Consider the circuit shown in the figure, In this circuit \(R = 1k\Omega\), and \(C = 1\mu F\). The input voltage is sinusoidal with a frequency of 50 Hz, represented as phasor with magnitude \({V_i}\) and phase angle 0 radian as shown in the figure. The output voltage is represented as a phasor with magnitude \({V_0}\) and phase angle \(\delta\) radian. What is the value of output phase angle \(\delta\) (in radian) relative to the phase angle of the input voltage?

This question was previously asked in

GATE EE 2015 Official Paper: Shift 1

Option 4 : \(- \frac{\pi }{2}\)

CT 1: Ratio and Proportion

3536

10 Questions
16 Marks
30 Mins

__ Concept__:

__Differential amplifier:__

The differential amplifier is to amplify the difference between the two signals received at the input

\({{\rm{V}}_{{\rm{out}}}} = - \frac{{{{\rm{R}}_{\rm{f}}}}}{{{{\rm{R}}_1}}}\left( {{{\rm{V}}_2} - {{\rm{V}}_1}} \right){\rm{\;assuming\;that\;}}{{\rm{R}}_1} = {{\rm{R}}_2}{\rm{\;and\;}}{{\rm{R}}_{\rm{f}}} = {{\rm{R}}_{\rm{g}}}{\rm{\;}}\)

Differential amplifiers are used in

- Inverting amplifiers
- Non-inverting amplifiers
- Summing amplifiers
- Instrumentation amplifiers
- Wheatstone Bridge Differential Amplifier

__Virtual ground__ -

The concept of the virtual ground is stated as if anyone of the i/p terminals is grounded physically the other i/p terminal will also be at ground potential even though, it is not grounded physically.

- One key feature of an Op-Amp is the differential input, and when put together in a circuit, this can form a virtual ground.
- The virtual ground concept is helpful for the analysis of Op Amps. This concept makes Op-Amp circuit analysis much easier.

__ Calculation__:

Given circuit can be re drawn as

using voltage Division Rule:

\(V_{NI} = \frac{R}{R + \frac{1}{Cs}} \times V_2\)

Since, V2 = 0

⇒ VNI = 0 V

From virtual Ground concept:

VI = VNI = 0

Applying KCL at inverting terminal

\(\frac{V_i ∠ 0 - V_I}{\frac{1}{Cs}}= \frac{V_I-V_0 ∠ δ}{R}\)

\(\frac{V_i ∠ 0^\circ}{\frac{1}{Cs}} = \frac{0- V_0 ∠ δ}{R}\)

V0∠δ = -RCs Vi ∠0

V0∠δ = -jRCω Vi < 0

V0∠δ = RCω Vi∠-90°

On comparing

\(\delta = - \frac{-\pi}{2}\)