For the circuit shown in Fig. 36 , V = V 0 sinωt . Calculate the current in each element of t

For the circuit shown in Fig. 36 , V = V 0 sinωt ....

Key Concepts

Sinusoidal Steady-State Analysis

Sinusoidal steady-state analysis is the study of AC circuits when the circuit has been operating long enough that transient effects have died out, and all sources are sinusoidal and operating at the same frequency. This analysis involves representing signals as sinusoids and using phasor techniques to analyze the circuit’s steady state response, including the computation of currents, voltages, and impedance.

Trial Solution Method

The trial solution method involves assuming a particular form for the solution—typically a sinusoidal function with unknown amplitude and phase—to solve the differential equations that describe the circuit. In the context of AC circuits, this approach leverages the fact that the circuit response will be sinusoidal (at the same frequency as the input) in the steady state, allowing for the determination of unknown parameters by substituting the trial solution into the circuit equations.

Phasor Analysis

Phasor analysis is a technique used in AC circuit analysis where sinusoidal functions are represented as rotating vectors (phasors) in the complex plane. This method simplifies the handling of circuit equations by converting differential equations in the time domain into algebraic equations in the frequency domain, enabling easier computation of amplitudes and phase shifts associated with currents and voltages.

Impedance in AC Circuits

Impedance is the generalization of resistance in AC circuits that accounts for both the resistive and reactive (inductive and capacitive) characteristics of circuit elements. Represented as a complex number, impedance describes not only the magnitude of opposition to current flow but also the phase relationship between voltage and current, which is crucial in determining the behavior of AC circuits.

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...

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