For a string of resistors connected in series across a DC source, which statement best describes how the source voltage appears across the individual resistors?

Introduction / Context:
Series resistor strings are used as voltage dividers in bias networks, measurement scaling, and reference generation. Knowing how voltage distributes across each resistor allows you to design predictable tap voltages and assess power dissipation.

Given Data / Assumptions:

• All resistors are in series, so the same current flows through each one.
• Ideal DC source; wiring resistances are negligible.
• Resistors are linear and within rated power limits.

Concept / Approach:
With the same current I through every series element, the drop on resistor Ri is Vi = I * Ri. Therefore, voltage division is proportional to resistance values: larger resistance gets a proportionally larger share of the total source voltage. The total obeys V_source = Σ Vi.

Step-by-Step Solution:
Let R_total = R1 + R2 + ... + Rn.Series current: I = V_source / R_total.Drop on Ri: Vi = I * Ri = (V_source / R_total) * Ri.Hence, Vi / V_source = Ri / R_total, showing proportional division.

Verification / Alternative check:
Choose numbers: V_source = 10 V, R1 = 2 kΩ, R2 = 3 kΩ. Then R_total = 5 kΩ, I = 10/5000 = 2 mA. Drops: V1 = 2 mA * 2 kΩ = 4 V; V2 = 2 mA * 3 kΩ = 6 V. The ratio V1:V2 = 4:6 equals R1:R2 = 2:3, confirming proportionality.

Why Other Options Are Wrong:

• Reduce the power to zero: Power is nonzero unless current is zero; each resistor dissipates P = I^2 * R.
• Cause the current to divide: Current is the same in series; division occurs in parallel networks.
• Increase the source voltage: Passive resistors cannot increase voltage; they only divide the applied voltage.

Common Pitfalls:

• Confusing series and parallel behaviors (current vs. voltage division).
• Ignoring tolerance and temperature effects that slightly alter the division ratio.

Final Answer:
divide the source voltage in proportion to their values