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 :
[b] The equivalent resistance of the circuit |
\[l=\frac{2E}{{{R}_{1}}+{{R}_{2}}+R}\] |
According to the question, |
\[-({{V}_{1}}-{{V}_{B}})\,=E-l\,{{R}_{2}}\] |
\[0=E-l\,{{R}_{2}}\] |
\[E=\,l{{R}_{2}}\] |
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