A) 25J
B) 50 J
C) \[{{H}_{2}}\]
D) 75 J
Correct Answer: C
Solution :
Impedance of the L-R circuit \[mg/m{{g}^{2+}}(0.01M)||Z{{n}^{2+}}(0.1M)/Zn\] \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.059}{2}{{\log }_{10}}\frac{[M{{g}^{2+}}]}{[Z{{n}^{2+}}]}\] \[=1.61V-\frac{0.059v}{2}{{\log }_{10}}\frac{0.01}{0.1}\] \[=\text{1}.\text{61V}+0.0\text{295V}=\text{1}.\text{6395V}\] \[({{C}_{7}}{{H}_{16}})=100\] Energy stored in the inductor \[\therefore \] \[{{n}_{hep}}=\frac{25}{100}\]You need to login to perform this action.
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