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The four arms of wheatstone bridge have the following resistanes,AB=100Ω, BC=10 Ω,CD=5 Ω, DA=60 Ω A Galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when 10V is maintained across AC.
The four arms of wheatstone  bridge have the following  resistanes,AB=100Ω, BC=10 Ω,CD=5 Ω, DA=60 Ω   A Galvanometer of 15 Ω resistance is connected across BD. Calculate the current through the galvanometer when 10V is maintained across AC.

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4 years ago Saurabh Kumar
2411 Points
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4 years ago
```								The four arms of a Wheatstone bridge have the following resistances:AB = 50Ω, BC = 5Ω, CD = 2Ω, DA = 25Ω. A galvanometer of 10Ω resistance is connected acrossBD. If a potential difference of 20V is maintained across AC, using Thevenin’s theorem, calculate the current through the galvanometer and draw the Thevenin and Norton Equivalent circuit
```
2 years ago
```							Considering the mesh ABDA, we have100I1 + 15Ig – 60I2 = 0or 20I1 + 3Ig – 12I2 = 0  ——(a)Considering the mesh BCDB, we have10 (I1 – Ig) – 5 (I2 + Ig) – 15Ig = 010I1 – 30Ig – 5I2 = 02I1 – 6Ig – I2 = 0 ——(b)Considering the mesh ADCEA,60I2 + 5 (I2 + Ig) = 1065I2 + 5Ig = 1013I2 + Ig = 2 ——(c)Multiplying Eq. (b) by 1020I1 – 60Ig – 10I2 = 0 ——(d)From Eqs. (d) and (a) we have63Ig – 2I2 = 0I2 = 31.5Ig ——(e)Substituting the value of I2 into Eq. [(c)], we get13 (31.5Ig ) + Ig = 2410.5 Ig = 2Ig = 4.87 mA.
```
6 months ago
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