In a circuit with a 6-volt battery having 1.2 ohms internal resistance feeding two 12-ohm lamps in series, what is the current through the circuit?

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Multiple Choice

In a circuit with a 6-volt battery having 1.2 ohms internal resistance feeding two 12-ohm lamps in series, what is the current through the circuit?

Current in a circuit is set by the total resistance the voltage has to push current through. The two lamps are in series, so they add: 12 + 12 = 24 ohms. Include the battery’s own internal resistance: 24 + 1.2 = 25.2 ohms total. With a 6-volt source, the current is I = V / R = 6 / 25.2 ≈ 0.238 A. So about 0.238 amps flow through the circuit, and the same current passes through both lamps and the internal resistor since it’s a series circuit. The other values would require a smaller or different total resistance, ignoring the extra 1.2 ohms of internal resistance or the second lamp in series.

Current in a circuit is set by the total resistance the voltage has to push current through. The two lamps are in series, so they add: 12 + 12 = 24 ohms. Include the battery’s own internal resistance: 24 + 1.2 = 25.2 ohms total. With a 6-volt source, the current is I = V / R = 6 / 25.2 ≈ 0.238 A. So about 0.238 amps flow through the circuit, and the same current passes through both lamps and the internal resistor since it’s a series circuit. The other values would require a smaller or different total resistance, ignoring the extra 1.2 ohms of internal resistance or the second lamp in series.

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