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In the Wheatstone's bridge (shown in figure) $X=Y$ and $A \gt B$. The direction of the current between $a b$ will be
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From $b$ to $a$
In the part cbd,
\(V_c-V_b=V_b-V_d \Rightarrow V_b=\frac{V_c+V_d}{2}\)
In the part cad,
\(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{a}}>\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}} \Rightarrow \frac{\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{d}}}{2}>\mathrm{V}_{\mathrm{a}} \Rightarrow \mathrm{V}_{\mathrm{b}}>\mathrm{V}_{\mathrm{a}}\)
\(V_c-V_b=V_b-V_d \Rightarrow V_b=\frac{V_c+V_d}{2}\)
In the part cad,
\(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{a}}>\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}} \Rightarrow \frac{\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{d}}}{2}>\mathrm{V}_{\mathrm{a}} \Rightarrow \mathrm{V}_{\mathrm{b}}>\mathrm{V}_{\mathrm{a}}\)
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