For the circuit shown find the voltage across each circuit...

For the circuit shown find the voltage across each circuit element and the total current in the circuit.\ Express your answers as phasors.\ $v(t)$\ 6$\Omega$ 120$mH$\ $2.5 \times 10^{-3} F$\ $v(t) = 10 \cos 100t$

Transcript

00:01 Everyone this is the problem based on ac circuit here it is given ac v is called to v not sign of omega t in the given circuit we have to find the current through register capacitor and inductor let us hear the solution reactance of capister is 1 upon omega c reactance of inductor is omega l and they will be in series the total reactance of the circuit will be one of they will be in panel so total reaction reactance of the circuit is burn upon xe minus one upon xl because they are out of phase by 180 degree so you can write burn upon take the lcm and simplify so you can write xc into xl that is xl minus xc so reactance of the circuit you will get xc into xc xl over xl minus xc so in the impedance of the circuit you will get root of r square plus x square substitute the value r square plus xc into xl over xl minus xc whole square so on simplifying impedance of the circuit can be written as r square into xl minus x c whole square plus x c square xl square divided by xl into xe so this is the impedance of the circuit now value of i r see the circuit if this is the current i r it will be ic and this will be i l and we can write i r to be to i c plus i n now i r can be written as v v upon z that is v0 sine of omega t upon j and there will be a phase difference of five so plus five here and value of five will be ten inverse of x upon r that is ten inverse of xl x c divided by xl minus x c upon r so this will be the current through register ir with phase difference and potential current across the capister will be v upon a reactance of capacitor voltage will be b not sign of omega t minus i r into r substitute the value.

03:59 So current through capacitor you will get v0 by xc sine of omega t minus ir.

04:24 Ir we know sine omega t plus 5 upon z into r and potential across inductor will be ir minus ic according to the church of the junction rule so you will get b0 into r sign of omega t plus 5 upon z minus b not xc sign of omega t minus sign of omega t plus 5 upon z into r...

Question

For the circuit shown find the voltage across each circuit element and the total current in the circuit.\ Express your answers as phasors.\ $v(t)$\ 6$\Omega$ 120$mH$\ $2.5 \times 10^{-3} F$\ $v(t) = 10 \cos 100t$

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