question_answer

A cell of emf E and internal resistance r is connected across a variable external resistance R. Plot graph to show variation of

(i) E with R

(ii) terminal potential difference of the cell if with R.

Or

Distance between the plates of a capacitor of capacitance C is d. A very thin metal sheet is placed as shown in the figure. Calculate the new capacitance.

Plate capacitor

Answer:

(i) The emf of a cell is independent of the external resistance R. Therefore, graph between E and R will be a straight line parallel to R-axis. (ii) Terminal potential difference of a cell             \[V=IR=\left( \frac{E}{R+r} \right)R=\frac{E}{\left( 1+\frac{r}{R} \right)}\] When the external resistance R increases, r/R decreases and therefore terminal potential difference of the cell increases. Therefore, graph between E and R will be a curve as shown below. When    R = 0, \[V=\frac{E}{1+(r/0)}=0\] When    R = r, \[V=\frac{E}{1+(r/r)}=\frac{E}{2}\] When    \[R=\infty ,\] ss Or Now, the capacitor will act as a combination of two capacitors connected in series. As,        \[C\propto \frac{1}{d}\] \[\therefore \]      \[{{C}_{1}}=4C\] and \[{{C}_{2}}=4C/3\] \[\therefore \] Equivalent capacitance of the combination, \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{1}{4C}+\frac{1}{\frac{4C}{3}}=\frac{1+3}{4C},\] \[\Rightarrow \]   \[\frac{1}{C'}=\frac{1}{C}\Rightarrow {{C}^{'}}=C\] Therefore, the capacitance will remain same.