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
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 IC is greater than IL , the supply current leads the supply voltage by "26\\degree"
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