Pulse-width modulation is used to provide a 60-Hz voltage...

Pulse-width modulation is used to provide a 60-Hz voltage across a series RL load with $R = 12 \Omega$ and $L = 20 \text{ mH}$. The dc supply voltage is 150 V. The amplitude of the 60-Hz voltage is to be 120 V. Use PSpice to obtain the current waveform in the load and the THD of the current waveform in the load. Use (a) bipolar PWM with $m_f = 21$, (b) bipolar PWM with $m_f = 41$, and (c) unipolar PWM with $m_f = 10$.

Transcript

00:01 Hi students in this question it is given that r is equal to 12 om l is equal to 20 million henry frequency is equal to 60 heads dc supply voltage is 150 volt and the amplitude of voltage heads is equal to 120 volt and we have to first of all we have to find out the amplitude modulation ratio so a amplitude modulation ratio and amplitude modulation ratio we have the equation for amplitude modulation ratio can be represented as amr as equal to amplitude of 60 heads voltage is given in the equation and divided by dc supply voltage dc supply voltage so we have the values for each amplitude of 60 hertz voltage and dc supply voltage it's nothing but amplitude of 60 has its voltage is 120 volt and divided by 150 is the ratio so you know unit so 0 .8 is the amplitude modulation ratio now we have to calculate the rms current at the load rms current at the load at the load so rms current can be represented as i rms is equal to amplitude modulation ratio amplitude modulation ratio into supply voltage supply voltage divided by root of square root of resistance square resistance can represent as r r square into two by frequency into inductance that is f into l whole square so irms equal to mr is 0 .8 into 150 150 divided by r square r is 12 ome total square plus 2 pi into f f is 60 60 into l l is 20 millie henry so 20 square that's equal to 9 .51 ampere so i rm s rmus value of the current at the load is 9 .51 ampere next we have to find out the average power average power absorbed by the load absorbed by the load so average power can be written as be average so it's nothing but a m r square into dc supply voltage dc supply voltage square divided by two into resistance that is r so it's equal to 0 .8 square into dc supply voltage is 150.

03:29 So 150 square divided by 2 into resistance is 12 oms.

03:33 So after the calculation fee average is equal to 800 volt...

Question

Please give Ace some feedback