How to find the average using integration?
My attempt
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4\$\begingroup\$ Please edit the question to clearly show your attempt at solving this question. Is this homework? \$\endgroup\$– AJNCommented Oct 1, 2023 at 15:48
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3\$\begingroup\$ Yes, I can. I'm not going to do your homework for you, though. Show an attempt first, then we might help. \$\endgroup\$– HearthCommented Oct 1, 2023 at 16:21
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\$\begingroup\$ @AJN Now can you verify ? \$\endgroup\$– Star LordCommented Oct 1, 2023 at 17:05
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\$\begingroup\$ @Hearth Now can you verify ? \$\endgroup\$– Star LordCommented Oct 1, 2023 at 17:06
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\$\begingroup\$ I'm voting to reopen because the attempt is worth guidance as to how it can be dealt with much easier. \$\endgroup\$– Kuba hasn't forgotten MonicaCommented Oct 1, 2023 at 17:10
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1 Answer
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It's a bridge rectifier which feeds load R5. The diodes have substantial series resistors which reduce the current by being in series with the load, too.
The diodes block voltages 1.2 volts and less, so you can find the no current time or actually angle zone where Vi is closer to zero than 1.2 volts. During that period Vab=0.
The rest of the time (or actually the rest of the angle period 2Pi) the diodes subtract 1.2 volts. The diode series resistors + the load resistance and Vi reduced by 1.2 volts define the current i and the load gets its portion of the total voltage =(i * R5).
The imbalance of the diode series resistances for positive and negative half cycles of Vi cause 2 different pulses to R5. To calculate the average integrate in pieces:
- the zero period around Vi=0 (that part is zero)
- the nonzero i area of the positive half cycle
- the nonzero i area of the negative half cycle
Divide the sum of the partial integration results with the full cycle length to get the average.
