Question

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?

Answer 1

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 2

Current, I = 1.35 A

Step-by-step 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 :

`P=I^2 R`

I is the current in a circuit

`I=\sqrt{(P)/(R)}`

`I=\sqrt{(55\ W)/(30\ \Omega)}`

I = 1.35 A

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