# Test: Frequency Response

## 20 Questions MCQ Test Analog Circuits | Test: Frequency Response

Description
Attempt Test: Frequency Response | 20 questions in 20 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study Analog Circuits for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
QUESTION: 1

### (Q.1-Q.3) A parallel resonant circuit has a resistance of 2k ohm and half power frequencies of 86 kHz and 90 kHz. Q. The value of capacitor is

Solution:

BW = 2p(90-86)k = 1/RC or C = 19.89 nF.

QUESTION: 2

### The value of inductor is

Solution:

w = (86 + 90)k/2 = 88 = (1/LC)(0.5) or L = 0.16 mH.

QUESTION: 3

### The quality factor is

Solution:

Q = w/BW = 176pK/8pk = 22.

QUESTION: 4

(Q.4-Q.5) A parallel resonant circuit has a midband admittance of 25 X 10(-3) S, quality factor of 80 and a resonant frequency of 200 krad s.4. The value of R (in ohm) is

Solution:

At mid-band frequency Y = 1/R or R = 1000/25 or 40 ohm.

QUESTION: 5

The value of C is

Solution:

Q = wRC or C = 80/(200 x 1000 x 40) or 10 µF.

QUESTION: 6

A parallel RLC circuit has R 1 k and C 1 F. The quality factor at resonance is 200. The value of inductor is

Solution:

Use Q = R (L/C)0.5.

QUESTION: 7

A parallel circuit has R = 1k ohm , C = 50 µF and L = 10mH. The quality factor at resonance is

Solution:

Use Q = R (L/C)0.5.

QUESTION: 8

A series resonant circuit has L = 1 mH and C = 10 F. The required R (in ohm) for the BW = 15 9 . Hz is

Solution:

Use BW = R/L.

QUESTION: 9

For the RLC parallel resonant circuit when R = 8k, L = 40 mH and C = 0.25 F, the quality factorQ is

Solution:

use Q = R (C/L)0.5.

QUESTION: 10

A series resonant circuit has an inductor L = 10 mH. The resonant frequency w = 10^6 rad/s and bandwidth is BW = 103 rad/s. The value of R and C will be

Solution:

Use wxw = 1/LC and BW = R/L.

QUESTION: 11

The maximum voltage across capacitor would be Solution:

Explanation : LC combination

3 = v1 + 10(v1 - 0.105v1)/125

= v1 = 100

Isc = 1100/125

= 0.8V

RTH = 3/0.8

= 3.75 ohm

Wo = 1/(LC)1/2

= 1000

Q = w0L/RTH

= (1000*4)/3.75

= 1066.67

|vc|max = Q*vTH

= 1066.67*3

= 3200V

QUESTION: 12

Find the resonnant frequency for the circuit.​ Solution: QUESTION: 13

Determine the resonant frequency of the circuit. Solution: QUESTION: 14

The value of C and A for the given network function is​ Solution: QUESTION: 15

H(w) = Vo/Vi = ?​  Solution: QUESTION: 16

H(w) = Vo/Vi = ?​ Solution: QUESTION: 17

The value of input frequency is required to cause a gain equal to 1.5. The value is Solution: QUESTION: 18

In the circuit shown phase shift equal to 45 degrees and is required at frequency w = 20 rad/s. The value of R (in kilo-ohm) is Solution: QUESTION: 19

For the circuit shown the input frequency is adjusted until the gain is equal to 0.6. The value of the frequency is​ Solution: QUESTION: 20

For a stable closed loop system, the gain at phase crossover frequency should always be:

Solution:

Phase crossover frequency is the frequency at which the gain of the system must be 1 and for a stable system the gain is decibels must be 0 db. Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code