# A question from Sedra/Smith Microelectronics

regarding question 1.12 from Sedra and Smith MicroElectronics book (2009 international edition): A transducer characterized by a voltage of 1V rms and a resistance of $1M\Omega$ is available to drive a $10 \Omega$ load. If connected directly, what voltage and power levels result at the load? if unity-gain (i.e $A_{v_o} =1$) buffer amplifier with $1M\Omega$ input resistance and $10\Omega$ output resistance is interposed between source and load, what do the output voltage and power levels become? For the new arrangement, find the voltage gain from source to load, and the power gain (both expressed in decibles).

I do get the first two answers, but I don't get the other answers.

So wev'e got: $V_L = 10\mu V_{rms} , \ P_L= 10^{-11} W$

I know that we have: $v_o = v_i \frac{R_L}{R_L + R_o}= v_i \frac{10}{10+10} =0.5 \cdot v_i$

Now if I am not mistaken, $v_i$ is the voltage of the transducer which is $\frac{1}{\sqrt 2}$, so unless I am mistaken here I should get $0.5\cdot \frac{1}{\sqrt 2}$, which is not $0.25V$, where did I go wrong here?

For the power I get:

$P_o = v_o ^2/R_o + v_i ^2 / R_o = (1/8+1/2)/10 = 6.25\cdot 10^{-2} W$ and again not as in the texbook which is $6.25\cdot 10^{-3} W$.

Again can someone please inform me what did I do wrong? perhaps I misused the formulas. 