A 15-mH inductor is connected to a standard electrical outlet (rms voltage 112 V, frequency f87 Hz). Determine the energy stored in the inductor at t 7.6 ms, assuming that this energy is zero at t 0. Answer suppose to be in Ji tried v(t) 112 sin (wt)but help please!
V L(di/dt), i (1/L)∫V(t)dt ; this is the current in an inductor If V 112 sin(2π87t) ∫112sin(2π87t)dt -0.2 cos(547t); this is the integral of V(t)dt i -13.3 cos(547t). this is the current in the inductor i -13.3 cos(547*0.0076) -13.26 amps at 7.6 milliseconds Energy in an inductor (1/2)(L)(i^2) E (1/2)(0.015)(-13.26)^2 1.32 joules
The energy goes up by 4 E ?LI? Energy in an inductor double the current, and because the I term is squared, the energy goes up by 2 squared. .
The previous answer was close, but not quite right. There are a couple of scenarios to consider. 1. Does the question assume that the inductor has been energized with the ac voltage source for a 'long time' prior to t 0? If so, then calculate the instantaneous current through the inductor at t 7.6 ms using the impedance formulas. Z j 2 * pi * f * L j 8.19956 ohms (where j sqrt(-1)) |I| |V| / |Z| 112 volts / 8.19956 ohms 13.659 amps rms Since energy 0 at t 0, then i(0) 0. Thus, we can write the equation for instantaneous current as i(t) sqrt(2) * 13.659 sin(2*pi*87 *t) amps. We know it is a sin() function without angular offset since i(t0) 0 At t 7.6 ms, i(t) 0.9895 amps And, E 0.5 * 0.015 * (0.9895)^2 0.007343 J. 2. The other scenario is similar to what the previous answerer gave, with the exception that you have to account for sqrt(2) in the conversion from rms to instantaneous values. Here, the assumption is that the voltage waveform is v(t) sqrt(2) * 112 * sin(2*pi*87 * t), and that a switch is closed at t0. The resulting current includes a large decaying quasi-dc component added to the ac component, to meet the initial condition that i(0-) i(0+) 0. I'm a bit too lazy to calculate that one right now, but I suspect that if you have no resistance in the circuit, it will be a non-decaying offset current, so pay careful attention to the integration step and be sure to add in the correct constant.
