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Determine the current flowing through th...

Determine the current flowing through the galvanometer (G) as shown in the figure.

Text Solution

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Applying Krchhoff.s is current law at P
` I_1 + I_2 = 2 " …(1)"`
Applying Kirchhoff.s voltage law for the closed loop PQSA, we get,
` 5 I_1 + 10I_g - 15 I_2 = 0`
devided by 5 we get,
` I_1 + 2 I_g - 3 I_2 = 0" ...(2)"`
Applying Kirchhoff.s Voltage law for the closed loop QRSQ, we get
`(I_1 - I_g) 10 - (I_2 + l_g) 20- 10I_g = 0`
` 10I_1 - 10I_g - 20 I_2 - 20 I_g - 10I_g = 0`
` 10 I_1 - 20 I_2 - 40 I_g = 0`
devided by 10 we get
` I_1 - 2I_2 - 4I_g = 0`
` I_1 - 4 I_g - 2I_2 = 0" ...(3)"`
From (3) and (2), we get
` - 5I_g + I_2 = 0 `
`:." "I_2 = 6Ig" ...(4)"`
From equation (1)
` I_2 = 2 - I_1" ... (5)"`
From (4) & (5) we get
` 2 - I_1 = 6I_g`
` -I_1 = 6I_g - 2`
` :." "I_1 = 2 - 6 I_g" ...(6)"`
Equation (2)
`I_1 + 2 I_g - 3I_2 = 0 " ...(2)" `
Substituting the equation (6), in the equation (2) we get
` 2 - 6 I_g + 2l_g + 2 l_g - 3l_2 = 0 " ...(7)"`
From (4) `I_2 = 6I_g " ...(8)"`
` 2- 6I_g + 2I_g - 3 (6I_g) = 0`
` 2 - 6I_g + 2 I_g - 18I_g = 0`
` 2 - 22I_g = 0`
`- 22I_g = -2`
`:." "I_g = (-2)/(-22) = 1/11 A`
` :." "l_g = 1/11 A`
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