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Please convert the below content to rendered equations without any errors.Let’s solve it **cleanly, step by step**, and keep everything symbolic until the end. I’ll also use the given simplification ( \omega = \dfrac{1}{RC} ). --- ## ? Circuit understanding All three elements are **in series**: [ \text{Vin} \rightarrow R_1 \rightarrow C \rightarrow R_2 \rightarrow \text{GND} ] Given: * (R_1 = R_2 = R) * (C_1 = C) * Input: (V_{in}) (sinusoidal, angular frequency ( \omega = \frac{1}{RC} )) --- # **A. Total impedance of the circuit** Impedances: * Resistor: ( Z_R = R ) * Capacitor: ( Z_C = \dfrac{1}{j\omega C} ) Total impedance: [ Z_{total} = R + \frac{1}{j\omega C} + R ] [ Z_{total} = 2R + \frac{1}{j\omega C} ] Now substitute ( \omega = \dfrac{1}{RC} ): [ \frac{1}{j\omega C} = \frac{1}{j(1/RC)C} = \frac{R}{j} = -jR ] [ \boxed{Z_{total} = R(2 - j)} ] --- # **B. Phasor diagram of total impedance** * Real part: (2R) (along +real axis) * Imaginary part: (-R) (downward, capacitive) Magnitude: [ |Z| = R\sqrt{2^2 + 1^2} = R\sqrt{5} ] Phase: [ \angle Z = \tan^{-1}\left(\frac{-1}{2}\right) \approx -26.6^\circ ] --- # **C. Expression for (V_1) (node across (R_2))** Voltage divider: [ V_1 = V_{in}\frac{R}{Z_{total}} ] [ V_1 = V_{in}\frac{R}{R(2 - j)} = \frac{V_{in}}{2 - j} ] Multiply numerator & denominator by conjugate: [ V_1 = \frac{V_{in}(2 + j)}{(2 - j)(2 + j)} = \frac{V_{in}(2 + j)}{5} ] [ \boxed{V_1 = \frac{2 + j}{5}V_{in}} ] --- # **D. Amplitude and phase of (V_1) (relative to (V_{in}))** Magnitude: [ |V_1| = \frac{\sqrt{2^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{5}}{5}|V_{in}| ] Phase: [ \angle V_1 = \tan^{-1}\left(\frac{1}{2}\right) \approx +26.6^\circ ] [ \boxed{ \begin{aligned} |V_1| &= \frac{\sqrt{5}}{5}|V_{in}| \ \angle V_1 &= +26.6^\circ \end{aligned}} ] --- # **E. Expression for (V_2) (node between (R_1) and (C))** Voltage across (R_2 + C): [ Z_{R_2 + C} = R + \frac{1}{j\omega C} = R - jR = R(1 - j) ] [ V_2 = V_{in}\frac{Z_{R_2 + C}}{Z_{total}} ] [ V_2 = V_{in}\frac{1 - j}{2 - j} ] Multiply by conjugate: [ V_2 = V_{in}\frac{(1 - j)(2 + j)}{5} ] [ (1 - j)(2 + j) = 3 - j ] [ \boxed{V_2 = \frac{3 - j}{5}V_{in}} ] --- # **F. Amplitude and phase of (V_2)** Magnitude: [ |V_2| = \frac{\sqrt{3^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{10}}{5}|V_{in}| ] Phase: [ \angle V_2 = \tan^{-1}\left(\frac{-1}{3}\right) \approx -18.4^\circ ] [ \boxed{ \begin{aligned} |V_2| &= \frac{\sqrt{10}}{5}|V_{in}| \ \angle V_2 &= -18.4^\circ \end{aligned}} ] --- # **G. Effect of increasing frequency** As ( \omega \uparrow ): * Capacitor impedance ( |Z_C| = \dfrac{1}{\omega C} \downarrow ) * Circuit becomes **more resistive** ### Effect on voltages: * **(V_1)**: * Amplitude **decreases** * Phase lead **reduces toward 0°** * **(V_2)**: * Amplitude **increases** * Phase lag **reduces toward 0°** ? At very high frequency: * Capacitor ≈ short circuit * Voltage divides mainly between two resistors --- ## ✅ Final Summary (Exam-ready) [ \boxed{ \begin{aligned} Z_{total} &= R(2 - j) \ V_1 &= \frac{2 + j}{5}V_{in}, \quad |V_1|=\frac{\sqrt{5}}{5}V_{in},\ \angle+26.6^\circ \ V_2 &= \frac{3 - j}{5}V_{in}, \quad |V_2|=\frac{\sqrt{10}}{5}V_{in},\ \angle-18.4^\circ \end{aligned}} ] If you want, I can: * Draw **phasor diagrams** * Convert this to **time-domain** * Show **low-frequency & high-frequency limits** Just tell me ?

