An alternating current of frequency $50 \,Hz$ has a peak value of $14.14 \,A$. The time taken by the alternating current to reach from zero to its maximum value and the root mean square (r.m.s.) value of the current are respectively:
- A$0.025 \,s, 5 \,A$
- B$0.005 \,s, 5 \,A$
- C$0.005 \,s, 10 \,A$
- D$0.025 \,s, 10 \,A$
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Similar Questions
The alternating voltage is given by $v=v_0 \sin \left(\omega t+\frac{\pi}{3}\right)$. When will the voltage be maximum for the first time?
DifficultMHT CET 2024
View SolutionAn alternating e.m.f. is given as $e = e_0 \sin \omega t$. In what time will the e.m.f. have half its maximum value if $e$ starts from zero?
$(T = \text{time period}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5)$
$(T = \text{time period}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5)$
$A$ bulb of resistance $280 \Omega$ is supplied with a $200 V$ $AC$ supply. What is the peak current?
Match the following current $\text{r.m.s.}$ values:
$(A) \ x_0 \sin \omega t$ $(i) \ x_0$ $(B) \ x_0 \sin \omega t \cos \omega t$ $(ii) \ \frac{x_0}{\sqrt{2}}$ $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ $(iii) \ \frac{x_0}{2 \sqrt{2}}$
| $(A) \ x_0 \sin \omega t$ | $(i) \ x_0$ |
| $(B) \ x_0 \sin \omega t \cos \omega t$ | $(ii) \ \frac{x_0}{\sqrt{2}}$ |
| $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ | $(iii) \ \frac{x_0}{2 \sqrt{2}}$ |
Medium
View SolutionFor a domestic $AC$ supply of $220 \,V$ at $50 \,cps$,the potential difference between the terminals of a two-pin electric outlet in a room is (in volt) given by
MediumKVPY 2011
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