The equations of voltage and current in an a.c.circuit are E = 220 2 sin ωt + π 6 i = 15 2 sin ωt - π 6 respectively. Calculate phase difference between E and i and power loss in the circuit.

We know that Impedance Z = E max I max ∴ Z = 220 2 15 2 = 14 . 65 Ω Average power dissipation is P = E max I max 2 cosϕ = 220 2 × 15 2 2 cos π 3 = 132 W

The equations of voltage and current in an a.c.circuit are E = 220 2 sin ωt + π 6 i = 15 2 sin ωt - π 6 respectively. Calculate phase difference between E and i and power loss in the circuit.

We know that Impedance Z = E max I max ∴ Z = 220 2 15 2 = 14 . 65 Ω Average power dissipation is P = E max I max 2 cosϕ = 220 2 × 15 2 2 cos π 3 = 132 W

What is meant by wattless current?

The current in an AC circuit is said to be Wattles Current when the average power consumed in such a circuit corresponds to Zero. Such current is also called as Ideal Current An AC circuit containing either capacitor or inductor current is said to wattless if the average power consumed in the circuit is zero. The formula used for calculating the wattles current is: ⇒ P = VIcosθ Where, P = Is the power in Watts I = Is the current in Amperes V = Is the voltage in Volts. When in the circuit only inductor or capacitor present in that condition the phase difference between voltage and current is π 2 So the power consumption can be calculated by above formula P = VI cos π 2 We know cos π 2 = 0 So power consumption is = P = VI ( 0 ) ⇒ P = 0 watt So we can clearly see that there is no power consumption when phase difference is π 2 So this is called wattles current

According to the given figure showing LCR circuit, the frequency of source is equal to the resonant frequency. find the reading of iii Ammeter A

As in given L-C-R series circuit the frequency of source is equal to the resonant frequency, hence resonance will occur in the circuit i.e X L will be equal to X C and since current in all the components is same, Therefore V L = V C The rms value of supply voltage is 200 V Reading of Voltmeter V 1 V 1 = V L = V C = 200 Volts The Voltage V 2 will give reading of potential difference across R = V R But here V = V R 2 + V L - V C 2 ∴ V R = V ∵ V L = V C ∴ Reading of Voltmeter V 2 = V R = Applied voltage = 200 volts The impedance of the circuit is Z = R 2 + X L - X C 2 = R ∵ X L = X C ∴ Reading of ammeter A = V Z = V R = 200 V 100 Ω = 2 A So, the reading of Voltmeter V 1 = 200 Volts , Voltmeter V 2 = 200 Volts , ammeter A = 2 A

According to the given figure showing LCR circuit, the frequency of source is equal to the resonant frequency. find the reading of ii Voltmeter V 2

As in given L-C-R series circuit the frequency of source is equal to the resonant frequency, hence resonance will occur in the circuit i.e X L will be equal to X C and since current in all the components is same, Therefore V L = V C The rms value of supply voltage is 200 V Reading of Voltmeter V 1 V 1 = V L = V C = 200 Volts The Voltage V 2 will give reading of potential difference across R = V R But here V = V R 2 + V L - V C 2 ∴ V R = V ∵ V L = V C ∴ Reading of Voltmeter V 2 = V R = Applied voltage = 200 volts The impedance of the circuit is Z = R 2 + X L - X C 2 = R ∵ X L = X C ∴ Reading of ammeter A = V Z = V R = 200 V 100 Ω = 2 A So, the reading of Voltmeter V 1 = 200 Volts , Voltmeter V 2 = 200 Volts , ammeter A = 2 A

According to the given figure showing LCR circuit, the frequency of source is equal to the resonant frequency. find the reading of i Voltmeter V 1

