SATHEE: Electro Magnetic Induction And Alternating Currents Question 52

Electro Magnetic Induction And Alternating Currents Question 52

Question: In the circuit shown below, the ac source has voltage $V = 20 \cos (ω t)$ volts with $ω$ = 2000 rad/sec. the amplitude of the current will be nearest to [AMU (Engg.) 2000]

Options:

A) $2 A$

B) $3.3 A$

C) $2 / \sqrt{5} A$

D) $\sqrt{5} A$

Show Answer

Answer:

Correct Answer: A

Solution:

$R = 6 + 4 = 10, Ω$

$X_{L} = ω L = 2000 \times 5 \times 10^{- 3} = 10, Ω$

$X_{C} = \frac{1}{ω C} = \frac{1}{2000 \times 50 \times 10^{- 6}} = 10, Ω$
$∴ Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = 10, Ω$

Amplitude of current $= i_{0} = \frac{V_{0}}{Z} = \frac{20}{10} = 2 A$

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Electro Magnetic Induction And Alternating Currents Question 52

Question: In the circuit shown below, the ac source has voltage V=20cos⁡(ωt) volts with ω = 2000 rad/sec. the amplitude of the current will be nearest to [AMU (Engg.) 2000]

Options:

Answer:

Solution:

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Question: In the circuit shown below, the ac source has voltage $V = 20 \cos (ω t)$ volts with $ω$ = 2000 rad/sec. the amplitude of the current will be nearest to [AMU (Engg.) 2000]