Approved Training Organization (ATO) Test 3 Alternating Current (AC) Practice

In a parallel circuit, the same voltage drives both branches, and the total current is the vector sum of the branch currents. The resistor draws current in phase with the voltage: I_R = V / R. The capacitor draws current that leads the voltage by 90 degrees: I_C = j ω C V. Adding these currents as phasors gives the total current as I_total = V × (1/R + j ω C). This quantity (1/R + j ω C) is the admittance Y of the parallel network, so I_total = V × Y. To get the magnitude, take the magnitude of the admittance: |Y| = sqrt((1/R)^2 + (ω C)^2). Therefore, the magnitude of the total current is |I_total| = V_rms × |Y| = V_rms × sqrt((1/R)^2 + (ω C)^2). This approach directly reflects how the resistor and capacitor currents combine in parallel. While you could also use the equivalent impedance Z_eq and write |I| = V_rms / |Z_eq| (since |Z_eq| = 1/|Y|), the admittance form is the most straightforward for parallel circuits because it sums the branch currents directly.