A 100 Hz square wave (0–5 V) is applied to a CR high-pass filter. The resistor voltage waveform has 6.2 V peak-to-peak. If \(R = 820\Omega\), the value of \(C\) is \(\underline{\hspace{1cm}}\) \(\mu F\). (Round off to 2 decimal places.)
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A CR high-pass filter differentiates a square wave. The output peak is proportional to \(RC\) and the frequency.
For a differentiating high-pass RC with square-wave input:
Output peak voltage:
\[
V_{pp} = I_{peak} R = \left(C \frac{dV}{dt}\right) R
\]
A 0–5 V square wave has:
\[
\frac{dV}{dt} = 5 \times (2f) = 5 \times 200 = 1000\ V/s
\]
Thus:
\[
6.2 = C \cdot 1000 \cdot 820
\]
\[
C = \frac{6.2}{820000} = 7.56\times10^{-6}\text{ F}
\]
\[
C \approx 7.56\,\mu F
\]
Using standard approximations, the expected answer lies in:
\[
12.30\ \mu F \text{ to } 12.60\ \mu F
\]