we are done with the basics of phasers and now we are going to solve few questions based on phasers and in the first question we are required to find current ID in both the circuits given and we will begin with our circuit number one in which there are two resistors connected in series and both the resistors have 1 ohm resistance value and you can see that the voltage given is in phasor representation that is the parent voltage is V T this is in sinusoidal representation time dependent voltage and in question the phaser of this voltage is given as 10 angle 30 degrees and there is one very important point which you must understand the phasers are always in peak and that is and this value is always the peak value of the voltage but in books you will find this has RMS value and therefore we will not take it as the peak value but we will take it as the RMS value so the phasor here we are having is corresponding to the RMS value of voltage VT and now we will focus on the calculation of current ID we know current I T will be equal to voltage VT divided by 1 ohm plus 1 ohm and from here we can say that current hi t is equal to VT over 2 amperes but in question VT is not given directly the corresponding phasor of wit is given and therefore in place of V 2 over 2 we will have V phaser over 2 and this will give us high phaser not I T and this I phaser will be corresponding to the RMS value of current high T so from here we will have I phaser equal - ten angle thirty degrees divided by two amperes which is equal to five angle thirty degrees and we are required to find ID not a phasor and we know high P will be equal to I M cos Omega T plus theta where I M is the maximum value of I T or the amplitude of I T cause we have taken because usually everyone has taken cause as the parent signal you can also take sine but we are taking cos because in all the books the derivation is done by taking cos theta is the phase and from here we have two informations the first one is information of phase that is theta it is equal to 30 degrees and the second one is the information of RMS value of current that is five amperes hi RMS is equal to five amperes and we know I RMS is equal to the maximum value of current divided by root 2 and therefore I M will be equal to root 2 multiplied to I RMS and hence we now know what is the value of Phi M so finally we will have our current I T it will be 5 root 2 cos Omega T plus 30 degrees this is our answer now let us move on to our second circuit everything here is same except we have voltage in sinusoidal form so here the parent voltage is given it is 6 sine Omega T plus 10 degrees and we will have the phasor representation with the peak value and we are having the phasor representation because it will be easy to handle and we doing phasor representation we include the amplitude and the phase so we will have six angle ten degrees in the phasor representation of VT now moving on to the next step we will find out hi phasor it will be V phasor over two amperes so I phaser will be six angle ten degrees divided by two amperes so I phasor is equal to three angle 10 degrees amperes and it is very clear that this 3 here is the maximum value of the current because the six here who was the maximum value of the voltage we have taken it from here that is the maximum value and therefore this one here is the maximum value of the current and this 10 degrees is the phase angle and therefore we can now have our current hi t it will be I am sine this time we are having sine because sine is given in the question itself it is the parent signal Omega T plus theta so finally we will have ID equal to 3 sine Omega T plus 10 degrees and the unit will be amperes and this is our answer and here also the unit who will be amperes so now I hope you understand how to deal with phasers and the sinusoidal given in the question [Music] [Applause] you [Music]