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What is Resonance? – Definition, LCR Circuit, Solved Examples

  • Last Updated : 28 Nov, 2021

Alternating Current is used in almost every electrical device operating in our lives right now. This form of electricity has many advantages over the traditional direct current form. Low power consumption, easy voltage switching are two of the major advantages of using alternating current over direct current. Now, these currents behave very differently when these currents are applied to different elements such as resistors, capacitors, and inductors. Circuits containing all these three elements are called LCR circuits. These circuits exhibit an interesting phenomenon called resonance. Let’s look at this phenomenon in detail. 

LCR Circuit

The figure below shows an example of an LCR circuit, in this circuit inductor, resistor and capacitor are connected to a voltage source. Let’s say that the voltage source is an alternating voltage source whose emf is given by, v = vmsin(ωt). If “q” is the charge on the capacitor and “i” is the current in the circuit at a particular time “t”, then in that case the equation using the Kirchoff’s rule is given by, 

L\frac{di}{dt} + ir + \frac{q}{C} = v

This equation should be solved to determine the instantaneous values of the current “i” and its phase relation with the alternating voltage. 

The current is given by, 

i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}

Comparing it with the analogy of resistance in a circuit, the impedance Z is given by, 

Z = \sqrt{R^2 + (X_C - X_L)^2}

Since the phasor I is always parallel to VR, the phase angle φ is the angle between VR and V, 

tan(\phi) = \frac{X_C - X_L}{R}

Resonance 

This is an interesting characteristic of LCR circuits. This phenomenon of resonance is common in the case of natural systems too, where these systems have a natural tendency to oscillate at a particular frequency. This frequency is called the system’s natural frequency. If these systems are driven by a source supplied energy at a natural frequency, then the amplitude of the oscillations is found to be large. This phenomenon is called resonance. 

For an RLC circuit the current is given by, 

i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}

with XC = 1/ωC and XL = ωL. 

In case the frequency is varied, then at a particular frequency, the impedance is minimum. 

XC = XL 

In this case, Z = \sqrt{R^2 + (0 - 0)^2} = R

XC = XL 

⇒ 1/ωC = ωL

⇒ ω2 = 1/LC

⇒ ω = 1/√(LC)

This frequency is called resonance frequency. 

At this point, im = vm/R

Sample Problems 

Question 1: A inductor of 100mH, capacitor of 100pF are connected to a voltage source in series with a resistance. If the system exhibits resonance. Find the frequency.

Answer: 

The resonance frequency is given by, 

 ω = 1/√(LC)

Given: 

L = 100mH 

C = 100pF 

Plugging the values in the equation, 

 ω = 1/√(LC)

⇒  ω = 1/√((100× 100) × 10-15

⇒  ω = 1013 Hz

Question 2: A inductor of 40mH, capacitor of 10pF are connected to a voltage source in series with a resistance. If the system exhibits resonance. Find the frequency. 

Answer: 

The resonance frequency is given by, 

ω = 1/√(LC)

Given: 

L = 40mH 

C = 10pF 

Plugging the values in the equation, 

 ω = 1/√(LC)

⇒  ω = 1/√((10× 40) × 10-15

⇒  ω = 5 × 1013 Hz

Question 3: An inductor of 20mH, a capacitor of 20pF, and a resistance of 100 ohms are connected in series to a voltage source of frequency 50Hz. Find the reactance of the circuit. 

Answer: 

The reactance of the circuit is given by, 

Z = \sqrt{R^2 + (X_C - X_L)^2}

Given: 

f = 50Hz 

L = 20 mH 

C = 20pF 

R = 100

ω = 2πf

⇒ ω = 2π(50)

⇒ ω = 100π

Plugging the values in the equation, 

X = ωL 

⇒ XL = 100π × 20 × 10-3 

⇒ XL = 2000π × 10-3 

⇒ XL =  6.28

⇒ XL = 6.28 Ohms

XC = 1/ωC 

⇒ XC = 1/100π × 20 × 10-3 

⇒ XC = 1/2000π × 10-3 

⇒ XC =  1/6.28

⇒ XC = 0.159 Ohms

Plugging the values in the equation, 

Z = \sqrt{R^2 + (X_C - X_L)^2}

⇒ Z = \sqrt{10000 + 37.46}

⇒ Z= 100.18 Ohms 

Question 4: Find the phase if the impedances and resistances are given below, 

R = 6, XL = 10, XC = 20 

Solution. 

 The phase angle is given by the equation, 

tan(\phi) = \frac{X_C - X_L}{R}

Given: 

R = 6, X = 10 and XC = 20 

Plugging the values in the equation, 

tan(\phi) = \frac{X_C - X_L}{R}

tan(\phi) = \frac{20 - 10}{6}

⇒ tan(φ) = 5/3 

⇒ φ = tan-1(5/3) 

Question 5: Find the phase if the impedances and resistances are given below, 

R = 10 , XL = 10, XC = 20 

Solution. 

 The phase angle is given by the equation, 

tan(\phi) = \frac{X_C - X_L}{R}

Given: 

R = 10 X = 10 and XC = 20 

Plugging the values in the equation, 

tan(\phi) = \frac{X_C - X_L}{R}

tan(\phi) = \frac{20 - 10}{10}

⇒ tan(φ) = 1

⇒ φ = tan-1(1)

⇒ φ = 45° 

Question 5: Find the rms value of the current in the circuit with a voltage source of v = 10sin(50t), if the impedances and resistances are given below, 

R = 5 , XL = 5, XC = 5

Solution. 

 The current is given by the equation, 

i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}

Given: 

R = 5, X = 5, XC = 5 and Vm = 10

Plugging the values in the equation, 

i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}

⇒ i_m = \frac{10}{\sqrt{5^2 + (5 - 5)^2}}

⇒ i_m = \frac{10}{\sqrt{5^2 + (0)^2}}

⇒ im = 2  

This was an example of resonance in a circuit. 


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