Question

A flashlight bulb with a potential difference of 4.5 v across it has a resistance of 8 ohms how much current is the bulb filament

Answer 1

Answer:

Current = 0.5625 A

Explanation:

As we know by Ohm's law that potential difference applied across the conductor and current flows in the conductor is given as

$V = i R$

here we know

$V = 4.5 Volts$

$R = 8 ohm$

now from the equation we know

$4.5 = 8 i$

$i = \frac{4.5}{8}$

$i = 0.5625 A$

Answer 2

Current flowing in bulb filament of flashlight bulb with a potential difference of 4.5 v and resistance of 8 ohms, is 0.5325 ampere.

What is the Ohm's law?

Ohm's law states that for a flowing current the potential difference of the circuit is directly proportional to the current flowing in it.

Thus,

$V\propto I$

Here, $V$ is the potential difference and $I$ is the current.

It can be written as,

$V=IR$

Here, $R$ is the resistance of the circuit.

Given information-

The potential difference of the flashlight bulb is 4.5 V.

The resistance of the flashlight is 8 ohms.

Put the values in the above formula to find the current flowing in the bulb filament,

$4.5=I\times8\\I=\dfrac{4.5}{8} \\I=0.5625\rm A$

Hence the value of the current flowing in the bulb filament is 0.5625 ampere.

Learn more about the Ohm's law here;

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