Fuse-limited dimmer design: in a 120 V lamp-dimming circuit with a 3 A fuse and a lamp resistance of 20 Ω, find the minimum rheostat resistance that prevents blowing the fuse.

Introduction / Context:
Series circuits with a protective fuse must ensure the total current stays at or below the fuse rating. For a lamp-dimmer using a rheostat in series with a lamp, choosing a sufficient rheostat resistance prevents overcurrent at maximum mains voltage.

Given Data / Assumptions:

• Supply voltage: 120 V (assume DC-equivalent or RMS for calculation).
• Fuse rating: 3 A (maximum allowable current).
• Lamp resistance: 20 Ω (assumed constant for this calculation).
• Rheostat in series with the lamp.

Concept / Approach:

The largest current occurs at the smallest total series resistance. To stay within 3 A, ensure R_total ≥ V / I_max. Since R_total = R_lamp + R_rheostat_min, solve for the minimum rheostat value that satisfies the fuse limit.

Step-by-Step Solution:

Verification / Alternative check:

Check current at this setting: I = V / (R_lamp + R_rheostat) = 120 / (20 + 20) = 120 / 40 = 3 A, which equals the fuse rating (no blow at steady state).

Why Other Options Are Wrong:

40 Ω would be the total required resistance, not the rheostat alone. 4 Ω or 2 Ω would yield total resistance below 40 Ω, causing I > 3 A and blowing the fuse.

Common Pitfalls:

Forgetting that total series resistance includes both lamp and rheostat; confusing fuse current limit with power rating; ignoring that lamp resistance can vary with temperature (here treated as constant by assumption).

Final Answer:

20 Ω