Lcr Circuit

Circuit 1

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20X_c%20%3D%20%5Cfrac%7B1%7D%7B%5Comega%20C%7D%20%3D%20100%20%5COmega

%5Ctherefore%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20Z_1%20%3D%20%5Csqrt%7B(100)%5E2%20%2B%20(100)%5E2%7D

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%3D%20100%20%5Csqrt%7B2%7D%20%5C%2C%20%5COmega

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cphi_1%20%3D%20cos%5E%7B-1%7D%20%5Cbigg(%20%5Cfrac%7BR_1%7D%7BZ_1%7D%5Cbigg)%20%3D%2045%5E%7B%5Ccirc%7D

In this circuit current leads the voltage.

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20I_1%20%3D%20%5Cfrac%7BV%7D%7BZ_1%7D%20%3D%20%5Cfrac%7B20%7D%7B100%20%5Csqrt%7B2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B5%20%5Csqrt%7B2%7D%7D%20A

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20V_%7B100%20%5COmega%7D%20%3D%20(100)%20I_1%20%3D%20(100)%20%5Cfrac%7B1%7D%7B5%20%5Csqrt%7B2%7D%7D%20V

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%3D%2010%20%5Csqrt%7B2%7D%20V

Circuit 2

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20X_L%20%3D%20%5Comega%20L%20%3D%20(100)(0.5)%20%3D%2050%20%5COmega

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20Z_2%20%3D%20%5Csqrt%7B(50)%5E2%20%2B%20(50)%5E2%7D%20%3D%2050%5Csqrt%7B2%7D%20%5COmega

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5Cphi_1%20%3D%20cos%5E%7B-1%7D%20%5Cbigg(%20%5Cfrac%7BR_1%7D%7BZ_1%7D%5Cbigg)%20%3D%2045%5E%7B%5Ccirc%7D

In this circuit voltage leads the current.

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20I_2%20%3D%20%5Cfrac%7BV%7D%7BZ_2%7D%20%3D%20%5Cfrac%7B20%7D%7B50%20%5Csqrt%7B2%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B5%7D%20A

%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20V_%7B50%20%5COmega%7D%20%3D%20(50)%20I_2%20%3D%2050%20%5Cbigg(%20%5Cfrac%7B%20%5Csqrt%7B2%7D%7D%7B5%7D%5Cbigg)%20%3D%2010%5Csqrt%7B2%7D%20V

Further,

I_1

and

I_2

have a mutual phase difference of

90%5E%7B%5Ccirc%7D

I%20%3D%20%5Csqrt%7BI%5E2%20%2C%20%2B%20I%5E2_2%7D%20%3D%200.34

To solve this problem, we use the concepts related to a series LCR circuit at resonance.

Given:
Resonant angular frequency, $%5Comega_0%20%3D%2010%5E5%20%5C%2C%20%5Ctext%7Brad%2Fs%7D$
Current amplitude at resonance, $I_0$
When $%5Comega%20%3D%208%20%C3%97%2010%5E4%20%5C%2C%20%5Ctext%7Brad%2Fs%7D$ , current amplitude is $0.05I_0$
Inductance, $L%20%3D%2050%20%5C%2C%20%5Ctext%7BmH%7D%20%3D%2050%20%C3%97%2010%5E%7B-3%7D%20%5C%2C%20%5Ctext%7BH%7D$

Steps:

1. Current amplitude relation for an LCR circuit is given by:
$I%20%3D%20%5Cfrac%7BI_0%7D%7B%5Csqrt%7B1%20%2B%20%5Cleft(Q%5E2%5Cleft(%5Cfrac%7B%5Comega_0%7D%7B%5Comega%7D%20-%20%5Cfrac%7B%5Comega%7D%7B%5Comega_0%7D%5Cright)%5E2%5Cright)%7D%7D$
Given $I%20%3D%200.05I_0$ , $%5Cfrac%7BI%7D%7BI_0%7D%20%3D%200.05$ :

$%5Cfrac%7BI_0%7D%7B%5Csqrt%7B1%20%2B%20%5Cleft(Q%5E2%5Cleft(%5Cleft(%5Cfrac%7B10%5E5%7D%7B8%C3%9710%5E4%7D%5Cright)%20-%20%5Cleft(%5Cfrac%7B8%C3%9710%5E4%7D%7B10%5E5%7D%5Cright)%5Cright)%5E2%5Cright)%7D%7D%20%3D%200.05I_0$
Simplify to find $Q$ :
$0.05%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1%20%2B%20Q%5E2(0.5625%20-%200.64)%5E2%7D%7D$
$Q%5E2(0.0775)%5E2%20%3D%20%5Cfrac%7B1%7D%7B0.05%5E2%7D%20-1$
Solving gives: $Q%20%3D%2044.721$

2. Quality factor $Q$ is also given by:
$Q%20%3D%20%5Cfrac%7B%5Comega_0%20L%7D%7BR%7D$
Calculate bandwidth, which is $%5CDelta%20%5Comega%20%3D%20%5Cfrac%7B%5Comega_0%7D%7BQ%7D$ :
$%5CDelta%20%5Comega%20%3D%20%5Cfrac%7B10%5E5%7D%7B44.721%7D%20%5Capprox%202237%20%5C%2C%20%5Ctext%7Brad%2Fs%7D$

3. Maximum current $I_0$ is given by the condition $%5Cfrac%7BV%7D%7BR%7D$ at resonance whe\) :
$I_0%20%3D%20%5Cfrac%7B45%7D%7B112%7D%20%5Capprox%200.401%20%5C%2C%20%5Ctext%7BA%7D%20%3D%20401%20%5C%2C%20%5Ctext%7BmA%7D$

4. Peak power dissipation $P_%7B%5Ctext%7Bmax%7D%7D%20%3D%20%5Cfrac%7BV%5E2%7D%7B2R%7D$ at resonance:
$%3D%20%5Cfrac%7B45%5E2%7D%7B2%20%5Ctimes%20112%7D%20%5Capprox%209.03%20%5C%2C%20%5Ctext%7BW%7D$

List-I	List-II
(P) I₀ in mA	(1) 44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in rad s⁻¹	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250

Matching with options:
P → 3 (401 mA approximated to 400 mA), Q → 1 (Q = 44.721 approximated to 44.4), R → 4 (Bandwidth = 2237 rad/s approximated to 2250 rad/s), S → 2 (Power = 9.03 approximated to 9 W)

Correct answer: P →3, Q→ 1, R →4, S →2

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Lcr Circuit

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs