SATHEE: Alternating Current - Result Question 15

Alternating Current - Result Question 15

15. An inductor $20 m H$ , a capacitor $50 μ F$ and a resistor $40 Ω$ are connected in series across a source of emf $= 10 \sin 340 t$ . The power loss in A.C. circuit is :

(a) $0.51 W$

(b) $0.67 W$

(d) $0.89 W$

[2016]

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Answer:

Correct Answer: 15. (a)

Solution:

(a) Given: $L = 20 m H; C = 50 μ F; R = 40 Ω$

$V = 10 \sin 340 t$

$∴ V_{runs} = \frac{10}{\sqrt{2}}$

$X_{C} = \frac{1}{ω C} = \frac{1}{340 \times 50 \times 10^{- 6}} = 58.8 Ω$

$X_{L} = ω L = 340 \times 20 \times 10^{- 3} = 6.8 Ω$

Impedance, $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$

$= \sqrt{40^{2} + (58.8 - 6.8)^{2}} = \sqrt{4304} Ω$

Power loss in A.C. circuit,

$\begin{aligned} P = i_{rms}^{2} R = (\frac{V_{rms}}{Z})^{2} R \\ = (\frac{10 / \sqrt{2}}{\sqrt{4304}})^{2} \times 40 = \frac{50 \times 40}{4304} ≃ 0.51 W \end{aligned}$

Alternating Current - Result Question 15

15. An inductor $20 m H$ , a capacitor $50 μ F$ and a resistor $40 Ω$ are connected in series across a source of emf $= 10 \sin 340 t$ . The power loss in A.C. circuit is :

Answer:

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Alternating Current - Result Question 15

15. An inductor 20mH, a capacitor 50μF and a resistor 40Ω are connected in series across a source of emf =10sin⁡340t. The power loss in A.C. circuit is :

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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15. An inductor $20 m H$ , a capacitor $50 μ F$ and a resistor $40 Ω$ are connected in series across a source of emf $= 10 \sin 340 t$ . The power loss in A.C. circuit is :