Dim Lamp in a Series–Parallel Circuit
A student builds a circuit to investigate why a lamp is dim. The battery has an emf of 6.0 V and an internal resistance of 0.50 Ω. A 12 Ω resistor is connected in series with a parallel combination of a 6.0 Ω resistor and a lamp. The lamp is marked '6.0 V, 3.0 W'. When the switch is closed, the lamp glows dimly. The student measures the current through the battery with an ammeter and the voltage across the lamp with a voltmeter. The student claims that the lamp is dim because the battery's internal resistance 'uses up' 6.0 V before the current reaches the lamp, and that removing the 12 Ω resistor will make the lamp operate at its rated power. The student also says that adding another resistor in parallel with the lamp will always increase the total current from the battery.
1. Calculate the resistance of the lamp from its rating, the equivalent resistance of the circuit, the current through the battery, and the actual power of the lamp.
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From the rating, R_lamp = V²/P = (6.0)²/3.0 = 12 Ω, assuming constant resistance. The 6.0 Ω resistor and the lamp are in parallel: 1/Rp = 1/6.0 + 1/12 = 3/12, so Rp = 4.0 Ω. Total resistance = 12 + 4.0 + 0.50 = 16.5 Ω. Battery current I = emf/R_total = 6.0/16.5 = 0.364 A. Terminal voltage = 6.0 − I r = 6.0 − 0.364 × 0.50 = 5.82 V. Voltage across the parallel section = I Rp = 0.364 × 4.0 = 1.45 V. Lamp current = 1.45/12 = 0.121 A. Lamp power = V²/R = (1.45)²/12 = 0.176 W. This is far below the rated 3.0 W, so the lamp is very dim.
2. Evaluate the student's claim that the internal resistance uses up 6.0 V and that removing the 12 Ω resistor will make the lamp operate at rated power.
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The internal resistance does not 'use up' the whole emf. It causes a voltage drop I r; here that is only 0.364 × 0.50 = 0.18 V. Most of the voltage drop is across the 12 Ω series resistor (0.364 × 12 = 4.37 V) and the parallel section (1.45 V). If the 12 Ω resistor is removed, total resistance becomes 4.0 + 0.50 = 4.5 Ω, current = 6.0/4.5 = 1.33 A, voltage across the parallel section = 1.33 × 4.0 = 5.33 V, lamp current = 5.33/12 = 0.444 A, and lamp power = 5.33²/12 = 2.37 W. The lamp is brighter but still not at the rated 3.0 W. The claim is therefore partly right about removing a cause of dimness, but wrong about the rated power and about the internal resistance being the main cause.
3. Explain whether adding another resistor in parallel with the lamp will always increase the total current from the battery.
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Adding a resistor in parallel with the lamp decreases the equivalent resistance of that parallel section, because 1/Rp increases. If the rest of the circuit is unchanged, the total resistance of the circuit decreases, so the battery current I = emf/R_total increases. That is true for an ideal battery and for a battery with fixed internal resistance, as long as the added resistor does not change the other resistances. It will not necessarily make the lamp brighter: the extra parallel branch draws current and lowers the voltage across the parallel section, so the lamp current may decrease. The student's statement about total current is generally correct, but the inference about lamp brightness is not.