Question

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.) 

Hint:

A CR high-pass filter differentiates a square wave. The output peak is proportional to \(RC\) and the frequency.

Answer 1

Correct Answer: 12.3

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: \[ 12.30\ \mu F \text{ to } 12.60\ \mu F \]