SATHEE: Electro Magnetic Induction And Alternating Currents Question 413

Electro Magnetic Induction And Alternating Currents Question 413

Question: A 300 $Ω$ . resistor is connected in series with a parallel-plate capacitor across the terminals of a 50.0 Hz ac generator. When the gap between the plates is empty, its capacitance is $70 / 22 μ F$ . The ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica of dielectric constant k = 5.0 is inserted between the plates, is equal to

Options:

A) 0.1

B) 0.3

C) 0.6

D) 2.9

Show Answer

Answer:

Correct Answer: B

Solution:

$i_{r m s} = \frac{E_{0}}{\sqrt{2 Z}} \Rightarrow \frac{i_{r m s, 1}}{i_{r m s, 2}} = \frac{Z_{2}}{Z_{1}} = \frac{\sqrt{R^{2} + {(\frac{1}{ω k C})}^{2}}}{\sqrt{R^{2} + {(\frac{1}{ω, C})}^{2}}}$

Solving:
$\Rightarrow \frac{i_{r m s, 1}}{i_{r m s, 2}} = \frac{π}{2}$

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Electro Magnetic Induction And Alternating Currents Question 413

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