SATHEE: Electro Magnetic Induction And Alternating Currents Question 414

Electro Magnetic Induction And Alternating Currents Question 414

Question: A resistance of 20 $Ω$ is connected to a source of an alternating potential V= 220 $\sin (100 π t)$ . The time taken by the current to change from the peak value to rms value, is

Options:

A) $0.2, s$

B) $0.25, s$

C) $0.25 \times 10^{- 3}, s$

D) $0.25 \times 10^{- 3}, s$

Show Answer

Answer:

Correct Answer: D

Solution:

Both V and I are in the same phase.

So, let us calculate the time taken by the voltage to change from peak value to rms value. Now 220 $\sin 100 π t_{1}$

Or $100 π t_{1} = \frac{π}{2}$

or $t_{1} = \frac{1}{200} s$ Again, $\frac{220}{\sqrt{2}} = 220 \sin 100 π t_{2}$

Or $\frac{1}{\sqrt{2}} = \sin 100 π t_{2}$ or $100 π t_{2} = \frac{3 π}{4}$

Or $t_{2} = \frac{3}{400} s$

Required time $= t_{2} - t_{1} = \frac{1}{400} s = 2.5 \times 10^{- 3} s$

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Electro Magnetic Induction And Alternating Currents Question 414

Question: A resistance of 20 Ω is connected to a source of an alternating potential V= 220 sin⁡(100πt) . The time taken by the current to change from the peak value to rms value, is

Options:

Answer:

Solution:

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Question: A resistance of 20 $Ω$ is connected to a source of an alternating potential V= 220 $\sin (100 π t)$ . The time taken by the current to change from the peak value to rms value, is