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

Q: 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......

c $i _{1}=\frac{40}{40+60}=0.4$ $i _{2}=\frac{40}{90+110}=\frac{1}{5}$ $v _{ B }+ i _{1}(40)- i _{2}(90)= v _{ D }$ $v _{ B }- v _{ D }=\frac{1}{5}(90)-\frac{4}{10} \times 40$ $v _{ B }- v _{ D }=18-16=2$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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......

A$4$
B$1$
C$2$
D$5$

JEE MAIN 2020, Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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