Question

Obtain the balancing  condition for the Wheatstone bridge arrangements as shown in Figure 4 below:

Answer 1

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.