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If you multiply the top and bottom of the equation by -j then you get the result shown. It is simply to get rid of the j on the bottom so it looks tidy
Problem 2
That is the equation for two parallel impedances, one of which is -j4000 the other is 4k. If you are not familiar with it, it comes from re-arranging 1/Zt = 1/Z1 +1/Z2
Problem-1 and Problem-2 expelanation is given by Keith1200s
But related to problem-3 equation of the voltage source is vs =750 cose(5000t) ....this equtation can be written in more general way of Vx= Vmax Cos( wt + phi) ....where phi is an angle now in our case values are vmax = 750 w=5000 and phi =0; so in the polar form it can be represented as 750 at an angle of 0 .....
The schematic shows a coil and capacitor in series.
The resistor may be a load (connected to a second-order low pass filter).
Or it may be a mathematical construct which has to do with finding out what frequency will cause 4000 ohms capacitor reactance.
Or will cause capacitor reactance to be the same as the coil reactance.
Or what load is needed to produce resonance for given values of coil and capacitor and a given frequency.
I think the above applies to your Note 3, which points to an equation as taking place at a zero degree phase. Zero phase change will happen only at the resonant frequency. When coil lag equals capacitor leading.
Hi moonnightingale!
First, (j)^2 = -1 => 1/j = -j, also the unit of impedance is Ohm.
Second one: (-j4000) is in parallel with 4kOhm, it likes resistor (-j4000 ohm) is in parallel to another 4kOhm <Because the concept impedance and resistor is quite same in this case>.
The third one: he can write 0 degrees it is clear that Vs = 0.750 cos(5000t + 0 degree) where 0 degree is the phase, and the point you highlighted in the last formula presents the phase of Vs (It is only the concept).
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