Series voltage summation: In a series circuit with five resistors, each resistor drops 6 V. What is the required source voltage to sustain this condition?

Introduction / Context:
This checks understanding of voltage division and Kirchhoff’s Voltage Law (KVL) in series networks. If each series element has a specified voltage drop, the source must supply the sum of those drops to maintain the same current through the chain under steady-state DC conditions.

Given Data / Assumptions:

• Five resistors in series.
• Voltage drop across each resistor is 6 V.
• Steady DC, ohmic behavior, and a single series path.

Concept / Approach:
By KVL, the algebraic sum of drops equals the source voltage (with sign convention). In a pure series chain, total source voltage equals the sum of individual drops regardless of exact resistance values, provided the same current flows through all elements.

Step-by-Step Solution:

Verification / Alternative check:
If the current is I and each resistor is R_i, then V_i = I * R_i = 6 V for each. Summing gives ΣV_i = I * ΣR_i = 30 V. Hence V_source = 30 V, independent of the specific R_i values as long as the drops are 6 V each.

Why Other Options Are Wrong:

• is 6 V: That would be true only for a single 6 V drop, not five in series.
• depends on the current flow / depends on the resistor values: Not here; the problem fixes each drop to 6 V, making the total uniquely 30 V.

Common Pitfalls:

• Confusing equal voltage drops with equal resistances; equal drops imply either equal R or carefully chosen values with the same current—either way, the sum is still 30 V.

Final Answer:
is 30 V