A Bulb, an Ammeter, and Battery Resistance
Students are given a nominally 1.5 V D-cell, a small incandescent bulb labeled 1.5 V, 0.30 A, an ammeter, and a voltmeter. With the battery disconnected, a high-impedance voltmeter reads 1.58 V. When the bulb is connected in series with the ammeter, the ammeter reads 0.32 A and the voltmeter across the bulb reads 1.12 V. The students calculate the bulb resistance as 3.5 Ω and the battery's internal resistance as (1.58 − 1.12)/0.32 = 1.44 Ω. They are surprised: the bulb label implies about 5 Ω, and the calculated internal resistance seems large for a D-cell. They wonder whether the ammeter resistance is corrupting the result and whether the bulb can be treated as an ohmic resistor. The teacher asks them to interpret the measurements, identify the most likely explanation, and design a test of their model.
1. Calculate the bulb power and resistance, and explain why the measured resistance differs from the label. How does ammeter resistance affect the internal-resistance estimate?
Read the worked answer
Bulb power P = VI = (1.12)(0.32) = 0.358 W, and its operating resistance is R_b = 1.12/0.32 = 3.5 Ω. The label implies R = 1.5/0.30 = 5.0 Ω at 0.30 A. The difference is not necessarily a contradiction: an incandescent filament is strongly non-ohmic because its resistance rises with temperature. At the lower voltage and power here, the filament is cooler and its resistance is lower than at the labeled operating point. The battery internal-resistance estimate r = (1.58 − 1.12)/0.32 = 1.44 Ω assumes the ammeter is ideal. If the ammeter has resistance R_A, the battery terminal voltage is V_bulb + I R_A, so the true internal resistance is smaller by R_A. A significant ammeter resistance would therefore inflate the 1.44 Ω value.
2. How would you test whether the bulb is ohmic and whether the ammeter resistance matters? Describe the measurements and the expected graphs.
Read the worked answer
Measure the ammeter resistance separately, with a multimeter or by comparing it with a known resistor, and either subtract it or use a low-resistance ammeter or shunt. To test the bulb, vary the applied voltage with a potentiometer or adjustable supply and record V and I at several points, including low and high voltages. Plot V versus I; a straight line through the origin means ohmic behavior, while a curve whose slope increases with I means resistance rises with temperature. Also measure the current immediately after switching on and after it stabilizes to see the filament heat up. A single V-I pair cannot establish ohmic behavior; the full characteristic curve can.
3. If the bulb is replaced by a 5.0 Ω resistor, predict the current and terminal voltage using the measured emf and internal resistance. Compare with the bulb result.
Read the worked answer
Assume E = 1.58 V and r = 1.44 Ω are constant and the 5.0 Ω resistor is ohmic. Then I = E/(R + r) = 1.58/(5.0 + 1.44) = 0.245 A. The terminal voltage is V = E − Ir = 1.58 − (0.245)(1.44) = 1.23 V, equal to IR = 1.23 V. The power in the resistor is I²R = 0.30 W. Compared with the bulb, the ohmic resistor draws less current and leaves a higher terminal voltage; the difference is a useful check on the internal-resistance model. If the measured current with the resistor differed substantially from 0.245 A, the assumed constant r would be questionable.