Question 242: AC fundamentals

Question

In an $$a.c.$$ circuit the voltage applied is $$E = {E_0}\,\sin \,\omega t.$$ The resulting current in the circuit is $$I = {I_0}\sin \left( {\omega t - {\pi  \over 2}} \right).$$ The power consumption in the circuit is given by

Accepted Answer

KEY CONCEPT :
We know that power consumed in a.c. circuit is given by,
$$P = {E_{rms}}{I_{rms}}\cos \phi $$
Here, $$E = {E_0}\sin \omega t$$
$$I = {I_0}\sin \left( {\omega t - {\pi  \over 2}} ight)$$
which implies that the phase difference, $$\phi  = {\pi  \over 2}$$
$$	herefore$$ $$P = {E_{rms}}.{I_{rms}}.\cos {\pi  \over 2} = 0$$
$$\left( {\,\,} ight.$$ as $$\left. {\,\,\cos {\pi  \over 2} = 0\,\,} ight)$$

Answer

$$P = \sqrt 2 {E_0}{I_0}$$

Answer

$$P = {{{E_0}{I_0}} \over {\sqrt 2 }}$$

Answer

$$P=zero$$

Answer

$$P = {{{E_0}{I_0}} \over 2}$$

AC fundamentals

In an $a.c. circuit the voltage applied is E = {E_0}\,\sin \,\omega t. The resulting current in the circuit is I = {I_0}\sin \left( {\omega t - {\pi \over 2}} \right).$ The power consumption in the circuit is given by

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