Non-Ohmic Filament in a Series Circuit
A small incandescent filament is described by a non-Ohmic relation V = kI² with k = 40 V/A² over its operating range. It is placed in series with a fixed resistor R = 10 Ω and an ideal 12 V battery. A student assumes the filament obeys Ohm's law with a resistance measured at low current (R_fil = 5 Ω) and predicts a current of 12/(10+5) = 0.80 A. The instructor points out that the filament heats up and its resistance changes, so the operating point must be found self-consistently.
1. Find the actual steady-state current and the voltage across the filament.
Read the worked answer
Kirchhoff's loop rule: 12 = IR + V_fil = 10I + 40I². Rearrange: 40I² + 10I − 12 = 0 → 20I² + 5I − 6 = 0. I = [−5 + √(25 + 480)]/40 = (−5 + 22.47)/40 ≈ 0.437 A. Then V_fil = 40(0.437)² ≈ 7.6 V, and V_R = 10(0.437) ≈ 4.4 V, summing to 12 V. The student's 0.80 A is wrong because it uses a resistance valid only at low current.
2. At this operating point, what is the filament's effective (differential vs. static) resistance, and how much power is dissipated in each element?
Read the worked answer
Static resistance R_static = V/I = 7.6/0.437 ≈ 17.4 Ω. Differential resistance dV/dI = 2kI = 2(40)(0.437) ≈ 35 Ω. Power in filament P_fil = V_fil·I = 7.6(0.437) ≈ 3.3 W; power in resistor P_R = I²R = (0.437)²(10) ≈ 1.9 W; total ≈ 5.2 W, matching P_battery = 12(0.437) ≈ 5.2 W. The distinction matters because a small-signal analysis of fluctuations would use the differential value, not the static one.