As in given L-C-R series circuit the frequency of source is equal to the resonant frequency, hence resonance will occur in the circuit i.e X L will be equal to X C and since current in all the components is same, Therefore V L = V C The rms value of supply voltage is 200 V Reading of Voltmeter V 1 V 1 = V L = V C = 200 Volts The Voltage V 2 will give reading of potential difference across R = V R But here V = V R 2 + V L - V C 2 ∴ V R = V ∵ V L = V C ∴ Reading of Voltmeter V 2 = V R = Applied voltage = 200 volts The impedance of the circuit is Z = R 2 + X L - X C 2 = R ∵ X L = X C ∴ Reading of ammeter A = V Z = V R = 200 V 100 Ω = 2 A So, the reading of Voltmeter V 1 = 200 Volts , Voltmeter V 2 = 200 Volts , ammeter A = 2 A

An alternating voltage of angular frequency ω , is applied across a series combination of a resistance R , an inductance L and a capacitance C . Write down the formula for the impedance of the circuit.

Resonant circuit: In L-C-R circuit, when the value of impedance is least, then the current becomes maximum this phenomenon is called resonance. In this state the frequency of the applied E.M.F. is equal to the natural frequency of the L-C circuit called resonant frequency and this circuit is called resonant electric circuit. In the figure an inductance L , a capacitance C and a resistance R are connected in series with an alternating voltage. At any instant alternating voltage is given as V=V ∘ sinωt . . . . . . . . . 1 Where V ∘ is the peak value of alternating voltage at any instant, if the current in the circuit be I , then. The potential difference across the inductance L is. V L = X L I . . . . . . . . ( 2 ) Similarly, potential difference across the capacitance C is V C = X C I . . . . . . . . . . . ( 3 ) And the potential difference across the resistance R is V R = RI . . . . . . . ( 4 ) Now V R and I are in the same phase while V L Leading ahead I by 90 ° and V C Logging behind I by 90 ° Hence, the angle between V L and V C is 180 ° i.e., V L and V C are in opposite phase. Now the resultant of V L and V C is V L − V C Then the angle between V L − V C and V R is 90 ° If the resultant of ( V L − V C ) and V R is V . Then according to the figure ( 2 ) by Pythagoras theorem. V 2 = ( V L − V C ) 2 + V R 2 . . . . . . . ( 5 ) from equation ( 2 ) , ( 3 ) , ( 4 ) and ( 5 ) V 2 = ( X L I − X C I ) 2 + ( RI ) 2 V 2 = I 2 ( X L − X C ) 2 + I 2 R V 2 I 2 = ( X L − X C ) 2 + R 2 V I = ( X L − X C ) 2 + R 2 . . . . . . . . . . ( 6 ) On comparing equation ( 6 ) with ohms law, the right-hand side of equation ( 6 ) shows that the effective resistance of this circuit which is called impedance of this circuit and denoted by Z L − C − R then, Z L − C − R = ( X L − X C ) 2 + R 2 . . . . . . . . 7 but, X L = ωL and X C = 1 ωC Z L − C − R = ( ωL − 1 ωc ) 2 + R 2 . . . . . . . . . . . 8 Equation ( 7 ) and ( 8 ) are the required expression for the impedance of the circuit.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 7 - Alternating Current>Uttar Pradesh Board-2019>Q 8

EASY

12th Uttar Pradesh Board

IMPORTANT

Earn 100

The equations of voltage and current in an a.c.circuit are

$E = 220 \sqrt{2} \sin (ωt + \frac{π}{6}) i = 15 \sqrt{2} \sin (ωt - \frac{π}{6})$

respectively. Calculate phase difference between $E$ and $i$ and power loss in the circuit.

Important Questions on Alternating Current

MEDIUM

12th Uttar Pradesh Board

IMPORTANT

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

Find:

$(i)$ the value of the phase difference between the current and the voltage in the given series LCR circuit, $(ii)$ if a capacitor $C_{1}$ is connected in parallel with the capacitor $C$ in order to make the power factor of the circuit unity $(1)$ , then find out the value of the capacitance of $C_{1}$ .

Question Image

EASY

12th Uttar Pradesh Board

IMPORTANT

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

Find:

$(ii)$ if a capacitor $C_{1}$ is connected in parallel with the capacitor $C$ in order to make the power factor of the circuit unity $(1)$ , then find out the value of the capacitance of $C_{1}$ .