Question

An ideal inductor ?=114mH is connected to a source whose peak potential difference is 75V.

a) If the frequency is 100Hz, what is the current at 5ms? What is the instantaneous power delivered to the inductor at this time? A W

b) At what frequency would the peak current be 2.5A?

Answer 1

a) The relation between voltage and charge in an inductor is

$V=L\frac{dI}{dt}$

Since it's directly connected to the source we can write

$V_{max}\sin 2\pi f t=L\frac{dI}{dt}$

$I= \frac{V_{max}}{L}\int\sin 2\pi f tdt=-\frac{V_{max}}{2\pi f L }\cos 2\pi f t$

$I(t=5ms)=-\frac{75V}{2\pi\times 100Hz\times 0.114H }\cos 2\pi \times 100Hz\times 5\times 10^{-3}s$

$I(t=5ms)=1.047A$

b) The peak current is just the amplitude of the I(t) function we just derived so:

$I_{max}=\frac{V_{max}}{2\pi f L }$

$f=\frac{V_{max}}{2\pi LI_{max}}=\frac{75V}{2\pi \times 0.114H\times 2.5A}=41.9Hz$