A battery, switch, resistor, and inductor are connected in series. When the switch is closed, the current rises to half its steady-state value in 2.0ms.How long does it take for the magnetic energy in the inductor to rise to half its steady-state value?My work:the time constant for the circuit is: L/R 2.9ms-I used the equation i(t) E/r ( 1 - e ^ -RT/L)It asks for time it takes for Magnetic Energy of inductor to rise to half its steady-state value.and the only equation i know that works is U .5LI^2 but i cant use it since i dont know L or I.Please help.
U (0.5) LI^2 (1) Question is to find time when it becomes U/2 Now from the above equation, you can find I when U becomes U/2. Let it be I' U/2 (0.5)LI'^2 (2) Taking ratio of (1) and (2), 2 (I/I')^2 I' I/√2 Now we have to find time when I becomes I/√2 given that I becomes I/2 in 2 minutes I/2 E/r [1 - e^(2/2.9)] and I/√2 E/r [1 - e^(t/2.9)] Taking ratio, √2 [1 - e^(t/2.9)] / [1 - e^(2/2.9)] 1 - e^(t/2.9) √2 [1 - e^(2/2.9)] - 1.40 e^(t/2.9) 2.40 t (2.9) ln (2.40) 2.54 minutes.
