For ac circuit shown in figure, R = 100 k Ω and C = 100 pF and the phase difference between Vin

Question

For ac circuit shown in figure, \(R = 100 k \ \Omega\) and C = 100 pF and the phase difference between \(V_{in}\) and \((V_B - V_A)\) is \(90 ^\circ\). The input signal frequency is \(10 ^x \ rad/sec\), where 'x' is

Answer 1

Answer is : (5)

Since both branch are identical. So phase difference between \(V_A\) and \(V_{in}\) and \(V_B\) and \(V_{in}\) are same but in opposite direction.

So, phase difference between \(V_{in}\) and \(V_A\) must be \(45 ^\circ\) as \(V_{in}\) and \(|V_A - V_D|\) has difference of \(90^ \circ\).

So, clearly \(|R| = (x_C)\)

\(\Rightarrow 100 \times 10^3 = \frac{10^{12}}{w \times 100}\)

\(\Rightarrow w = 10^5 \ rad/s\)