In the AC circuit shown in the original figure, R = 100 kΩ and C = 100 pF. The phase difference between $V_in$ and ( $V_B-V_A)$ is 90°. The input angular frequency is 10ˣ rad/s. Find x.
Your answer

Question

In the AC circuit shown in the original figure, R = 100 kΩ and C = 100 pF. The phase difference between $V_in$ and ( $V_B-V_A)$ is 90°. The input angular frequency is 10ˣ rad/s. Find x.

Your answer

G1 Physics · Accepted Answer

From the phasor geometry in the source solution, the required 90° phase condition gives θ + θ = 90° ⇒ θ = 45°. For a series RC branch, tanθ = X C / R . X_C/R. X C ​ / R . At θ=45°, X C X_C X C ​ = R. Since X C X_C X C ​ = 1/(ωC), we have 1/(ωC) = R ⇒ ω = 1/(RC). Given R = 100 kΩ = 10⁵ Ω, C = 100 pF = 10⁻¹⁰ F. Thus, ω = 1/(10⁵×10⁻¹⁰) = 10⁵ rad/s. Therefore 10ˣ = 10⁵ and x = 5.

In the AC circuit shown in the original figure, R = 100 kΩ and C = 100 pF. The phase difference between V i n V_in V i ​ n and ( V B − V A ) V_B-V_A) V B ​ − V A ​ ) is 90°. The input angular frequency is 10ˣ rad/s. Find x.

In the AC circuit shown in the original figure, R = 100 kΩ and C = 100 pF. The phase difference between V i n V_in V i n and ( V B − V A ) V_B-V_A) V B − V A ) is 90°. The input angular frequency is 10ˣ rad/s. Find x.