A) \[R=\frac{{{R}_{2}}({{R}_{1}}+{{R}_{2}})}{({{R}_{2}}-{{R}_{1}})}\]
B) \[R={{R}_{2}}-{{R}_{1}}\]
C) \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\]
D) \[R=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{2}}-{{R}_{1}})}\]
Correct Answer: B
Solution :
The equivalent resistance of the circuit \[l=\frac{2E}{{{R}_{1}}+{{R}_{2}}+R}\] According to the question, \[-({{V}_{1}}-{{V}_{B}})\,=E-l\,{{R}_{2}}\] \[0=E-l\,{{R}_{2}}\] \[E=\,l{{R}_{2}}\]You need to login to perform this action.
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