A) 5:1
B) 5:4
C) 3:4
D) 3:2
Correct Answer: D
Solution :
Case I. when cell are connected in series. \[{{\operatorname{E}}_{1}}+{{E}_{2}}=\operatorname{K}\times 50\] ?.(1) Case II. When cell are connected opposite to each other. \[{{\operatorname{E}}_{1}}-{{E}_{2}}=\operatorname{K}\times 10\] ?.(2) \[\left( 1 \right)\div \left( 2 \right)\] \[\frac{{{E}_{1}}+{{E}_{2}}}{{{E}_{1}}-{{E}_{2}}}=\frac{K\times 50}{K\times 10}\Rightarrow \frac{5}{1}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{2}\]You need to login to perform this action.
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