A) \[\frac{R-X}{2Y}\]
B) \[\frac{X-R}{Y+R}\]
C) \[\frac{Y}{X}+R\]
D) \[\frac{X-R}{R}\]
E) \[\left( \frac{X-Y}{Y} \right)R\]
Correct Answer: E
Solution :
Let r be internal resistance of the cell. The current through the circuit \[I=\frac{E}{(R+r)}\] Substituting \[E=x,\]we get \[I=\frac{X}{R+r}\] Potential difference across \[R\] is \[V=IR\] when \[V=Y,\] then \[I=\frac{X}{R+r}\] \[\Rightarrow \] \[Y=\frac{XR}{R+r}\] which on solving, gives \[r=\left( \frac{X-Y}{Y} \right)R\]You need to login to perform this action.
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