Question:
A $16 \Omega$ wire is bend to form a square loop. A $9 \mathrm{~V}$ supply having internal resistance of $1 \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is ____________$\times 10^{-1} \mathrm{~V}$
Solution:
here assume current as
By KVL in outer loop
$9-12 i-4 i=0$
$16 i=9$
$8 i=\frac{9}{2}=4.5$
$=45 \times 10^{-1}$
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