The general divider formula can be used to determine voltage drop across series elements.

The general divider formula can be used to determine voltage drop across series elements.

• A. True
• B. False
• C. Only for parallel circuits
• D. Only with a single resistor

Answer: (A)

In a series chain, the same current flows through every element and the total source voltage is the sum of all individual drops. The general voltage-divider rule lets you find the drop across any one element by comparing that element’s impedance to the total. Specifically, the voltage across a chosen element is the source voltage times its impedance divided by the sum of all impedances in the series: V_element = V_source × Z_element / (Z1 + Z2 + ... + Zn). In DC circuits with resistors, this becomes V_element = V_source × R_element / (R1 + R2 + ... + Rn). In AC circuits, you use impedances Z, which may be complex; the same idea applies using phasors. This works for any number of series elements and always adds up to the total supply. It doesn’t apply to parallel circuits, where voltages across branches are the same and currents split instead. Example: three resistors in series, 3 Ω, 4 Ω, 5 Ω, across 12 V yield the drop across the 4 Ω resistor as 12 × 4/(3+4+5) = 4 V.

In a series chain, the same current flows through every element and the total source voltage is the sum of all individual drops. The general voltage-divider rule lets you find the drop across any one element by comparing that element’s impedance to the total. Specifically, the voltage across a chosen element is the source voltage times its impedance divided by the sum of all impedances in the series: V_element = V_source × Z_element / (Z1 + Z2 + ... + Zn). In DC circuits with resistors, this becomes V_element = V_source × R_element / (R1 + R2 + ... + Rn). In AC circuits, you use impedances Z, which may be complex; the same idea applies using phasors. This works for any number of series elements and always adds up to the total supply. It doesn’t apply to parallel circuits, where voltages across branches are the same and currents split instead. Example: three resistors in series, 3 Ω, 4 Ω, 5 Ω, across 12 V yield the drop across the 4 Ω resistor as 12 × 4/(3+4+5) = 4 V.