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The four arma os a Wheatstone bridge ha...

The four arma os a Wheatstone bridge have the following resistances:
`AB =100Omega, BC =10 Omega, CD=5Omega and DA=60Omega`

A galvanometer of `15Omega` resistacne is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Text Solution

Verified by Experts

Applying loop rule for closed loop BADB, we ge
`100 I_(1) - 60 I_(2) + 15 I_(g) = 0`
`therefore 20 I_(1) - 12 I_(2) + 3I_(g) = 0 " " ` .... (1)
(Where `I_(g) ` = current passing through galvanometer )
Applying loop rule for closed loop BCDE, WE get
` -10 (I_(1) - I_(8)) + 5 (I_(2) + I_(g)) + 15 I_(8) = 0 `
` therefore - 10 I_(1) + 10I_(g) + 5I_(2) + 5I_(g) + 15 I_(g) = 0`
`therefore - 10 I_(1) + 5I_(2) + 30 I_(g) = 0`
Dividing by - 5 , we get,
`2I_(1) - I_(2) - 6I_(g) = 0 " " ` ... (2)
Applying KVL for closed loop ADCEA, we get
`- 60 I_(2) - 5 (I_(2) + I_(g)) = - 10`
`therefore - 65 I_(2) - 5I_(g) = - 10`
Dividing by - 5 , we get
`13 I_(2) + I_(g) = 2 " "` ...(3)
Multiplying equation (2) by 10 , we get
`20 I_(1) - 10 I_(2) - 60 I_(g) = 0 " " `.. (4)
Subtracting equation (4) from equation (1),
` - 2I_(2) + 63 I_(g) = 0`
`therefore I_(2) = (63)/(2) I_(g) " " ` .... (5)
From equation (3) and (5),
`13((63)/(2) I_(g)) + I_(g) = 2 `
`therefore 819I_(g) + 2I_(g) = 4 `
`therefore 821I_(g) = 4 `
`therefore I_(g) = (4)/(821) = 0.00487 A = 4.87 xx 10^(-3) `A
= 4.87 mA
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