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What current (in A) flows through the bulb of a 3.00 V flashlight when its hot resistance is 3.90 Ω?

User Balaji Viswanath
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1 Answer

4 votes
4 votes

Given:

The voltage across the bulb of the flashlight is: V = 3.00 v

The resistance offered by the bulb of the flashlight is: R = 3.90 Ω

To find:

The amount of current flowing through the bulb.

Step-by-step explanation:

The relation between voltage V, current I, and the resistance R is given by Ohm's law. The expression for Ohm's law is:


V=IR

Rearranging the above equation, we get:


I=(V)/(R)

Substituting the values in the above equation, we get:


\begin{gathered} I=\frac{3.00\text{ v}}{3.90\text{ ^^^^2126}} \\ \\ I=0.77\text{ A} \end{gathered}

Final answer:

The current flowing through the bulb of the flashlight is 0.77 Ampere.

User Mustapha El Kojji
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