z1= (3+6j) and z2= (4-3j) for this impedance in series with supply voltage of 120 v. determine a magnitude current and its phase angle relative to voltage

Question

all method need to be illustrated

Answer 1

The combined impedance z=z1+z2=7+3j. Magnitude of z, or |z|=√7^2+3^2=√(49+9)=√58=7.6158 ohms (Ω) approx.

The j-component is reactance (X=3Ω) and the other component is resistance (R=7Ω).

A right-angled triangle represents the impedance where z is the hypotenuse. Hence |z| as above.

The voltage V=120 (AC volts) and V=IR (Ohm's Law) applies. I=120/R for the resistance=120/7 and 120/X for the reactance X. The phase angle is given by tan(p)=X/R=3/7, so p=23.2º. This is because the reactance current is 90 degrees out of phase with the voltage while the pure resistance is in-phase. In the triangle X/R is the tangent of the phase difference. Magnitude of the current is 120/√58=15.76 amps approx.