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PHXII07:ALTERNATING CURRENT

356149 An $A C$ voltage $V = 20 \sin 200 π t$ is applied to a series $L C R$ circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is

1

50 W

2

200 W

3

173.2 W

4

21.6 W^{-}

Explanation:

$V = 20 \sin 200 π t, I = 10 \sin (200 π t + \frac{π}{3})$
Average power dissipated in a cycle is given by
$P = V_{r m s} I_{r m s} \cos ϕ$
Where $\cos ϕ =$ Power factor
$P = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ$
Here, $V_{0} = 20, I_{0} = 10; ϕ = \frac{π}{3}$
$P = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \cos \frac{π}{3} = 50 W$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356150 Calculate the value of inductance which should be connected in series with a capacitance of $5 μ F$ and a resistance of $10 Ω$ and an $a . c$ source of $50 H z$ , so that the power factor of the circuit is unity.

1

2.028 H

2

2.029 H

3

2.026 H

4

2.01 H

Explanation:

$L = ?, C = 5 μ F = 5 \times 10^{- 6} F, R$
$= 10 Ω, v = 50 H z$
Power factor is unity, when $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$ or $L = \frac{1}{ω^{2} C}$
$\Rightarrow L = \frac{1}{(2 π \times 50)^{2} \times 5 \times 10^{- 6}}$
$\Rightarrow L = 2.026 H$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356151 A series $L R$ circuit is connected to an ac source of frequency $ω$ and the inductive reactance is equal to $2 R$ . A capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$ . The ratio of the new power factor to the old one is :

1

\sqrt{\frac{3}{2}}

2

\sqrt{\frac{2}{5}}

3

\sqrt{\frac{2}{3}}

4

\sqrt{\frac{5}{2}}

Explanation:

The initial Power factor is
$ϕ_{1} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R)^{2}}} = \frac{R}{\sqrt{5} R}$
The final Power factor is
$= \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R - R)^{2}}} = \frac{R}{\sqrt{2} R}$
$∴ \frac{Final power factor}{Initial power factor} = \frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}} = \sqrt{\frac{5}{2}}$

JEE - 2013

PHXII07:ALTERNATING CURRENT

356152 A inductor 20 $m H$ , a capacitor $100 μ F$ and a resistor $5 Ω$ are connected in series across a source of emf, $V = 10 \sin 314 t$ . The power loss in the circuit is

1

2.74 W

2

1.13 W

3

0. 43 W

4

0. 79 W

Explanation:

$L = 20 m H$
$X_{L} = ω L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$C = 100 μ F$
$X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 10^{- 6}} = \frac{10^{4}}{314} = 31.84 Ω$
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{{(50)}^{2} + {(25)}^{2}} = 25 \sqrt{5}$
The average power loss is
$P = V_{r m s} \times I_{R M S} \times \cos ϕ$
$= V_{r m s} \times \frac{V_{r m s}}{Z} \times \frac{R}{Z} = \frac{{(V_{r m s})}^{2} \times R}{Z^{2}} = \frac{{(\frac{10}{\sqrt{2}})}^{2} \times 50}{{(25 \sqrt{5})}^{2}}$
$= 0.8 W$

NEET - 2018

PHXII07:ALTERNATING CURRENT

356149 An $A C$ voltage $V = 20 \sin 200 π t$ is applied to a series $L C R$ circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is

1

50 W

2

200 W

3

173.2 W

4

21.6 W^{-}

Explanation:

$V = 20 \sin 200 π t, I = 10 \sin (200 π t + \frac{π}{3})$
Average power dissipated in a cycle is given by
$P = V_{r m s} I_{r m s} \cos ϕ$
Where $\cos ϕ =$ Power factor
$P = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ$
Here, $V_{0} = 20, I_{0} = 10; ϕ = \frac{π}{3}$
$P = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \cos \frac{π}{3} = 50 W$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356150 Calculate the value of inductance which should be connected in series with a capacitance of $5 μ F$ and a resistance of $10 Ω$ and an $a . c$ source of $50 H z$ , so that the power factor of the circuit is unity.

