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Question
Obtain the balancing condition for the Wheatstone bridge arrangements as shown in Figure 4 below:
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Solution
Let `I_3` and `I_4` be the currents in resistors Q and S respectively . Let `I_g` be the current through galvanometer. For balanced condition,
`I_g = 0`
Applying junction law at ‘b’ we get
`I_1 = I_3 + I_g`
`because I_g = 0 , I_1 = I_3` ....(i)
Applying junction law at ‘d’, we get
`I_2 + I_g = I_4`
`because I_g = 0 , I_2 = I_4` ....(ii)
Applying loop law in the loop abda, we get
`-I_1·P - I_g·Q + -I_2·R = 0`
⇒ `-I_1P + I_2R = 0` (`because I_g = 0`)
⇒ `I_1P = I_2R`
⇒ `P/R = I_2/I_1` ....(iii)
Applying loop law in the loop bcdb, we get
`-I_3·Q + I_4·S + I_g·6 = 0`
⇒ `-I_3·Q + I_4·S + 0 = 0 (because I_g =0)`
⇒ `-I_3Q = I_4S`
⇒ `Q/S = I_4/I_3`
⇒ `Q/S = I_2/I_1` ...(iv) [using eq.(i) and (ii)]
From eq. (iii) and (iv), `P/ R = Q/s`
⇒ `P/Q = R/S`
This is the balanced condition.
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