Question:

Please convert the below content to rendered equations without any errors.Let’s solve it **cleanly, step by step**, and keep everything symbolic until the end. I’ll also use the given simplification ( \omega = \dfrac{1}{RC} ). --- ## ? Circuit understanding All three elements are **in series**: [ \text{Vin} \rightarrow R_1 \rightarrow C \rightarrow R_2 \rightarrow \text{GND} ] Given: * (R_1 = R_2 = R) * (C_1 = C) * Input: (V_{in}) (sinusoidal, angular frequency ( \omega = \frac{1}{RC} )) --- # **A. Total impedance of the circuit** Impedances: * Resistor: ( Z_R = R ) * Capacitor: ( Z_C = \dfrac{1}{j\omega C} ) Total impedance: [ Z_{total} = R + \frac{1}{j\omega C} + R ] [ Z_{total} = 2R + \frac{1}{j\omega C} ] Now substitute ( \omega = \dfrac{1}{RC} ): [ \frac{1}{j\omega C} = \frac{1}{j(1/RC)C} = \frac{R}{j} = -jR ] [ \boxed{Z_{total} = R(2 - j)} ] --- # **B. Phasor diagram of total impedance** * Real part: (2R) (along +real axis) * Imaginary part: (-R) (downward, capacitive) Magnitude: [ |Z| = R\sqrt{2^2 + 1^2} = R\sqrt{5} ] Phase: [ \angle Z = \tan^{-1}\left(\frac{-1}{2}\right) \approx -26.6^\circ ] --- # **C. Expression for (V_1) (node across (R_2))** Voltage divider: [ V_1 = V_{in}\frac{R}{Z_{total}} ] [ V_1 = V_{in}\frac{R}{R(2 - j)} = \frac{V_{in}}{2 - j} ] Multiply numerator & denominator by conjugate: [ V_1 = \frac{V_{in}(2 + j)}{(2 - j)(2 + j)} = \frac{V_{in}(2 + j)}{5} ] [ \boxed{V_1 = \frac{2 + j}{5}V_{in}} ] --- # **D. Amplitude and phase of (V_1) (relative to (V_{in}))** Magnitude: [ |V_1| = \frac{\sqrt{2^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{5}}{5}|V_{in}| ] Phase: [ \angle V_1 = \tan^{-1}\left(\frac{1}{2}\right) \approx +26.6^\circ ] [ \boxed{ \begin{aligned} |V_1| &= \frac{\sqrt{5}}{5}|V_{in}| \ \angle V_1 &= +26.6^\circ \end{aligned}} ] --- # **E. Expression for (V_2) (node between (R_1) and (C))** Voltage across (R_2 + C): [ Z_{R_2 + C} = R + \frac{1}{j\omega C} = R - jR = R(1 - j) ] [ V_2 = V_{in}\frac{Z_{R_2 + C}}{Z_{total}} ] [ V_2 = V_{in}\frac{1 - j}{2 - j} ] Multiply by conjugate: [ V_2 = V_{in}\frac{(1 - j)(2 + j)}{5} ] [ (1 - j)(2 + j) = 3 - j ] [ \boxed{V_2 = \frac{3 - j}{5}V_{in}} ] --- # **F. Amplitude and phase of (V_2)** Magnitude: [ |V_2| = \frac{\sqrt{3^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{10}}{5}|V_{in}| ] Phase: [ \angle V_2 = \tan^{-1}\left(\frac{-1}{3}\right) \approx -18.4^\circ ] [ \boxed{ \begin{aligned} |V_2| &= \frac{\sqrt{10}}{5}|V_{in}| \ \angle V_2 &= -18.4^\circ \end{aligned}} ] --- # **G. Effect of increasing frequency** As ( \omega \uparrow ): * Capacitor impedance ( |Z_C| = \dfrac{1}{\omega C} \downarrow ) * Circuit becomes **more resistive** ### Effect on voltages: * **(V_1)**: * Amplitude **decreases** * Phase lead **reduces toward 0°** * **(V_2)**: * Amplitude **increases** * Phase lag **reduces toward 0°** ? At very high frequency: * Capacitor ≈ short circuit * Voltage divides mainly between two resistors --- ## ✅ Final Summary (Exam-ready) [ \boxed{ \begin{aligned} Z_{total} &= R(2 - j) \ V_1 &= \frac{2 + j}{5}V_{in}, \quad |V_1|=\frac{\sqrt{5}}{5}V_{in},\ \angle+26.6^\circ \ V_2 &= \frac{3 - j}{5}V_{in}, \quad |V_2|=\frac{\sqrt{10}}{5}V_{in},\ \angle-18.4^\circ \end{aligned}} ] If you want, I can: * Draw **phasor diagrams** * Convert this to **time-domain** * Show **low-frequency & high-frequency limits** Just tell me ?

