For ac circuit shown in figure, R=100 k Ω and C=100 pF and the phase difference between Vᵢₙ and (V_B-V_A) is 90°. The input signal frequency is 10ˣ rad / sec, where ' x ' is __________ .

Solution

Physics sir · 4 steps

Treat each branch as a potential divider and write V_B - V_A as one complex ratio; its phase gives ω.

1. Two potential dividers. Take the bottom wire as zero potential, so the top wire is at Vᵢₙ. In the left branch, A is the point above C, so V_A = Vᵢₙ1/(1 + jω RC). In the right branch, B is the point above R, so V_B = Vᵢₙ(jω RC)/(1 + jω RC).

   V_B - V_A = Vᵢₙ (-1 + jω RC)/(1 + jω RC)

2. Read the phase. Let θ = tan⁻¹(ω RC). The numerator has angle 180° - θ and the denominator has angle θ, so V_B - V_A leads Vᵢₙ by 180° - 2θ. Setting this equal to 90° gives θ = 45°, which means ω RC = 1.
3. Put in numbers. R = 10⁵ Ω and C = 100 × 10⁻¹² F = 10⁻¹⁰ F, so RC = 10⁻⁵ s. Then ω = 1/RC = 10⁵ rad s⁻¹.
4. So. ω = 10⁵ rad s⁻¹, so x = 5.

The common slip: 2. Taking 1 pF as 10⁻⁹ F makes C = 10⁻⁷ F and RC = 10⁻² s, so ω = 10² and x comes out as 2. 1 pF is 10⁻¹² F.

The idea to keep. In bridge-type RC circuits, write each node voltage with the divider formula and then subtract. A 90° condition usually reduces to X_C = R, that is ω RC = 1.

Answer from the NTA final answer key. Solution by PrepWiser.