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JEE Mains · Physics · STD 12 - 3. current electricity

પરિપથમાં \(40\  \Omega, 60\  \Omega, 90\  \Omega\) અને \(110\  \Omega\) જોડેલા છે. \(AC\) વચ્ચે \(40\, V\) ની બેટરી જોડેલ છે તો \(BD\) વચ્ચે વૉલ્ટેજ ......... \(V\)

  1. A \(4\)
  2. B \(1\)
  3. C \(2\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

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