A) \[\frac{V}{3}\]
B) \[\frac{V}{2}\]
C) \[\frac{2}{9}V\]
D) \[\frac{3}{4}V\]
Correct Answer: C
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 \[\Rightarrow \,\,\frac{f(2)+3}{2}\le 5\]You need to login to perform this action.
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