Use the worked example above to help you solve this problem. A 12.0-V battery is

Question

Use the worked example above to help you solve this problem. A 12.0-V battery is connected in series to a 24.9-? resistor and a 5.26-H inductor. (a) Find the maximum current in the circuit. A (b) Find the energy stored in the inductor at this time. J (c) How much energy is stored in the inductor when the current is changing at a rate of 1.56 A/s? Your response differs from the correct answer by more than 10%. Double check your calculations. J EXERCISE HINTS: GETTING STARTED | I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. For the same circuit, find the energy stored in the inductor when the rate of change of current is 1.09 A/s. J

Explanation / Answer

depends on the circuit. Are they in series or parallel? Does "max current" include turn on transients? Xc = 1/(2pfC) 1.06 µF at 222 Hz has a reactance of 676 ohms. If they are in series, Impedance Z = v(R² + X²) = 677 ohms I = 5.1/677 = 0.00754 amps For such a situation as given, you have to solve the differential equation L (di / dt) + R i = V The solution of the equation gives i = i(max) { 1 - e^(- R t / L )} ...........(1) where i = current at any instant when it is rising. i(max) = steady value of the current after a sufficiently long time = V / R Energy stored in the inductor when the current reaches the maximum value i(max) E = (1/2) L (i(max))² ............ (2) i(max) = 24 V / 9 Ohm = 8 / 3 A, L = 6 H Substitute in the above equation (2) and calculate E Time constant = L / R When t = L / R , substituting in equation (1) i = i(max) { 1 - e^(- 1)} { e = 2.718...} = (8 / 3) * { 1 - (1 / e)} A = (8 / 3) { 1 - 0.37} A = 0.63 * (8 / 3) A

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