In a simple AC voltage divider with series impedances Z1 and Z2 driven from a source Vs, the output voltage across Z2 is:

• A. Vs * Z2/(Z1+Z2)
• B. Vs * Z1/(Z1+Z2)
• C. Vs * (Z1+Z2)/Z2
• D. Vs * Z1*Z2/(Z1+Z2)

Answer: (A)

In a simple AC voltage divider, the two impedances are in series, so the same current flows through both. The current from the source is I = Vs/(Z1 + Z2). The output across the second impedance is the voltage drop across Z2, which is Vout = I × Z2 = Vs × Z2/(Z1 + Z2). This holds for complex impedances as well; if you want the magnitude, use |Vout| = |Vs| × |Z2|/|Z1 + Z2| and the phase is the angle of Z2/(Z1 + Z2). The other expressions correspond to different quantities: the voltage across the first impedance would be Vs × Z1/(Z1 + Z2), and Z1Z2/(Z1 + Z2) is the parallel combination of the two impedances, not the voltage across Z2.