In the following circuit, $RC$ is much larger than the input period. Assume diode is ideal and $R$ is large. The dc output voltage across $R$ will be .............. V. (Specify answer up to one digit after the decimal point.) 
Step 1: Note that $RC \gg T$.
Thus capacitor charges to peak voltage and discharges very slowly β typical peak detector.
Step 2: Input is 24 Vrms.
Peak voltage = $24\sqrt{2} = 33.94\ \text{V}$.
Step 3: Ideal diode.
No drop across diode, so capacitor charges to full peak voltage.
Step 4: DC output across $R$.
Since discharge is negligible,
$V_{out} \approx V_{peak} = 33.94 \approx 34.0\ \text{V}$.




In the circuit shown in the figure, both OPAMPs are ideal. The output for the circuit \(V_{out}\) is: 