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

Question

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

Infinity Learn · Accepted Answer

Both B and I are in the same phase. So, let us calculate the time taken by the voltage to change from peak value of rms value.Now, $$220=220sin100$$$$π$$$$t1$$ .Or $$100$$$$π$$$$t1=$$π$$2$$ or $$t1=1200sAgain$$, $$2002=200sin100$$$$π$$$$t2$$ $$or12=sin100$$$$π$$$$t2o$$ r $$100$$$$π$$$$t2=$$π$$4Or$$ $$t2=1400sRequired$$ time $$=t1$$−$$t2=1200$$−$$1400=2$$−$$1400=1400s=2.5$$×$$10$$−$$3s$$

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

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

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