Question

A flashlight uses batteries that add up to 6.0 V and has a power output of 0.51 W.(a) How much current is drawn from the batteries?______ A(b) What is the effective resistance of the flashlight? _______Ω

Answer 1

a)

We know that the power is given by:

`P=VI`

where V is the voltage and I is the current. In this case the voltage is 6 V and the power is 0.51 W; plugging these values and solving for the current we have that:

`\begin{gathered} 0.51=6I \\ I=(0.51)/(6) \\ I=0.085 \end{gathered}`

Therefore, the current is 0.085 A.

b)

Using Ohm's law we have that:

`\begin{gathered} R=(6)/(0.085) \\ R=70.59 \end{gathered}`

Therefore, the resistance is 70.59 Ohms