Question 362: 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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