A single-phase voltage source $v_s=325 \sin (2 \pi 50 t) \vee$ delivers a current, $i=12 \sin (2 \pi 50 t) +9 \sin (2 \pi 50 t)$ A to a load.

The load power factor, correct up to two decimal places, is

An electrical component has voltage drop $v=V_m \sin (\omega t)$, when the current through it is $i=I_m \sin (\omega t-\theta)$. What is the average power dissipated over a half cycle corresponding to $\omega$ ?

The electrical network shown has an independent voltage source $(10 \mathrm{~V})$ and a current source ( $1 \mathrm{u}(\mathrm{t}) \mathrm{mA}$ ).

The voltage across the capacitor at time instants (in seconds) $t=0^{+}, t=0.50$, and $t=\infty$, respectively, is:

The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.

$$ \begin{array}{|c|c|} \hline \text { Terminal voltage }\left(v_t\right) & \text { Terminal current }\left(i_t\right) \\ \hline 18 \mathrm{~V} & -0.5 \mathrm{~A} \\ \hline 30 \mathrm{~V} & 0.5 \mathrm{~A} \\ \hline 36 \mathrm{~V} & 1.0 \mathrm{~A} \\ \hline \end{array} $$

The correct choice for the parameters ( $\mathrm{I}_{\mathrm{N}}, \mathrm{R}_{\mathrm{N}}$ ) of the Norton equivalent circuit shown in Figure (b) is: