A 6-volt battery with 1.2 ohms internal resistance is connected across two 12-ohm bulbs in parallel. What is the total current drawn from the battery?

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Multiple Choice

A 6-volt battery with 1.2 ohms internal resistance is connected across two 12-ohm bulbs in parallel. What is the total current drawn from the battery?

When two identical bulbs are in parallel, their combined resistance is half of one bulb: 12 Ω in parallel with 12 Ω equals 6 Ω. The battery has internal resistance of 1.2 Ω in series with that load, so the total circuit resistance is 1.2 Ω + 6 Ω = 7.2 Ω. Using Ohm’s law, the current drawn from the battery is I = V / R = 6 V / 7.2 Ω ≈ 0.833 A, about 0.83 A. This also means the voltage across the bulbs is V_load ≈ 0.833 A × 6 Ω ≈ 5.0 V, so each bulb draws about 5 V / 12 Ω ≈ 0.417 A, totaling the same 0.83 A.

When two identical bulbs are in parallel, their combined resistance is half of one bulb: 12 Ω in parallel with 12 Ω equals 6 Ω. The battery has internal resistance of 1.2 Ω in series with that load, so the total circuit resistance is 1.2 Ω + 6 Ω = 7.2 Ω. Using Ohm’s law, the current drawn from the battery is I = V / R = 6 V / 7.2 Ω ≈ 0.833 A, about 0.83 A. This also means the voltage across the bulbs is V_load ≈ 0.833 A × 6 Ω ≈ 5.0 V, so each bulb draws about 5 V / 12 Ω ≈ 0.417 A, totaling the same 0.83 A.

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