For a series LCR circuit I = 100 sin ( 100 π t - π 3 ) mA and V = 100 sin ( 100 π t ) V , then calculate the resistance and reactance of the circuit.

The current I = 0 . 1 sin 100 π t - π 3 A and voltage V = 100 sin 100 π t V , The phase difference between current and voltage is π 3 and current is lagging. The impedance Z = V 0 I 0 ⇒ Z = 100 0 . 1 ⇒ Z = 1000 Ω From phasor diagram, the phase angle tan ϕ = X L - X C R ⇒ tan π 3 = X L - X C R ⇒ X L - X C = 3 R , The impedance Z = R 2 + ( X L - X C ) 2 ⇒ 1000 = R 2 + 3 R 2 ⇒ R = 500 Ω , The reactance of the circuit X L - X C = 500 3 Ω .

For a series LCR circuit I = 100 sin ( 100 π t - π 3 ) mA and V = 100 sin ( 100 π t ) volt , then find average power loss.

The current I = 0 . 1 sin 100 π t - π 3 A and voltage V = 100 sin 100 π t V The phase difference between current and voltage is π 3 and current is lagging. ⇒ ϕ = π 3 The peak current I 0 = 0 . 1 A The RMS current I RMS = I 0 2 = 0 . 1 2 A , The peak voltage V 0 = 100 V The RMS voltage V RMS = 100 2 = 50 2 V , The average power P avg = V RMS I RMS cos ϕ P a v g = 50 2 × 0 . 1 2 cos π 3 ⇒ P avg = 2 . 5 W

In the given circuit R = 100 Ω . If voltage leads current by 60 ∘ then find current supply by source.

The voltage applied is V = 200 sin 100 π t V Voltage leads current by π 3 , therefore the current I = I 0 sin 100 π t - π 3 A , The phase difference between current and voltage is π 3 and current is lagging . From the Phasor diagram, the phase angle tan ϕ = X L - X C R ⇒ tan π 3 = X L - X C R ⇒ X L - X C = 3 × 100 The impedance Z = R 2 + ( X L - X C ) 2 Z = ( 100 ) 2 + ( 100 3 ) 2 ⇒ Z = 200 Ω , The impedance Z = V 0 I 0 ⇒ I 0 = 200 200 ⇒ I 0 = 1 A Therefore, the current I RMS = I 0 2 = 1 2 ⇒ I RMS = 1 2 A

E M B I B E

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 3>Q 6

MEDIUM

NEET

IMPORTANT

Earn 100

Find the phase difference between voltage and current in a series LCR circuit at half-power frequencies.

Important Questions on Alternating Current

HARD

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 4>Q 4

In a series LCR circuit as shown in the figure. Find heat developed in $80$ seconds.

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 4>Q 6

For a series LCR circuit

I = 100 \sin (100 π t - \frac{π}{3}) mA

and

V = 100 \sin (100 π t) V,

then calculate the resistance and reactance of the circuit.

MEDIUM

NEET

IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 4>Q 7

For a series LCR circuit

I = 100 \sin (100 π t - \frac{π}{3}) mA

and

V = 100 \sin (100 π t) volt,

then find average power loss.

EASY

NEET

IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 4>Q 8

The source voltage and current in the circuit are represented by the following equations -

E = 110 \sin (ω t + \frac{π}{6}) V, I = 5 \sin (ω t - \frac{π}{6}) A

Find Impedance of circuit.

EASY

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 5>Q 1

The source voltage and current in the circuit are represented by the following equations-

$E = 110 \sin (ω t + \frac{π}{6}) V, I = 5 \sin (ω t - \frac{π}{6}) A$

Find power factor with nature.

HARD

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>BEGINNER'S BOX - 5>Q 2

In the given circuit $R = 100$ $Ω$ . If voltage leads current by $60^{\circ}$ then find current supply by source.

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>Exercise (Conceptual Questions)>Q 1

Why is the core of a transformer laminated?

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IMPORTANT

Beta Question Bank for Medical: Physics>Chapter 28 - Alternating Current>Exercise (Conceptual Questions)>Q 2

A step down transformer is used to reduce the main supply of 220 V to 11 V. If the primary coil draws a current of 5A and the current in secondary coil 90A, What is the efficiency of the transformer?

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