joshuaratcliffe
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The power in a lightbulb is given by the equation P=I^2 R, where I is the current flowing through the lightbulb and R is the resistance of the lightbulb. What is the current in a circuit that has a resistance of 30.0 Ω and a power of 55.0 W?

Respuesta :

AL2006
The power DISSIPATED BY a lightbulb is given by the equation P=I² R,
where I is the current flowing through the lightbulb and R is the resistance
of the lightbulb.

If a circuit has a resistance of 30 Ω and it dissipates 55 W of power in the form of heat or light that it never gets back, then ...

          P  =  I² R

        55 W  =  I² · 30 Ω 

            I²  =  55 W / 30 Ω 

            I   =  √(55/30)  =  √1.833   =   1.35 Amperes

Answer:

Current, I = 1.35 A

Explanation:

Given that,

Resistance of circuit, R = 30 ohms

Power of the circuit, P = 55 W

The power in a light bulb is given by the equation is :

[tex]P=I^2 R[/tex]

I is the current in a circuit

[tex]I=\sqrt{\dfrac{P}{R}}[/tex]

[tex]I=\sqrt{\dfrac{55\ W}{30\ \Omega}}[/tex]

I = 1.35 A

So, the current in a circuit is 1.35 A. Hence, this is the required solution.

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