# 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?

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1. 0
2. $$\pi$$
3. $$\frac{\pi }{2}$$
4. $$- \frac{\pi }{2}$$

## Answer (Detailed Solution Below)

Option 4 : $$- \frac{\pi }{2}$$
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CT 1: Ratio and Proportion
3536
10 Questions 16 Marks 30 Mins

## Detailed Solution

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

$$\delta = - \frac{-\pi}{2}$$
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