A standard filament lamp consumes 100,W when | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The power consumed by the lamp is given by $P = V_{rms} 	imes I_{rms}$.Given $P = 100\,W$ and $V_{rms} = 200\,V$.Therefore,the root mean square c

Vedclass · Accepted Answer

$0.707$

Vedclass · Answer

(A) The power consumed by the lamp is given by $P = V_{rms} 	imes I_{rms}$.Given $P = 100\,W$ and $V_{rms} = 200\,V$.Therefore,the root mean square current $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{200} = 0.5\,A$.The peak current $I_{peak}$ is related to the $RMS$ current by the formula $I_{peak} = I_{rms} 	imes \sqrt{2}$.Substituting the value of $I_{rms}$,we get $I_{peak} = 0.5 	imes \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.707\,A$.

$A$ standard filament lamp consumes $100\,W$ when connected to $200\,V$ $AC$ mains supply. The peak current through the bulb will be $........\,A$.

Similar Questions

An alternating e.m.f. is given by $e = e_{0} \sin(\omega t)$. In what time will the e.m.f. reach half its maximum value,if $e$ starts from zero? ($T$ = Time period)

The average value of alternating current $I = I_0 \sin \omega t$ in the time interval $[0, \frac{\pi}{\omega}]$ is:

An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

An $ac$ generator produces an output voltage $E = 170 \sin(377t) \text{ volts}$,where $t$ is in seconds. The frequency of the $ac$ voltage is......$Hz$.

$A$ periodic voltage $V$ varies with time $t$ as shown in the figure. $T$ is the time period. The $r.m.s.$ value of the voltage is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module