Fuse Margin and Resistor Substitution in a DC Circuit
A technician builds a 12.0 V DC circuit. A 60 Ω resistor and a 40 Ω resistor are connected in parallel; that parallel pair is in series with a 10 Ω resistor rated 2 W. A 0.50 A fuse protects the entire circuit. The technician measures the supply at 12.0 V and assumes the resistors are within tolerance. The work order asks whether the fuse has enough margin, how much power the 10 Ω resistor dissipates, and whether a 4.7 Ω, 1/2 W resistor can be substituted for the 10 Ω unit to draw more current. No other components are in the circuit.
1. Find the total resistance and total current. Is the 0.50 A fuse adequately rated for normal operation?
Read the worked answer
For the parallel pair:
- R_p = (60 × 40) ÷ (60 + 40) = 2,400 ÷ 100 = 24 Ω.
Add the series resistor:
- R_total = 24 + 10 = 34 Ω.
- I = V ÷ R = 12.0 ÷ 34 = 0.353 A.
The fuse is 0.50 A, so normal current is about 71% of the fuse rating. That leaves 0.147 A of margin. The fuse is not blown by normal current, and the margin is reasonable for a simple circuit, though a technician should still check whether the fuse is intended to protect the wiring or the components.
2. How much power does the 10 Ω resistor dissipate, and how does that compare with its 2 W rating?
Read the worked answer
Use P = I²R with I = 0.3529 A:
- P = (0.3529)² × 10 ≈ 0.1246 × 10 = 1.25 W.
The resistor is rated 2 W, so it is dissipating about 1.25 W, or roughly 62% of its rating. That is within rating but not a large margin.
Check the total power as a sanity check:
- P_total = V × I = 12.0 × 0.3529 ≈ 4.24 W.
- The parallel pair dissipates 4.24 − 1.25 ≈ 2.99 W, which is consistent with the 8.47 V across the parallel pair.
3. Can the 10 Ω resistor be replaced with a 4.7 Ω, 1/2 W resistor to draw more current?
Read the worked answer
With 4.7 Ω in series:
- R_total = 24 + 4.7 = 28.7 Ω.
- I = 12.0 ÷ 28.7 ≈ 0.418 A.
That is still below the 0.50 A fuse, but the margin is much smaller: only about 0.082 A.
Now check the resistor's power:
- P = I²R = (0.418)² × 4.7 ≈ 0.1748 × 4.7 ≈ 0.82 W.
The substitute is rated 1/2 W = 0.50 W, so it would dissipate about 0.82 W, well above its rating. It would overheat and could fail. The substitution is not acceptable merely because the fuse would survive; the resistor's power rating is the limiting factor.