ch31-p088 - 88. (a) The amplifier is connected across the...

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dP dx Rr xR rx R avg = + ε 2 3 bg . This is zero for xrR == = // 1000 10 100 ΩΩ b g . We note that for small x, P avg increases linearly with x , and for large x it decreases in proportion to 1/ x . Thus x = r / R is indeed a maximum, not a minimum. Recalling x = ( N p / N s ) 2 , we conclude that the maximum power is achieved for /1 0 ps NN x = = . 88. (a) The amplifier is connected across the primary windings of a transformer and the resistor R is connected across the secondary windings. (b) If I s is the rms current in the secondary coil then the average power delivered to R is PI R s avg = 2 . Using s I = () / p NNI , we obtain P IN N R pp s avg = F H G I K J 2 Next, we find the current in the primary circuit. This is effectively a circuit consisting of a generator and two resistors in series. One resistance is that of the amplifier (
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ch31-p088 - 88. (a) The amplifier is connected across the...

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