Fault diagnosis in a series chain: five resistors are in series across a 6 V battery. Measured voltage is zero across all resistors except R3. What voltage appears across R3?

Introduction / Context:
In a healthy series circuit with current flowing, voltages divide among all resistors proportionally. If a meter shows zero volts across every element except one, this indicates a fault such as an open circuit. This question tests reasoning about abnormal measurements and where the full source voltage appears under an open condition.

Given Data / Assumptions:

• Series chain: five resistors labeled R1–R5.
• Source: 6 V battery.
• Measurements: 0 V across all except R3; nonzero only across R3.
• Assume ideal meter (no loading).

Concept / Approach:

If a resistor is open, current in the series loop becomes zero. With zero current, there is no voltage drop across intact resistors (V = I * R = 0). The open element takes the entire source voltage because it is the interruption point separating the two battery terminals via the series chain.

Step-by-Step Solution:

Verification / Alternative check:

Kirchhoff’s laws: with I = 0, V across each intact resistor is 0; loop equation reduces to V_source = V_open. Only the open bears the entire 6 V.

Why Other Options Are Wrong:

1.2 V or 0.6 V imply current and normal division; 0 V contradicts the observation that only across R3 a voltage is present.

Common Pitfalls:

Assuming nonzero current despite the readings; forgetting that an open in series halts current everywhere.

Final Answer:

6 V