Saved Bookmarks
| 1. |
Consider a wheatstone bridge PQRS as shown in Fig. 7.19 where current I is in the circuit of four resistances 10 ,20 ,30, and 40 Omega. Find the ratio of the heat generated in the four arms PQ,QR, PS, and SR. |
|
Answer» Solution :Given ` R_(1) = 10 omega , R_(2) = 20 Omega , R_(3) = 20 Omega , and R_(4) = 40 Omega`. Now, `(R_(1))/( R_(2)) = ( 10) / ( 20) = (1) /(2)` and `(R_(3))/(R_(4)) = ( 20) /( 40) = (1)/( 2)` ltbgt `:. (R_(1))/(R_(2)) = (R_(3))/(R_(4))` Hence , wheatstone bridge is balanced . Now as the bridge is balanced , no current will flow through arm `QS`. Let `I_(1) and I_(2)` be the currents flowing in arms `PQ and PS`. respectively. Then POTENTIAL differences across `P and Q` is equal to potential difference across `P and S`. That is , `I_(1) XX 10 = I_(2) xx 20 or I_(1) = 2I_(2)` Therefore, heat produced in arm `PQ` is `H_(1) xx I_(1)^(2) xx 10 = 40 I_(2)^(2) J` Also heat produced in arm `QR` is `H_(2)= I_(1)^(2) xx 20 = 80 I_(2)^(2) J` Similarly , heat produced in arm `PS` is `H_(3) = I_(2)^(2) xx 20 = 20 I_(2)^(2) J` And heat produced in arm `SR` is `H_(4) = I_(2)^(2) xx 40 = 40 I_(2)^(2) J` `:. H_(1) : H_(2) : H_(3) : H_(4) = 40 I_(2)^(2): 80 I_(2)^(2) : 20 I_(2)^(2) : 40 I_(2)^(2) = 2 : 4 : 1 : 2`, which is the required ratio. 
|
|