Mixed series–parallel network: A 12 kΩ, 15 kΩ, and 22 kΩ series string is in series with the parallel of two 10 kΩ resistors. With a 75 V source, what is the approximate current through the 15 kΩ resistor?

Introduction / Context:
When multiple series resistors are combined with a parallel subnetwork in series, the same current passes through all series elements. Finding the current requires computing the overall series equivalent first, then applying Ohm’s law to the entire path. This problem reinforces correct series/parallel reduction and unit handling in kilohms.

Given Data / Assumptions:

• Series resistors: 12 kΩ, 15 kΩ, 22 kΩ.
• Parallel pair: two 10 kΩ → equivalent 5 kΩ.
• Source voltage: 75 V.
• Ideal components, DC steady state.

Concept / Approach:

First reduce the parallel pair (10 kΩ ‖ 10 kΩ = 5 kΩ). Then add all series values to get total resistance. The current through each series element (including the 15 kΩ resistor) is I = V / R_total. No further division is needed because series currents are identical.

Step-by-Step Solution:

Verification / Alternative check:

Power sanity: P_total = V * I ≈ 75 * 0.00139 ≈ 0.104 W. Individual drops: V_15k ≈ I * 15k ≈ 20.8 V (consistent).

Why Other Options Are Wrong:

14 mA and 50 mA are orders of magnitude too large; 5 mA requires a much smaller total resistance than present.

Common Pitfalls:

Forgetting to reduce the 10 kΩ pair to 5 kΩ; treating kΩ as Ω; assuming different currents in series elements.

Final Answer:

1.4 mA