The potential difference between points $A$ and $D$ in the given circuit is:-

\frac{2}{3} V

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\frac{8}{9} V

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\frac{4}{3} V

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2 V

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Solution

The correct option is A $\frac{2}{3} V$
$\begin{matrix} V_{A} - V_{C} = \frac{5}{10 + 5} \times 2 = \frac{2}{3} V - - - - (i) V_{D} - V_{C} = \frac{10}{10 + 5} \times 2 = \frac{4}{3} V - - - - (i) s u b t r a c i n g (i) f r o m (i i) w e g e t V_{D} - V_{C} - V_{A} + V_{C} = V_{A D} = (\frac{4}{3} - \frac{2}{3}) V o r, V_{D} - V_{A} = V_{A D} = \frac{2}{3} V ∴ V_{A D} = \frac{2}{3} V H e n c e, o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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The potential difference between points A and D in the given circuit is:-

The correct option is A 23VVA−VC=510+5×2=23V−−−−(i)VD−VC=1010+5×2=43V−−−−(i)subtracing(i)from(ii)wegetVD−VC−VA+VC=VAD=(43−23)Vor,VD−VA=VAD=23V∴VAD=23VHence,optionAisthecorrectanswer.

The potential difference between points $A$ and $D$ in the given circuit is:-