- #1
kahwawashay1
- 95
- 0
problem: switch of LR circuit is closed at time = 0; what is ratio of inductor's self-induced emf ( E(L) ) to battery's emf ( E(bat) ) at t = 2τ?
my solution:
switch is closed so current begins to build up according to equation:
i = (E(bat)/R) (1-e^(-t/τ) )
multiplying R on both sides and plugging in for t = 2τ:
E(L) = E(bat) (1-e^(-2) )
E(L) / E(bat) = 1-e^(-2) ≈ 0.865
But my book says the answer is 0.135, which is just the value of e^(-2), which you would obtain if you used the equation for current decay
i = (E(bat)/R) (e^(-t/τ))
instead of the equation for current rise. But the current in this case is clearly rising. So is my book wrong or am I wrong?
my solution:
switch is closed so current begins to build up according to equation:
i = (E(bat)/R) (1-e^(-t/τ) )
multiplying R on both sides and plugging in for t = 2τ:
E(L) = E(bat) (1-e^(-2) )
E(L) / E(bat) = 1-e^(-2) ≈ 0.865
But my book says the answer is 0.135, which is just the value of e^(-2), which you would obtain if you used the equation for current decay
i = (E(bat)/R) (e^(-t/τ))
instead of the equation for current rise. But the current in this case is clearly rising. So is my book wrong or am I wrong?