In a circuit with two series impedances Z1 and Z2, what is the expression for the voltage across Z1 (V1)?

In a circuit with two series impedances Z1 and Z2, what is the expression for the voltage across Z1 (V1)?

• A. V1 = V_source * Z1 /(Z1 + Z2)
• B. V1 = V_source * Z2 /(Z1 + Z2)
• C. V1 = V_source * Z1 Z2 /(Z1 + Z2)
• D. V1 = V_source * (Z1 - Z2) /(Z1 + Z2)

Answer: (A)

In a series circuit, the same current flows through both impedances, and the voltage across each part is proportional to its impedance. Using the voltage divider idea, the current is I = V_source/(Z1 + Z2). The voltage across Z1 is V1 = I × Z1, which gives V1 = V_source × Z1/(Z1 + Z2). This shows that the share of the source voltage across Z1 depends on how large Z1 is relative to the total impedance. The other expressions don’t match the series-division rule: putting Z2 in the numerator would give the voltage across Z2, not Z1; Z1Z2/(Z1+Z2) would be the parallel-equivalent impedance times the source voltage, not the actual voltage across Z1 in a series circuit; and (Z1 − Z2)/(Z1 + Z2) isn’t a valid way to express a voltage drop in this setup.

In a series circuit, the same current flows through both impedances, and the voltage across each part is proportional to its impedance. Using the voltage divider idea, the current is I = V_source/(Z1 + Z2). The voltage across Z1 is V1 = I × Z1, which gives V1 = V_source × Z1/(Z1 + Z2). This shows that the share of the source voltage across Z1 depends on how large Z1 is relative to the total impedance.

The other expressions don’t match the series-division rule: putting Z2 in the numerator would give the voltage across Z2, not Z1; Z1Z2/(Z1+Z2) would be the parallel-equivalent impedance times the source voltage, not the actual voltage across Z1 in a series circuit; and (Z1 − Z2)/(Z1 + Z2) isn’t a valid way to express a voltage drop in this setup.