• # question_answer In the shown circuit, all three capacitor are identical and have capacitance C each. Each resistor has resistance of R. An ideal cell of emf V is connected as shown. Then the potential difference across capacitor ${{C}_{3}}$ in steady state is:              A) $\frac{V}{3}$                                   B) $\frac{V}{2}$ C) $\frac{2}{9}V$                 D) $\frac{3}{4}V$

Solution :

The current through branch ABC and GFED in steady state is V/3 R each If we consider $={{105}^{\text{o}}}$ then $\vec{V}=(2,\,-1,\,1)\,+\lambda (1,\,2,\,-1)$ and $\frac{\vec{V}.\vec{r}}{|\vec{r}|}=\frac{4}{\sqrt{6}}\,\Rightarrow \,\,\frac{(\vec{p}.\vec{r})+\lambda (\vec{q}.\vec{r})}{|\vec{r}|}\,=\frac{4}{\sqrt{6}}$ Now the sum of charges on the plates towards point H of the capacitors is zero. $\therefore \,\frac{-1+\lambda (5)}{\sqrt{6}}\,=\frac{4}{\sqrt{6}}$$\therefore \,\,\lambda =1$ $\therefore \,\,\frac{f(2)-f(0)}{2-0}\,=f'(c)$ Therefore, the p.d. across $c\in (0,\,2)$ is $\Rightarrow \,\,\frac{f(2)+3}{2}\le 5$

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