A 900 V source is applied to three series resistors. Two measured drops are 300 V and 280 V. What must the third series voltage drop be so that the drops sum to the source?

Introduction / Context:Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltage rises and drops around any closed loop is zero. In a simple series circuit, the sum of the individual resistor drops equals the source voltage. This question checks your ability to use KVL to find a missing drop from the others and the source.

Given Data / Assumptions:

• Source voltage Vs = 900 V.
• Measured drops across two resistors: V1 = 300 V and V2 = 280 V.
• All elements are in series; polarities consistent with passive sign convention.

Concept / Approach:For series elements, Vs = V1 + V2 + V3. Solve for the unknown V3 by rearranging: V3 = Vs - (V1 + V2). This uses only KVL and requires no knowledge of individual resistances or current.

Step-by-Step Solution:Write KVL sum: Vs = V1 + V2 + V3.Substitute known values: 900 = 300 + 280 + V3.Compute partial sum: 300 + 280 = 580 V.Solve for V3: V3 = 900 − 580 = 320 V.Therefore, the third drop is 320 V.

Verification / Alternative check:Add all three: 300 + 280 + 320 = 900 V, exactly matching the source, confirming correctness.

Why Other Options Are Wrong:

• 30 V or 270 V: Do not bring the sum to 900 V; violate KVL.
• 900 V: Would imply zero drop on the other resistors, contradicting the given measurements.

Common Pitfalls:

• Arithmetic mistakes in subtracting from the source voltage.
• Forgetting that series drops always add to the source when polarities are consistent.

Final Answer:320 V