Question 349: AC fundamentals

Question

A group of lamps having total power rating of $1000 \mathrm{~W}$ is supplied by an AC voltage of $E=200 \sin \left(310 t+60^{\circ}\right)$, the rms value of current flowing through the circuit is

Accepted Answer

Power of lamp $(P)=1000 \mathrm{~W}$
AC voltage $(E)=200 \sin \left(310 t+60^{\circ}ight)$
Comparing with $E=E_0 \sin (\omega t+\phi)$,
we get, $E_0=$ peak value of voltage
$\phi=$ phase difference
rms value of supply voltage,
$E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}$
$E_{\mathrm{rms}}=\frac{200}{\sqrt{2}}=100 \sqrt{2}$
Average power is given by,
$\begin{aligned}
P =V_{\mathrm{rms}} I_{\mathrm{rms}} \cos \phi \
1000 =(100 \sqrt{2}) I_{\mathrm{rms}}\left(\frac{1}{2}ight) \quad\left(\because \cos \phi=\cos 60^{\circ}=\frac{1}{2}ight) \
I_{\mathrm{rms}} =10 \sqrt{2} \mathrm{~A}
\end{aligned}$

Answer

$10 \mathrm{~A}$

Answer

$5 \sqrt{2} \mathrm{~A}$

Answer

$20 \mathrm{~A}$

Answer

$10 \sqrt{2} \mathrm{~A}$

AC fundamentals

A group of lamps having total power rating of 1000 \mathrm{~W} is supplied by an AC voltage of E=200 \sin \left(310 t+60^{\circ}\right), the rms value of current flowing through the circuit is

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