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