A battery, switch, resistor, and inductor are connected in series. When the swit

Question

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 1.0 ms. Find how long it takes for the magnetic energy in the inductor to rise to half its steady-state value by following the questions below? What is the magnetic energy stored in the inductor in terms of the current I (t), the self- inductance L, and time t? What is the current I (t) in terms of the resistance R, the self-inductance L, time t and the current at t = infinity I ? What is the magnetic energy stored the resistance R, the self-inductance L, time t and the current at t = infinity I ? Find the time tH at which the magnetic energy stored in the inductor is half its steady- state value.

Explanation / Answer

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.

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