Question:

Please convert the below content to rendered equations without any errors.Got it ? thanks for the correction. Let’s redo the solution **cleanly** with the **correct component roles**. --- ## Given data (corrected) * **Series resistor**: ( R = ? ) (to be found) * **Shunt resistor**: ( R_o = 100,\text{k}\Omega ) * **Capacitor**: [ C = 100,\text{nF} = 100 \times 10^{-9},\text{F} ] * **Frequency**: [ f = 1,\text{kHz}, \quad \omega = 2\pi f = 6283,\text{rad/s} ] * **Output voltage** is taken at the node where ( R_o ) and ( C ) are connected to ground * **Required phase difference** between ( v_{in} ) and ( v_{out} ): [ 45^\circ ] --- ## Step 1: Capacitor impedance [ Z_C = \frac{1}{j\omega C} ] [ Z_C = \frac{1}{j(6283)(100\times10^{-9})} = -j1592,\Omega ] --- ## Step 2: Parallel combination ( R_o \parallel C ) [ Z_p = R_o \parallel Z_C ] Since: [ R_o = 100,\text{k}\Omega \gg |Z_C| = 1592,\Omega ] ? **The shunt resistor has negligible effect**, so: [ Z_p \approx Z_C = \frac{1}{j\omega C} ] --- ## Step 3: Transfer function [ H(j\omega) = \frac{v_{out}}{v_{in}} = \frac{Z_p}{R + Z_p} ] With ( Z_p \approx \frac{1}{j\omega C} ): [ H(j\omega) = \frac{1}{1 + j\omega RC} ] --- ## Step 4: Phase of the transfer function [ \angle H = -\tan^{-1}(\omega RC) ] We want: [ |\angle H| = 45^\circ ] So: [ \tan^{-1}(\omega RC) = 45^\circ ] [ \Rightarrow \omega RC = 1 ] --- ## Step 5: Solve for ( R ) [ R = \frac{1}{\omega C} ] [ R = \frac{1}{(6283)(100\times10^{-9})} ] [ R \approx 1592,\Omega ] --- ## ✅ Final Answer [ \boxed{R \approx 1.6,\text{k}\Omega} ] --- ### ? Key insight (important for exams!) Because [ R_o \gg \frac{1}{\omega C}, ] the circuit behaves like a **simple RC low-pass filter**, and **45° phase shift occurs when**: [ \boxed{\omega RC = 1} ] If you want, I can: * Show a **phasor diagram** * Verify numerically including ( R_o ) * Or give you a **one-line shortcut rule** for phase-shift RC circuits ?