Series network with a shunt across one element: Three 10 kΩ resistors are in series, but one of the 10 kΩ resistors has a 20 kΩ resistor connected in parallel across it. With a 24 V source, what is the total current drawn by the circuit?

Introduction / Context:
This question blends series–parallel reduction with Ohm’s law. Recognizing which elements are in parallel and which are in series is crucial. After simplifying the parallel branch, we add the remaining series resistances and compute the total current from the source voltage.

Given Data / Assumptions:

• Resistors: three of 10 kΩ nominal; across one of these, a 20 kΩ resistor is placed in parallel.
• Source voltage: 24 V (DC).
• Ideal components and wiring; ignore tolerances and loading elsewhere.

Concept / Approach:
First, reduce the parallel pair: R_p = (10 kΩ * 20 kΩ) / (10 kΩ + 20 kΩ) = 200 / 30 kΩ ≈ 6.667 kΩ. Then, since the remaining two 10 kΩ resistors are simply in series with this equivalent, sum them to find the total resistance R_total. Finally, apply I_total = V / R_total and convert to microamperes or milliamperes as appropriate.

Step-by-Step Solution:

Verification / Alternative check:
Quick ratio check: 24 V across ~26.7 kΩ should be slightly under 1 mA, consistent with 900 µA. A back-check of power P = V * I ≈ 24 * 0.0009 = 0.0216 W across the network is also reasonable.

Why Other Options Are Wrong:

• 9 mA and 90 mA: Would require total resistances of only a few kilo-ohms or a few hundred ohms, not ≈ 26.7 kΩ.
• 800 µA: Close but results from rounding the equivalent too aggressively; the correct computed value is ~900 µA.

Common Pitfalls:

• Adding 10 kΩ and 20 kΩ instead of taking their parallel combination.
• Forgetting to keep units in kΩ consistently during arithmetic.

Final Answer:
900 µA