Find V_ab and Is. 

(a) Real Power, $P=v_{rms}{i}_{rms}\cos \left({\theta }_v-{\theta }_i\right)=\frac{50}{\sqrt{2}}\times \frac{2}{\sqrt{2}}\cos \left\{\left(-20°\right)-70°\right\}=0W$

Reactive Power, $Q=v_{rms}{i}_{rms}sin({\theta }_v-{\theta }_i)=\frac{50}{\sqrt{2}}\times \frac{2}{\sqrt{2}}\sin {\left\{\left(-20°\right)-70\right\}}^{ }=-50VAR$

Apparent Power, $S=v_{rms}{i}_{rms}=\frac{50}{\sqrt{2}}\times \frac{2}{\sqrt{2}}=50VA$

(b) Power factor, $pf=cos({\theta }_v-{\theta }_i)=cos\left\{\right(-20°)-70°\}=cos(-90°)=0$

(c) Impedance, $Z=\frac{V_m}{I_m}\angle \left({\theta }_v-{\theta }_i\right)=\frac{50}{2}\angle \left\{\left(20°\right)-70\right\}=25\angle -90°=-j25\Omega$

(d) As the reactive power is negative (Q = - 50VAR) and the impedance has a negative imaginary part (- j25) the load is purely capacitive (the current leads the voltage by 90$°$ .