Arrive at the expression for the impedance of a series LCR circuit using phasor diagram method and hence write the expression for the current through the circuit.

Consider a series LCR circuit connected to an AC source. Applied AC voltage is V = V m sin ω t Let the current through the series circuit be I = I m sin ω t + ϕ Where ϕ is the phase difference between voltage and current. Let I → be the phasor representing current and V R → , V L → , V C → and V → be the phasors representing voltage across R , L , C and source respectively. Let's assume that V C → > V L → V R = I R V L = I X L V C = I X C According to Kirchoff's loop rule V L + V C + V R = V The phasor relation is represented as V L → + V C → + V R → = V → Applying Pythagorean theorem to the right angle triangle, V 2 = V R 2 + V C - V L 2 V 2 = I 2 R 2 + X C - X L 2 V = I R 2 + X C - X L 2 1 2 I = V R 2 + X C - X L 2 I = V Z Where, Z is called impedance of the circuit. Z = R 2 + X C - X L 2

Derive an expression for the impedance of a series LCR circuit, when an AC voltage is applied to it.

Reactance: The effective opposition offered by the component of the A.C. circuit to the flow of current in the circuit is referred to be Reactance. Impedance: The effective opposition offered by combination of resistors, inductors and capacitors to the flow of current is referred to be Impedance. Instantaneous current in L - C - R circuit: Suppose, I be the root-mean-square value of current flowing through all the circuit elements. Potential difference across Inductor L : V L = I X L ............................ 1 Potential difference across Capacitor C : V C = I X C ............................ 2 Potential difference across Resistor R : V R = I R ............................ 3 Since, V R and I are in phase. So, V R is represented by O A in the direction of I . The current lags behind the V L , by angles of π 2 . So, V L is represented by O B perpendicular to the direction of I . The current leads V C by angle of π 2 . So, V C is represented by O F perpendicular to the direction of I . Suppose, V L > V C . Then, resultant of V R and V L - V C is given by O H O H = O A 2 + O D 2 ⇒ V = V R 2 + V L - V C 2 ............................. 4 Substituting equation 1 , 2 and 3 in equation 4 we get, V = I 2 R 2 + I X L - I X C 2 ⇒ V I = R 2 + X L - X C 2 I = V R 2 + X L - X C 2 Here, Z = R 2 + X L - X C 2 is the impedance. Now, tan ϕ = V L - V C V R = I X L - X C I R = X L - X C R ⇒ ϕ = tan - 1 X L - X C R Hence the expression for impedance in series L - C - R has been established.

Mention the three types of energy loss in a transformer.

The three types of energy loss in a transformer are: Loss due to heating of coil, due to increase temperature the resistance also increases. Loss due to Eddy current, due to production of eddy current the impedance increases. Hysteresis loss is due to the magnetization saturation in the core of the transformer.

Mention one power loss in transformer.

The power loss in transformer is due to: 1. The resistance of windings – The low resistance copper cable used for the windings remains resistant and thus leads to heat loss. 2. Leakage of flux – If the core design is not good then the flux produced by the primary coil may not all be connected to the secondary coil. 3. Eddy currents loss – The varying magnetic field not only induces secondary coil currents but also iron core currents themselves. In the iron core, these currents flow in small circles and are termed as eddy currents. 4. Hysteresis – This is because of the repeated iron core magnetisation and demagnetisation induced by the alternating input current. By using alloys such as silicon steel, this can be reduced.

If the peak value of a.c current is 4 . 24 A . What is its root-mean-square value?

We are given, The peak value of current, I 0 = 4 . 24 A I r m s = I 0 2 I r m s = 4 . 24 2 = 2 . 9981 A

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Karnataka Board-2020>Q 6

MEDIUM

12th Karnataka Board

IMPORTANT

Earn 100

How does capacitive reactance vary with frequency?

Important Questions on Alternating Current

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Karnataka Board-2020>Q 7

Mention two sources of energy loss in the transformer.

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Karnataka Board-2020>Q 8

A series LCR circuit contains a pure inductor of inductance $5.0 H$ , a capacitor of capacitance $20 μF$ and a resistor of resistance $40 Ω$ . Find the resonant frequency of the circuit.

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Karnataka Board-2020>Q 9

A series LCR circuit contains a pure inductor of inductance $5.0 H$ , a capacitor of capacitance $20 μ F$ and a resistor of $40 Ω$ . Calculate the Quality factor (Q-factor) of the circuit.

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Karnataka Board-2020>Q 10

A series LCR circuit contains a pure inductor of inductance $5.0 H$ , a capacitor of capacitance $20 μ F$ and a resistor of $40 Ω$ . What is the impedance at resonant condition?

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