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Loaded Unbalanced Bridge

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Question:

Cramster Expert

dhara

(Cramster SME)

Cramster In-House Subject Matter Expert

Date Posted: 7/25/2008 12:58:32 AM Status: Live

Asker's Rating: Helpful

Response:

Applying Kirchoff's law to the first loop,we get

40i₁ + (i₁ - i₂)60 + 20i₃ = 0

5i₁ - 3i₂ + i₃ = 0 ------(1)

For the second loop,we get

(i₁ - i₂) * 60 + 70i₄ + 50i₅ = 0

or 6i₁ - 6i₂ + 7i₄ + 5i₅ = 0 -------(2)

For the loop ABCDA,we get

40i₁ + 70i₄ + 20 = 0

or 4i₁ + 7i₄ = -2 -------(3)

For the loop ADCA,we get

20i₃ + 50i₅ + 20 = 0

or 2i₃ + 5i₅ = -2 ------(4)

Solving the above four equations,we get the values of i₁,i₂,i₃,i₄ and i₅.The voltage at point B is

(60i₂) and the potential at point D is (20i₃ - 50i₅).

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