A) 24.13 J/sec
B) 2.413 J/sec
C) 0.2413 J/sec
D) 2413 J/sec \[({{Z}_{Cu}}=6.6\times {{10}^{-4}}gm/C\] and \[{{Z}_{Ag}}=11.2\times {{10}^{-4}}\,gm/C)\]
Correct Answer: A
Solution :
The current taken by the silver voltameter \[{{I}_{1}}=\frac{m}{Zt}=\frac{1}{11.2\times {{10}^{-4}}\times 30\times 60}=0.496\,\,A\] and by copper voltameter \[{{I}_{2}}=\frac{1.8}{6.6\times {{10}^{-4}}\times 30\times 60}=1.515\,\,A\] Total current \[I=({{I}_{1}}+{{I}_{2}})=2.011\,\,A\] Power \[P=IV=2.011\times 12=24.132\,\,J/\sec \]
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