Voltage division in a series network: Resistors of 3.9 kΩ, 7.5 kΩ, and 5.6 kΩ are connected in series across a 34 V DC source. What is the voltage drop across the 7.5 kΩ resistor?

Introduction / Context:Voltage division is a cornerstone concept in series circuits. When several resistors share a series current, each drop is proportional to its resistance. This problem exercises the divider formula and careful arithmetic in kilo-ohm units under a DC supply.

Given Data / Assumptions:

• Series resistors: 3.9 kΩ, 7.5 kΩ, 5.6 kΩ.
• Supply voltage: 34 V DC.
• Ideal resistors and wires (no extra drops).

Concept / Approach:For a series string, the same current I flows through all resistors. The voltage drop across any one resistor R_i is v_i = V_total * (R_i / R_total). First compute the total resistance, then apply the proportion for the 7.5 kΩ element.

Step-by-Step Solution:

Verification / Alternative check:Find current I = V / R_total = 34 / 17k = 2 mA. Then v_7.5k = I * 7.5k = 0.002 * 7500 = 15 V. Both methods agree.

Why Other Options Are Wrong:34 V is the whole supply, not a single drop. 7.8 V and 11.2 V correspond to other ratios in the network, not 7.5 kΩ. 2.9 V is not consistent with the series proportion.

Common Pitfalls:Forgetting units (kΩ vs Ω) and mixing formulas; double-check arithmetic when dividing by totals.

Final Answer:15 V