RC network shown in the given figure can provide maximum theoretical phase shift of

Explanation:

The given circuit consists of three RC networks connected in cascade.

In a passive RC  circuit, when the input is connected to a capacitor and the output voltage is taken across a resistor, then the circuit is known as a high pass RC circuit.

Frequency Domain Analysis:

The output voltage is calculated using the voltage division rule as:

\({{V}_{o}}\left( jω \right)=V_i(jω)× \frac{R}{\frac{{1}}{jω C}+R}\)

\({{V}_{o}}\left( jω \right)=V_i(jω )× \frac{Rjω}{1+Rjω}\)

\(\frac{V_0}{V_1}=\frac{Rjω}{1 \ + \ Rjω}\)

Putting s = jω

\(\frac{V_0}{V_1}=\frac{Rs}{1 \ + \ Rs}\)

Taking phase shift of above equation:

∠ϕ = 90° - tan^-1(Rs)  ---(1)

A single high pass RC circuit provides a phase shift of less than 90° at a non-zero frequency.

For maximum theoretical phase shift put R = 0 in the equation (1):

∠ϕ = 90°

Hence for three RC high pass networks connected in cascade, the total phase shift would be:

∠ϕ = 90° × 3 = 270° 

Hence option (3) is the correct answer.

Important Points

For ω = 0, i.e. at low frequencies, the output is:

\({{V}_{o}}\left(0 \right)=0\)

For ω = ∞, i.e. at high frequencies, the output is:

\({{V}_{0}}\left(\infty \right)=V_i(\infty)\)

We conclude that for low frequencies the output is 0 and for high frequencies, we have an output equal to the input. This is the function of a high-pass filter that passes only high frequencies and does not allow low frequencies to appear at the output.​