1

2.028 H

2

2.029 H

3

2.026 H

4

2.01 H

Explanation:

$L = ?, C = 5 μ F = 5 \times 10^{- 6} F, R$
$= 10 Ω, v = 50 H z$
Power factor is unity, when $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$ or $L = \frac{1}{ω^{2} C}$
$\Rightarrow L = \frac{1}{(2 π \times 50)^{2} \times 5 \times 10^{- 6}}$
$\Rightarrow L = 2.026 H$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356151 A series $L R$ circuit is connected to an ac source of frequency $ω$ and the inductive reactance is equal to $2 R$ . A capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$ . The ratio of the new power factor to the old one is :

1

\sqrt{\frac{3}{2}}

2

\sqrt{\frac{2}{5}}

3

\sqrt{\frac{2}{3}}

4

\sqrt{\frac{5}{2}}

Explanation:

The initial Power factor is
$ϕ_{1} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R)^{2}}} = \frac{R}{\sqrt{5} R}$
The final Power factor is
$= \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R - R)^{2}}} = \frac{R}{\sqrt{2} R}$
$∴ \frac{Final power factor}{Initial power factor} = \frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}} = \sqrt{\frac{5}{2}}$

JEE - 2013

PHXII07:ALTERNATING CURRENT

356152 A inductor 20 $m H$ , a capacitor $100 μ F$ and a resistor $5 Ω$ are connected in series across a source of emf, $V = 10 \sin 314 t$ . The power loss in the circuit is

1

2.74 W

2

1.13 W

3

0. 43 W

4

0. 79 W

Explanation:

$L = 20 m H$
$X_{L} = ω L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$C = 100 μ F$
$X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 10^{- 6}} = \frac{10^{4}}{314} = 31.84 Ω$
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{{(50)}^{2} + {(25)}^{2}} = 25 \sqrt{5}$
The average power loss is
$P = V_{r m s} \times I_{R M S} \times \cos ϕ$
$= V_{r m s} \times \frac{V_{r m s}}{Z} \times \frac{R}{Z} = \frac{{(V_{r m s})}^{2} \times R}{Z^{2}} = \frac{{(\frac{10}{\sqrt{2}})}^{2} \times 50}{{(25 \sqrt{5})}^{2}}$
$= 0.8 W$

NEET - 2018

PHXII07:ALTERNATING CURRENT

356149 An $A C$ voltage $V = 20 \sin 200 π t$ is applied to a series $L C R$ circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is

1

50 W

2

200 W

3

173.2 W

4

21.6 W^{-}

Explanation:

$V = 20 \sin 200 π t, I = 10 \sin (200 π t + \frac{π}{3})$
Average power dissipated in a cycle is given by
$P = V_{r m s} I_{r m s} \cos ϕ$
Where $\cos ϕ =$ Power factor
$P = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ$
Here, $V_{0} = 20, I_{0} = 10; ϕ = \frac{π}{3}$
$P = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \cos \frac{π}{3} = 50 W$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356150 Calculate the value of inductance which should be connected in series with a capacitance of $5 μ F$ and a resistance of $10 Ω$ and an $a . c$ source of $50 H z$ , so that the power factor of the circuit is unity.

1

2.028 H

2

2.029 H

3

2.026 H

4

2.01 H

Explanation:

$L = ?, C = 5 μ F = 5 \times 10^{- 6} F, R$
$= 10 Ω, v = 50 H z$
Power factor is unity, when $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$ or $L = \frac{1}{ω^{2} C}$
$\Rightarrow L = \frac{1}{(2 π \times 50)^{2} \times 5 \times 10^{- 6}}$
$\Rightarrow L = 2.026 H$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356151 A series $L R$ circuit is connected to an ac source of frequency $ω$ and the inductive reactance is equal to $2 R$ . A capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$ . The ratio of the new power factor to the old one is :

1

\sqrt{\frac{3}{2}}

2

\sqrt{\frac{2}{5}}

3

\sqrt{\frac{2}{3}}

4

\sqrt{\frac{5}{2}}

Explanation:

The initial Power factor is
$ϕ_{1} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R)^{2}}} = \frac{R}{\sqrt{5} R}$
The final Power factor is
$= \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R - R)^{2}}} = \frac{R}{\sqrt{2} R}$
$∴ \frac{Final power factor}{Initial power factor} = \frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}} = \sqrt{\frac{5}{2}}$

JEE - 2013

PHXII07:ALTERNATING CURRENT

356152 A inductor 20 $m H$ , a capacitor $100 μ F$ and a resistor $5 Ω$ are connected in series across a source of emf, $V = 10 \sin 314 t$ . The power loss in the circuit is

