question_answer

A cell of bmfX is connected across a resistor R. The potential difference across the wire is measured as \[Y.\]The internal resistance of the cell should be

A)  \[\frac{X-Y}{R}\]                             

B)  \[\frac{(X-Y)R}{Y}\]

C)  \[\frac{(X-Y)Y}{R}\]                       

D)  \[(X-Y)R\]

Correct Answer: B

Solution :

                 Let \[I\] be the current through the circuit, then from Ohms law                                                 Given,   E = X, \[\therefore \]  \[I=\frac{X}{R+r}\]                           ... (i) The potential difference across resistor is                 \[{{V}_{R}}=IR\]                                ... (ii) Given    \[{{V}_{R}}=Y\], From Eqs. (i) and (ii), we get                 \[Y=\frac{X\,R}{R+r}\] \[\Rightarrow \]               \[Yr=(X-Y)R\] \[\Rightarrow \]               \[r=\frac{(X-Y)}{Y}R\]