Saved Bookmarks
| 1. |
Find the equivalent resistance between the points A and B of the circuit shown in the fig. |
|
Answer» Solution :Supposing a SOURCE is connected between the terminals A and B . The current DISTRIBUTION is shown in fig. At junction A , `i= i_1 + i_2` Resistance between A and B , `R_(AB)= V/i = V/(i_1 +i_2)` In close loop ACDA, -`5i_1 - 5i_3 + 10i_2 =0` or `-i_1- i_3 + 2i_2 =0"".....(i)` In close loop C B D E `-10(i_1 - i_3) + 5(i_2 + i_3)+ 5i_3 =0` or `-2i_i + i_2 + 4i_3 =0""....(3)` Now in close loop A C B E F A `-5i_1 - 10(i_1- i_3) + V= 0 ` or ` -3i_1 + 2i_3 = - V/5 "".......(iii)` From equations (i) and (iii), we GET `-5i_1 + 4i_2 = - V/5""......(iv)` From equations (ii) and (iii), we get `4i_1 + i_2 = (2V)/(5)""...(v)` Solving equations (iv) and (v) , we get `i_1 = (9V)/(105)` `i_2 = (6V)/(105)` `i_3= (V)/(35)` `R_(AB) = (V)/(i_1 + i_2)=(V)/(((9V)/(105) + (6V)/(105)))+ 7 Omega` |
|