0 0 votes For the circuit shown in the figure, the voltage and current expressions are $v(t) = E_1 \sin (\omega t) + E_3 \sin (3 \omega t)$ and $i(t)=I_1 \sin (\omega t – \phi _1) + I_3 \sin (3 \omega t – \phi _3) + I_5 \sin (5 \omega t).$ The average power measured by the Wattmeter is $\dfrac{1}{2} E_1 I_1 \cos \phi _1 \\$ $\dfrac{1}{2} [E_1 I_1 \cos \phi _1 + E_1 I_3 \cos \phi _3 + E_1I_5] \\$ $\dfrac{1}{2} [E_1 I_1 \cos \phi _1 +E_3 I_3 \cos \phi _3 ] \\$ $\dfrac{1}{2} [E_1 I_1 \cos \phi _1 + E_3 I_1 \cos \phi _1]$ Electric Circuits gate2012-ee electric-circuits power-systems + – Andrijana3306 201 views answer comment Share Follow 0 reply Please log in or register to add a comment.
