Four resistances $40 \ \Omega, 60\ \Omega, 90\ \Omega$ and $110\ \Omega$ make the arms of a quadrilateral $A,B,C,D$. Across $AC$ is a battery of emf $40\, V$ and internal resistance negligible. The potential difference across $BD$ is $V$ is......
JEE MAIN 2020, Medium
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$i _{1}=\frac{40}{40+60}=0.4$
$i _{2}=\frac{40}{90+110}=\frac{1}{5}$
$v _{ B }+ i _{1}(40)- i _{2}(90)= v _{ D }$
$v _{ B }- v _{ D }=\frac{1}{5}(90)-\frac{4}{10} \times 40$
$v _{ B }- v _{ D }=18-16=2$
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