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In network shown in figure below the potential difference between \mathrm{A} and \mathrm{B} is

Option: 1

2 \mathrm{V}


Option: 2

-0.5 \mathrm{V}


Option: 3

0.5 \mathrm{V}


Option: 4

-2 \mathrm{V}


Answers (1)

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The distribution of current is shown in fig. Keeping in view that the inflow and out flow of current in a cell must be same. Applying the loop rule to left and right loops.


2 \mathrm{i}_{1}+6 \mathrm{i}_{1}=4 \text { or } 2 \mathrm{i}_{1}=0.5 \mathrm{~A}
3 \mathrm{i}_{1}+1 \mathrm{i}_{2}=6 \text { or } \mathrm{i}_{2}=1.5 \mathrm{~A}

\mathrm{~V}_{\mathrm{AB}}=\sum \text { ir }-\sum e

(=-2 \times 0.5+3 \times 1.5-4=-0.5 \mathrm{V})
\mathrm{V}_{\mathrm{AB}}=-0.5 \mathrm{V}
 

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Gaurav

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