A 3-volt flashlight uses a bulb with 60-ohm resistance. What current flows through the flashlight?

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Question

Using Ohm's Law (I = V / R), calculate the current in a circuit with a voltage of 3 V and a resistance of 60 Ω.

Answer

0.05 Amperes (or 50 milliamps).

Question

f the resistance of the flashlight bulb were halved to 30 Ω, what would the new current be?

Answer

0.1 Amperes (I = 3V / 30Ω = 0.1 A).

Question

What is the power (in watts) dissipated by the 60-ohm bulb when operating at 3 volts? (Use P = V * I or P = V²/R)

Answer

0.15 Watts (P = 3V * 0.05A = 0.15 W, or P = (3V)² / 60Ω = 9/60 = 0.15 W).

Question

Rearrange Ohm's Law to solve for resistance. What would the resistance be if a current of 0.5 A flowed from a 3V battery?

Answer

6 Ohms (R = V / I = 3V / 0.5A = 6Ω).

Question

In the original problem, which answer choice (0.5 A) would result from a common calculation error? What wrong operation was likely performed?

Answer

The error is multiplying voltage and resistance (3V * 60Ω = 180, then misplacing decimal) instead of dividing them.

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Is the unit for electrical current in this problem the Ampere?

A 3-volt flashlight with a 60-ohm bulb draws a current of 0.05 amperes.