Saurabh Koranglekar
Last Activity: 5 Years ago
Dear student
IS drawn from the supply is equal to the vector sum of the resistive, inductive and capacitive current, not the mathematic sum of the three individual branch currents, as the current flowing in resistor, inductor and capacitor are not in same phase with each other; so they cannot be added arithmetically.
Apply Kirchhoff’s current law, which states that the sum of currents entering a junction or node, is equal to the sum of current leaving that node we get,
sqr( Is) = sqr( Ir) + sqr( I inductor - Ic)
Regards