The circuit below shows a 6 V battery, an ammeter reading 3 A, and a lamp. What is the electrical power dissipated by the lamp?
Question 1 Report
The circuit below shows a 6 V battery, an ammeter reading 3 A, and a lamp.
What is the electrical power dissipated by the lamp?
Answer Details
Electrical power is calculated using the equation \(P = V \times I\), where \(P\) is the power in watts, \(V\) is the potential difference in volts, and \(I\) is the current in amperes. The circuit shows a 6 V battery and an ammeter reading 3 A, with a single lamp as the only component:
\[P = 6 \times 3 = 18 \text{ W}\]
Since the lamp is the only component in the circuit (apart from the battery and ammeter, which has negligible resistance), the full 6 V of the battery acts across the lamp, and the full 3 A flows through it. The power dissipated by the lamp is therefore 18 W. Other options such as 2 W (which would be \(V/I\), an incorrect formula) or 9 W (which would be \(V + I\) or some other erroneous combination) do not result from the correct application of the power equation.
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