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Asked by: jack
Created at: 2026-02-27 09:00:42

Final Answer:

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Below is your content with all equations rendered in LaTeX, step by step, cleanly andically. --- # ? Circuit Understanding All three elements are **in series**: \[ \text{Vin} \rightarrow R_1 \rightarrow C \rightarrow R_2 \rightarrow \text{GND} \] Given: - \( R_1 = R_2 = R \) - \( C_1 = C \) - Input: \( V_{in} \) (sinusoidal), angular frequency \( \omega = \frac{1}{RC} \) --- # A. Total Impedance of the Circuit Impedances: - Resistor: \( Z_R = R \) - Capacitor: \( Z_C = \dfrac{1}{j\omega C} \) Total impedance: \[ Z_{total} = R + \frac{1}{j\omega C} + R \] \[ Z_{total} = 2R + \frac{1}{j\omega C} \] Substitute \( \omega = \dfrac{1}{RC} \): \[ \frac{1}{j\omega C} = \frac{1}{j \left(\frac{1}{RC}\right)C} = \frac{R}{j} = -jR \] So, \[ \boxed{Z_{total} = R(2 - j)} \] --- # B. Phasor Diagram of Total Impedance - Real part: \( 2R \) (along +real axis) - Imaginary part: \( -R \) (downward, capacitive) Magnitude: \[ |Z| = R\sqrt{2^2 + 1^2} = R\sqrt{5} \] Phase: \[ \angle Z = \tan^{-1}\left(\frac{-1}{2}\right) \approx -26.6^\circ \] --- # C. Expression for \( V_1 \) (Node Across \( R_2 \)) Voltage divider: \[ V_1 = V_{in}\frac{R}{Z_{total}} \] \[ V_1 = V_{in}\frac{R}{R(2 - j)} = \frac{V_{in}}{2 - j} \] Multiply numerator and denominator by complex conjugate: \[ V_1 = \frac{V_{in}(2 + j)}{(2 - j)(2 + j)} = \frac{V_{in}(2 + j)}{5} \] \[ \boxed{V_1 = \frac{2 + j}{5}V_{in}} \] --- # D. Amplitude and Phase of \( V_1 \) (Relative to \( V_{in} \)) Magnitude: \[ |V_1| = \frac{\sqrt{2^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{5}}{5}|V_{in}| \] Phase: \[ \angle V_1 = \tan^{-1}\left(\frac{1}{2}\right) \approx +26.6^\circ \] \[ \boxed{ \begin{aligned} |V_1| &= \frac{\sqrt{5}}{5}|V_{in}| \\ \angle V_1 &= +26.6^\circ \end{aligned} } \] --- # E. Expression for \( V_2 \) (Node Between \( R_1 \) and \( C \)) Voltage across \( R_2 + C \): \[ Z_{R_2 + C} = R + \frac{1}{j\omega C} = R - jR = R(1 - j) \] \[ V_2 = V_{in}\frac{Z_{R_2 + C}}{Z_{total}} \] \[ V_2 = V_{in}\frac{1 - j}{2 - j} \] Multiply by conjugate: \[ V_2 = V_{in}\frac{(1 - j)(2 + j)}{5} \] \[ (1 - j)(2 + j) = 1 \cdot 2 + 1 \cdot j - j \cdot 2 - j \cdot j = 2 + j - 2j + 1 = 3 - j \] \[ \boxed{V_2 = \frac{3 - j}{5}V_{in}} \] --- # F. Amplitude and Phase of \( V_2 \) Magnitude: \[ |V_2| = \frac{\sqrt{3^2 + 1^2}}{5}|V_{in}| = \frac{\sqrt{10}}{5}|V_{in}| \] Phase: \[ \angle V_2 = \tan^{-1}\left(\frac{-1}{3}\right) \approx -18.4^\circ \] \[ \boxed{ \begin{aligned} |V_2| &= \frac{\sqrt{10}}{5}|V_{in}| \\ \angle V_2 &= -18.4^\circ \end{aligned} } \] --- # G. Effect of Increasing Frequency As \( \omega \uparrow \): - Capacitor impedance \( |Z_C| = \dfrac{1}{\omega C} \downarrow \) - Circuit becomes **more resistive** ### Effect on Voltages - **\( V_1 \)**: - Amplitude **decreases** - Phase lead **reduces toward °** - **\( V_2 \)**: - Amplitude **increases** - Phase lag **reduces toward °** At very high frequency: - Capacitor acts almost like a short circuit - Voltage divides mainly between the two resistors --- # ✅ Final Summary (Exam-Ready) \[ \boxed{ \begin{aligned} Z_{total} &= R(2 - j) \\ V_1 &= \frac{2 + j}{5}V_{in}, \quad |V_1| = \frac{\sqrt{5}}{5}V_{in},\ \angle +26.6^\circ \\ V_2 &= \frac{3 - j}{5}V_{in}, \quad |V_2| = \frac{\sqrt{10}}{5}V_{in},\ \angle -18.4^\circ \end{aligned} } \] --- **If you need:** - Phasor diagrams - Time-domain conversion - Low-frequency & high-frequency limits Just ask.

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