Question

Three branches, possessing a resistance of 50ohm, an inductance of 0.15H and a capacitor of 100µF respectively, are connected in parallel across a 100V, 50Hz supply. Calculate

a) The current in each branch

b) The supply current

• c) The phase angle between the supply current and the supply voltage.

Answer 1

a)

$I_R=U/R=100/50=2$ A

$I_L=\frac{U}{2\pi fL}=\frac{100}{2\pi \cdot50\cdot 0.15}=2.12$ A

$I_C=2\pi fCU=2\pi \cdot50\cdot100\cdot10^{-6}\cdot100=3.14$ A

b)

$I=\sqrt{I_R^2+(I_C-I_L)^2}=\sqrt{2^2+(3.14-2.12)^2}=2.24$ A

c)

$cos\phi=I_R/I=2/2.24=0.893$

$\phi=26\degree$

Since I_C​ is greater than I_L​ , the supply current leads the supply voltage by $26\degree$