The output voltage of a step-down transformer is measured to be $24 \text{ V}$,when connected to a $12 \text{ W}$ light bulb. The value of the peak current is . . . . . . .
- A$2\sqrt{2} \text{ A}$
- B$\sqrt{2} \text{ A}$
- C$2 \text{ A}$
- D$\frac{1}{\sqrt{2}} \text{ A}$
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Similar Questions
Match the following:
Currents $r.m.s.$ values $(A) \ x_0 \sin \omega t$ $(i) \ x_0$ $(B) \ x_0 \sin \omega t \cos \omega t$ $(ii) \ \frac{x_0}{\sqrt{2}}$ $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ $(iii) \ \frac{x_0}{2\sqrt{2}}$
| Currents | $r.m.s.$ values |
|---|---|
| $(A) \ x_0 \sin \omega t$ | $(i) \ x_0$ |
| $(B) \ x_0 \sin \omega t \cos \omega t$ | $(ii) \ \frac{x_0}{\sqrt{2}}$ |
| $(C) \ x_0 \sin \omega t + x_0 \cos \omega t$ | $(iii) \ \frac{x_0}{2\sqrt{2}}$ |
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View Solution$A$ $10\, \Omega$ resistance is connected across $220\, V - 50\, Hz$ $AC$ supply. The time taken by the current to change from its maximum value to the rms value is $....\, ms$.
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View SolutionThe $r.m.s.$ value of potential difference $V$ shown in the figure is
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View SolutionThe $ rms $ value of current in a $ 50 \,Hz $ $ AC $ circuit is $ 6 \,A $. The average value of $ AC $ current over a complete cycle is:
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