Fuse Rating in a Series–Parallel Circuit
A student assembles a series–parallel circuit for a practical. A battery of emf 12.0 V and internal resistance 0.50 Ω is connected through a 3.0 A fuse to a parallel combination of a 6.0 Ω resistor and a 3.0 Ω resistor. The fuse is a thin wire that melts when the current through it exceeds 3.0 A. The student first closes the switch and observes that the fuse blows immediately. A lab partner suggests replacing the fuse with a 5.0 A one and continuing; another warns that the internal resistance will then dissipate too much power. The connecting wires have negligible resistance. The battery is treated as an ideal emf in series with its internal resistance, and the resistors obey Ohm's law.
1. Calculate the current that would flow with the 3.0 A fuse in place and explain why it blows.
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Equivalent resistance of 6.0 Ω parallel with 3.0 Ω: R_p = (6×3)/(6+3) = 18/9 = 2.0 Ω. Total resistance = R_p + r = 2.0 + 0.50 = 2.5 Ω. Current from battery = emf/R_total = 12.0/2.5 = 4.8 A. This current passes through the fuse, and 4.8 A > 3.0 A, so the fuse melts almost immediately. The student's observation is expected; the circuit cannot run with a 3.0 A fuse.
2. With a 5.0 A fuse, find the terminal voltage, the current in each resistor, the power wasted in the internal resistance, and the efficiency.
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With a 5.0 A fuse, 4.8 A is below the rating, so the fuse holds. Terminal voltage V_T = emf − Ir = 12.0 − (4.8)(0.50) = 12.0 − 2.4 = 9.6 V. This is the voltage across the parallel combination. Current in 6.0 Ω: 9.6/6.0 = 1.6 A. Current in 3.0 Ω: 9.6/3.0 = 3.2 A; sum = 4.8 A, as required. Power wasted in internal resistance: P_r = I²r = (4.8)²(0.50) = 11.52 W. Total power from the emf: P_total = emf × I = 12.0 × 4.8 = 57.6 W. Useful power in resistors = 57.6 − 11.52 = 46.08 W. Efficiency = 46.08/57.6 = 0.80, or 80%. The partner is right that internal dissipation is significant, about 20% of the total.
3. If the 3.0 Ω resistor is replaced by a 1.0 Ω resistor while the 5.0 A fuse is in place, determine whether the fuse holds and what new current flows.
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Replace 3.0 Ω with 1.0 Ω. New parallel resistance: R_p' = (6×1)/(6+1) = 6/7 = 0.857 Ω. Total resistance = 0.857 + 0.50 = 1.357 Ω. Current = 12.0/1.357 = 8.84 A. This is well above the 5.0 A fuse rating, so the fuse will blow. If a larger fuse were used, the battery would be delivering about 8.8 A, and internal power would rise to (8.84)²(0.50) ≈ 39 W, which is much larger than before. The circuit should not be run with the 1.0 Ω resistor and the 5.0 A fuse; either the resistor value or the fuse rating must be reconsidered, and the battery's maximum current rating should also be checked.