mathrm{1}.\mathrm{11}−\mathrm{1}\mid \\ $$$$=\mathrm{0}.\mathrm{11}{A} \\ $$$$\mathrm{relative}\:\mathrm{error}=\frac{\mathrm{error}}{{I}_{\mathrm{reading}} }×\mathrm{100\%} \\ $$$$=\frac{\mathrm{0}.\mathrm{11}}{\mathrm{1}}×\mathrm{100\%} \\ $$$$=\mathrm{11\%} \\ $$$$\mathrm{The}\:\mathrm{reading}\:\mathrm{error}\:\mathrm{of}\:\mathrm{the}\:\mathrm{electrict} \\ $$$$\mathrm{meer}\:\mathrm{is}\:\mathrm{0}.\mathrm{11}{A},\mathrm{relative}\:\mathrm{error}\:\mathrm{11\%} \\ $$

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