Using a 6 V Battery with Two Resistors and a Bulb
Students have a 6.0 V ideal battery, a 12 Ω resistor R₁, a 6.0 Ω resistor R₂, an ammeter, a voltmeter, and a bulb marked “3.0 V, 0.30 A”. They first connect R₁ and R₂ in series across the battery (circuit X). They then connect R₁ and R₂ in parallel across the battery (circuit Y). The bulb is not used in either circuit at first. Assume the bulb’s resistance is constant at its rated value and that connecting wires have negligible resistance. The students want to use the same battery and resistors to make the bulb glow at its rated voltage and current, without buying new parts.
1. For circuits X and Y, calculate the total resistance, the current through the battery, and the current through each resistor.
Read the worked answer
Circuit X (series):
- R_total = 12 + 6.0 = 18 Ω.
- Battery current = 6.0 / 18 = 0.333 A.
- The same 0.333 A passes through both R₁ and R₂.
Circuit Y (parallel):
- 1/R_total = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12, so R_total = 4.0 Ω.
- Battery current = 6.0 / 4.0 = 1.5 A.
- Current through R₁ = 6.0 / 12 = 0.50 A.
- Current through R₂ = 6.0 / 6.0 = 1.0 A.
The branch currents add to 1.5 A, as expected.
2. Calculate the bulb’s resistance at its rated values. If the bulb is connected in place of R₂ in circuit X, find the bulb’s voltage and current. Is it at rated brightness?
Read the worked answer
Bulb resistance = V/I = 3.0 / 0.30 = 10 Ω.
With the bulb replacing R₂ in series with R₁:
- Total resistance = 12 + 10 = 22 Ω.
- Current = 6.0 / 22 = 0.273 A.
- Bulb voltage = 0.273 × 10 = 2.73 V.
The bulb receives about 2.7 V and 0.27 A, not 3.0 V and 0.30 A. Its power is about 0.75 W instead of the rated 0.90 W, so it glows dimmer than normal. It is not at rated brightness.
3. The students try putting the bulb in parallel with R₁ across the 6.0 V battery. Explain why this is unsafe, and determine whether any series/parallel combination of R₁ and R₂ alone can put exactly 3.0 V across the bulb. Support with calculations.
Read the worked answer
In parallel with R₁, the bulb is connected directly across the full 6.0 V. Its current would be 6.0 / 10 = 0.60 A, twice the rated current, and its power would be 6.0 × 0.60 = 3.6 W, four times the rated 0.90 W. The bulb would very likely burn out.
To operate the bulb at 3.0 V and 0.30 A from a 6.0 V battery, the rest of the circuit must drop 3.0 V at 0.30 A, so it must provide an equivalent resistance of 3.0 / 0.30 = 10 Ω in series.
Using R₁ and R₂ alone, the available equivalent resistances are 12 Ω, 6.0 Ω, 18 Ω (series), and 4.0 Ω (parallel). None equals 10 Ω. Networks where the bulb shares current with one resistor also fail to give exactly rated values; the closest simple arrangement is R₂ in series with the parallel combination of the bulb and R₁, which gives about 2.86 V across the bulb, still below 3.0 V. Thus exact rated operation is not possible with only these two resistors.