 
                 
                InterviewSolution
 Saved Bookmarks
    				| 1. | Derive an expression for the total magnetic energy stored in two coils with inductances `L_(1)` and `L_(2)` and mutual inductance `M`, when the currents in the coils are `I_(1)` and `I_(2)`, respectively. | 
| Answer» When the currents are increaseing in the circuit, we have emfs `E_(1) = - L_(1) (dI_(1))/(dt) +- M (dI_(2))/(dt)`,`E_(2) = - L_(2) (dI_(2))/(dt) +- M(dI_(1))/(dt)` where `+-` sign appears conistently in both equations and depends on the geometry of the coils and the sense of current. Work done in pushing charges `dq_(1)` and `dq_(2)` through each circuit. respectively, is `dW = - E_(1) dq - E_(2) dq_(2)` `= L_(1) (dI_(1))/(dt) dq_(1) bar+ M (dI_(2))/(dt)dq_(1) + L_(2) (dI_(2))/(dt) dq_(2) bar+ M (dI_(1))/(dt) dq_(2)` `I_(1) = (dq_(1))/(dt),dq_(1) = I_(1)dt, I_(2) = (dq_(2))/(dt),dq_(2) = I_(2) dt` `dW = L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+(MI_(1) dI_(2) + MI_(2) dI_(1))` `= L_(1)I_(1)dI_(1) + L_(2)I_(2)dI_(2) bar+Md(I_(1)I_(2))` On integating the above expression from `0` to final current, we have `U = int dW = L_(1) int_(0)^(I_(1)) I_(1) dI_(1) + L_(2) int_(0)^(I_(2)) I_(2) dI_(2) bar+ M int_(0)^((I_(1)I_(2))) d(I_(1)I_(2))` `= (1)/(2) L_(1)L_(1)^(2) + (1)/(2) L_(2)I_(2)^(2) bar+MI_(1)I_(2)` | |