Saved Bookmarks
| 1. |
Three resistors R_1, R_2 and R_3 are to be combined to form an electrical circuit as shown in the figure. It is found that when R_1 R_2 and R_3 are put respectively in positions A, B and C, the effective resistance of the circuit is 70 OmegaWhen R_2, R_3 and R_1 are put respectively in position A, B and C the effective resistance is 35Omega and when R_3, R_1 and R_2 are respectively put in the position A, B and C, the effective resistance is 42 OmegaIf R_(1), R_(2) and R,_(3)are put in series, the effective resistance will be |
|
Answer» `210Omega` `(R_(2)R_(3)+R_(3)R_(1)+R_(1)R_(2))/(R_(2)+R_(3))35ldots(2)` `(R_(3)R_(1)+R_(1)R_(2)+R_(2)R_(3))/(R_(3)+R_(1))=42ldots(3)` `(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))=k` Then the system of three equation transforms to`R_(1)+R_(2)=(k)/(70)` `R_(2)+R_(3)=(k)/(35)R_(3)+R_(1)=(k)/(42)` Solving, this we get`R = k//210, R = k//105, R = 2k//105` implies`R_(1): R_(2): R_(3) = 1:2:4` So, let `R_(1) = x, R_(2) = 2X` and `R_(3) = 4x` PUTTING these VALUED in equation 1, we get`X = 15` `R_(1) = 15, R_(2) = 30, R_(3) = 60` |
|