SATHEE: Electro Magnetic Induction And Alternating Currents Question 174

Electro Magnetic Induction And Alternating Currents Question 174

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

Options:

A) 0.51 W

B) 0.67 W

C) 0.76 W

D) 0.89 W

Show Answer

Answer:

Correct Answer: A

Solution:

Given: L=20mH; $C = 50 μ F$ ;

$R = 40 Ω$

$V = 10 \sin 340 t$

$∴ V_{r u n s} = \frac{10}{\sqrt{2}}$

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

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

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, $P = i_{r m s}^{2} R = {(\frac{V_{r m s}}{Z})}^{2}$

$R = {(\frac{10 / \sqrt{2}}{\sqrt{4304}})}^{2} \times 40$

$= \frac{50 \times 40}{4304} ≃ 0.51 W$

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Electro Magnetic Induction And Alternating Currents Question 174

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

Options:

Answer:

Solution:

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