1

2.74 W

2

1.13 W

3

0. 43 W

4

0. 79 W

Explanation:

$L = 20 m H$
$X_{L} = ω L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$C = 100 μ F$
$X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 10^{- 6}} = \frac{10^{4}}{314} = 31.84 Ω$
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{{(50)}^{2} + {(25)}^{2}} = 25 \sqrt{5}$
The average power loss is
$P = V_{r m s} \times I_{R M S} \times \cos ϕ$
$= V_{r m s} \times \frac{V_{r m s}}{Z} \times \frac{R}{Z} = \frac{{(V_{r m s})}^{2} \times R}{Z^{2}} = \frac{{(\frac{10}{\sqrt{2}})}^{2} \times 50}{{(25 \sqrt{5})}^{2}}$
$= 0.8 W$

NEET - 2018

PHXII07:ALTERNATING CURRENT

356149 An $A C$ voltage $V = 20 \sin 200 π t$ is applied to a series $L C R$ circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is

1

50 W

2

200 W

3

173.2 W

4

21.6 W^{-}

Explanation:

$V = 20 \sin 200 π t, I = 10 \sin (200 π t + \frac{π}{3})$
Average power dissipated in a cycle is given by
$P = V_{r m s} I_{r m s} \cos ϕ$
Where $\cos ϕ =$ Power factor
$P = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ$
Here, $V_{0} = 20, I_{0} = 10; ϕ = \frac{π}{3}$
$P = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \cos \frac{π}{3} = 50 W$

JEE - 2024

PHXII07:ALTERNATING CURRENT

356150 Calculate the value of inductance which should be connected in series with a capacitance of $5 μ F$ and a resistance of $10 Ω$ and an $a . c$ source of $50 H z$ , so that the power factor of the circuit is unity.

1

2.028 H

2

2.029 H

3

2.026 H

4

2.01 H

Explanation:

$L = ?, C = 5 μ F = 5 \times 10^{- 6} F, R$
$= 10 Ω, v = 50 H z$
Power factor is unity, when $X_{L} = X_{C}$
$ω L = \frac{1}{ω C}$ or $L = \frac{1}{ω^{2} C}$
$\Rightarrow L = \frac{1}{(2 π \times 50)^{2} \times 5 \times 10^{- 6}}$
$\Rightarrow L = 2.026 H$
So, correct option is (3).

PHXII07:ALTERNATING CURRENT

356151 A series $L R$ circuit is connected to an ac source of frequency $ω$ and the inductive reactance is equal to $2 R$ . A capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$ . The ratio of the new power factor to the old one is :

1

\sqrt{\frac{3}{2}}

2

\sqrt{\frac{2}{5}}

3

\sqrt{\frac{2}{3}}

4

\sqrt{\frac{5}{2}}

Explanation:

The initial Power factor is
$ϕ_{1} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R)^{2}}} = \frac{R}{\sqrt{5} R}$
The final Power factor is
$= \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{R}{\sqrt{R^{2} + (2 R - R)^{2}}} = \frac{R}{\sqrt{2} R}$
$∴ \frac{Final power factor}{Initial power factor} = \frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}} = \sqrt{\frac{5}{2}}$

JEE - 2013

PHXII07:ALTERNATING CURRENT

356152 A inductor 20 $m H$ , a capacitor $100 μ F$ and a resistor $5 Ω$ are connected in series across a source of emf, $V = 10 \sin 314 t$ . The power loss in the circuit is

1

2.74 W

2

1.13 W

3

0. 43 W

4

0. 79 W

Explanation:

$L = 20 m H$
$X_{L} = ω L = 314 \times 20 \times 10^{- 3} = 6.28 Ω$
$C = 100 μ F$
$X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 10^{- 6}} = \frac{10^{4}}{314} = 31.84 Ω$
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{{(50)}^{2} + {(25)}^{2}} = 25 \sqrt{5}$
The average power loss is
$P = V_{r m s} \times I_{R M S} \times \cos ϕ$
$= V_{r m s} \times \frac{V_{r m s}}{Z} \times \frac{R}{Z} = \frac{{(V_{r m s})}^{2} \times R}{Z^{2}} = \frac{{(\frac{10}{\sqrt{2}})}^{2} \times 50}{{(25 \sqrt{5})}^{2}}$
$= 0.8 W$

NEET - 2018

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