
An AC circuit has\[R{\text{ }} = {\text{ }}100\Omega \], \[C{\text{ }} = {\text{ }}2\mu F\]and\[L{\text{ }} = {\text{ }}80{\text{ }}mH\]connected in series. The quality factor of the circuit is
1. \[20\]
2. $2$
3. $0.5$
4. $400$
Answer
161.7k+ views
Hint: Q factor or also called as quality factor of a circuit is nothing but a picture of efficiency of the circuit. It is ratio of total energy stored per cycle to energy dissipated per cycle. For an inductor, it can be explained as the ratio of its inductive reactance ${X_C}$ to resistance i.e. $Q = \frac{{{X_C}}}{R}$. If all energy is in the stored form and no energy is dissipated then it is the ideal case and such a circuit is an ideal circuit. Quality factor is a dimensionless quantity.
Complete answer:
Given that resistance, capacitance and inductance connected in series and values of each are as follows
Capacitance, \[C{\text{ }} = {\text{ }}2\mu F = 2{(10)^{ - 6}}F\]
Inductance, \[L{\text{ }} = {\text{ }}80{\text{ }}mH = 80{(10)^{ - 3}}H\]
Step 1: In resonance condition, capacitance reactance= inductance reactance.
${X_L} = {X_C}$
Step 2: Divide both sides by Resistance R
\[\frac{{{X_L}}}{R} = \frac{{{X_C}}}{R}\]
\[\omega L = \frac{1}{{\omega C}}\]
\[{\omega ^2} = \frac{1}{{LC}}\]
\[{\omega ^2} = \frac{1}{{80{{(10)}^{ - 3}}2{{(10)}^{ - 6}}}}\]
\[{\omega ^2} = 6.25{(10)^6}\]
Step 3: Taking square root
\[\omega = 2500\]
Step 4: Quality factor can be calculated by the formula
$Q = \frac{{\omega L}}{R}$
$Q = \frac{{2500(80){{(10)}^{ - 3}}}}{{100}}$
$Q = 2$
Hence the correct option is (2) and the correct value of quality factor Q is 2.
Note: Capacitance can be simply defined as storing of energy in the form of electric field whereas inductance is storing of energy in form of magnetic field. Quality factor is defined for resonance. Unit of inductance is henry(H) and unit of capacitance is farad(F). While performing calculation, micro farad and milli henry must be converted to farad and henry.
Significance of quality factor can be compared to that of a Carnot engine in thermodynamics. It is used to describe damping in a circuit.
Complete answer:
Given that resistance, capacitance and inductance connected in series and values of each are as follows
Capacitance, \[C{\text{ }} = {\text{ }}2\mu F = 2{(10)^{ - 6}}F\]
Inductance, \[L{\text{ }} = {\text{ }}80{\text{ }}mH = 80{(10)^{ - 3}}H\]
Step 1: In resonance condition, capacitance reactance= inductance reactance.
${X_L} = {X_C}$
Step 2: Divide both sides by Resistance R
\[\frac{{{X_L}}}{R} = \frac{{{X_C}}}{R}\]
\[\omega L = \frac{1}{{\omega C}}\]
\[{\omega ^2} = \frac{1}{{LC}}\]
\[{\omega ^2} = \frac{1}{{80{{(10)}^{ - 3}}2{{(10)}^{ - 6}}}}\]
\[{\omega ^2} = 6.25{(10)^6}\]
Step 3: Taking square root
\[\omega = 2500\]
Step 4: Quality factor can be calculated by the formula
$Q = \frac{{\omega L}}{R}$
$Q = \frac{{2500(80){{(10)}^{ - 3}}}}{{100}}$
$Q = 2$
Hence the correct option is (2) and the correct value of quality factor Q is 2.
Note: Capacitance can be simply defined as storing of energy in the form of electric field whereas inductance is storing of energy in form of magnetic field. Quality factor is defined for resonance. Unit of inductance is henry(H) and unit of capacitance is farad(F). While performing calculation, micro farad and milli henry must be converted to farad and henry.
Significance of quality factor can be compared to that of a Carnot engine in thermodynamics. It is used to describe damping in a circuit.
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