# Question 4.SP.25: The amplifier of Example 4.7 has plate current iP = IP + ip ......

The amplifier of Example 4.7 has plate current
$i_P = I_P + i_p = 8 + \cos ωt \text{mA}$

Determine   (a) the power delivered by the plate supply voltage $V_{PP}$,    (b) the average power delivered to the load $R_L$, and   (c) the average power dissipated by the plate of the triode. (d) If the tube has a plate rating of 2 W, is it being properly applied?

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(a)   The power supplied by the source $V_{PP}$ is found by integration over a period of the ac waveform:
$P_{P P} = \frac{1}{T} \int_0^T V_{P P} i_P d t = V_{P P} I_P = (300)\left(8 \times 10^{-3}\right) = 2.4 W$

(b$P_L = \frac{1}{T} \int_0^T i_P^2 R_L d t = R_L \left(I_P^2 + I_p^2\right) = 10 \times 10^3 \left[\left(8 \times 10^{-3}\right)^2 + \left(\frac{1 \times 10^{-3}}{\sqrt{2}}\right)^2\right] = 0.645 W$

(c)    The average power dissipated by the plate is
$P_P = P_{P P} – P_L = 2.4 – 0.645 = 1.755 W$

(d)   The tube is not properly applied.    If the signal is removed (so that $i_P = 0$), then the plate dissipation increases to $P_P = P_{P P} = 2.4 W$, which exceeds the power rating.

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