Electrical and Electronic Engineering tutors

This is a constructed paper exercise in an ideal DC circuit, not instructions to build or test hardware. Assume an ideal 5 V source, ideal resistors and no resistance in connecting wires. V means volts; Ω means ohms; 1 kΩ = 1,000 Ω.

Start with an upper resistor R1 = 1 kΩ in series with a lower resistor R2 = 1 kΩ across the source. Define the output voltage across R2, relative to its lower end. With no load connected across the output, the divider rule gives:

Vout = 5 V × R2 / (R1 + R2) = 5 V × 1,000 Ω / 2,000 Ω = 2.5 V.

Now suppose the circuit model includes a 1 kΩ load across R2. A tempting answer is still 2.5 V because the original two resistors have equal values. The missed step is that the load is in parallel with R2: both share the same two nodes. Their effective resistance is:

Rlower = (1,000 Ω × 1,000 Ω) / (1,000 Ω + 1,000 Ω) = 500 Ω.

The upper resistor is now in series with that effective 500 Ω lower branch, so:

Vout = 5 V × 500 Ω / (1,000 Ω + 500 Ω) = 5/3 V ≈ 1.67 V.

The lower branch resistance has fallen, so it receives a smaller share of the source voltage. A current check gives about 3.33 mA through the upper resistor and 1.67 mA through each lower branch: the incoming current equals the two outgoing currents. Here mA means milliamperes.

The improvement is in the explanation as well as the number: “I included the load in the equivalent resistance before applying the divider rule.” For independent practice, change the load resistance on paper and first predict whether the output moves towards or away from the unloaded 2.5 V. Then calculate and explain the change.

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