Mixed-series/parallel network: Compute the total resistance when R1 = 7 kΩ is in series with a parallel combination of R2 = 20 kΩ, R3 = 36 kΩ, and R4 = 45 kΩ. Show the correct equivalent of the parallel branch added to R1.

Introduction / Context:
This problem checks your ability to simplify a resistor network that combines both parallel and series connections. Such reductions are routine in circuit analysis before applying Ohm's law, Thevenin equivalents, or power calculations. The key is to first collapse the parallel branch to a single equivalent resistance and then add the series resistor.

Given Data / Assumptions:

• R1 = 7 kΩ in series with a parallel network.
• Parallel network: R2 = 20 kΩ, R3 = 36 kΩ, R4 = 45 kΩ.
• All components are ideal resistors; temperature and tolerance effects ignored.

Concept / Approach:
For parallel resistors, use the conductance-sum rule. The parallel equivalent is given by 1/Rp = 1/R2 + 1/R3 + 1/R4. After finding Rp, add R1 in series: R_total = R1 + Rp. Keeping units consistent (kΩ) avoids mistakes.

Step-by-Step Solution:

Verification / Alternative check:
If you replace the three parallel resistors by 10 kΩ and measure the total with an ohmmeter across the two ends of the whole string, you would read approximately 17 kΩ, confirming the arithmetic.

Why Other Options Are Wrong:

• 4 kΩ: Much too small; parallel of 20, 36, 45 kΩ cannot be below the smallest 20 kΩ by that much, and series with 7 kΩ would increase the total.
• 41 kΩ / 108 kΩ: These ignore the strong reduction due to the parallel branch and/or wrongly add all values.

Common Pitfalls:
Accidentally adding parallel resistances directly (incorrect) or forgetting to convert units consistently. Always compute the parallel equivalent first, then add series resistances.

Final Answer:
17 kΩ