If I make a simple circuit with a 5 volt battery and a single light bulb then I can expect the voltage across the lightbulb to be 5 volts. If I add another five volt battery to the circuit I can expect the voltage across the bulb to be_______. I can also expect the current through the bulb to________.

Question

BlackTom AI · Accepted Answer

Question restatement: A simple circuit initially has a 5-volt battery powering a single light bulb. If a second identical 5-volt battery is added in series, what is the expected voltage across the bulb, and what happens to the current through the bulb?

Option 1: 2.5 volts, decrease
- This would imply the voltage is halved and the current decreases. In a series arrangement, voltages from identical batteries add, not divide, so halving the bulb’s voltage is incorrect. Therefore this option misapplies how voltages combine in series.

Option 2: 20 volts, decrease
- Two 5 V batteries in series sum to 10 V, not 20 V. The “20 volts” figure would require either a different voltage source or a misinterpretation of series vs parallel connections. Additionally, if the voltage were 20 V, the current would typically increase, not decrease, given the same bulb resistance, so this pair of claims is inconsistent with basic circuit relationships.

Option 3: 5, volts, stay the same (but the batteries will last longer)
- This option suggests the voltage remains at 5 V. However, adding a second identical battery in series increases total voltage to 10 V, not 5 V, so the voltage claim is incorrect. Even if the voltage were unchanged, the current would change according to Ohm’s law; in this setup the current would not stay the same because the source voltage changes.

Option 4: 10 volts, increase
- Here, the voltage across the bulb becomes 10 V because the two 5 V batteries in series add their voltages. Ohm’s law states I = V/R, so with the bulb’s resistance roughly constant, doubling the applied voltage doubles the current (increase), assuming the bulb’s resistance remains approximately constant under the new operating point. This aligns with the basic idea that series batteries sum their voltages and that current through a fixed resistor increases with higher voltage.

Additional clarifications:
- In a series circuit, you add voltages: V_total = V1 + V2 = 5 V + 5 V = 10 V. The bulb’s resistance does not change significantly due to the modest voltage change, so the current roughly scales with the voltage: I ∝ V. Therefore, the current increases when you double the supply voltage.
- Battery life consideration: When you connect identical batteries in series, the voltage doubles but the capacity (in ampere-hours, Ah) stays the same. Since the bulb draws more current at the higher voltage, the overall runtime would not increase and may even decrease compared to the single-battery case, depending on how the bulb’s resistance behaves with temperature. This nuance explains why a claim like “stay the same (batteries last longer)” is not correct in this scenario.

Summary note on correctness mismatch:
- The only option that correctly matches both the new voltage (10 V) and the expected increase in current (due to higher supply voltage) is the one stating '10 volts, increase.' The other options misstate the voltage addition in series, mischaracterize how current responds to voltage for a resistive load, or mix in incorrect implications about battery life.

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