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What is the current in a 4.90 Ω resistor connected to a battery that has a 0.200 Ω internal resistance if the terminal voltage of the battery is 9.00 V?

Respuesta :

olumideolawoyin

Answer:

Current = 1.8 A

Explanation:

E = terminal voltage of the battery = 9v

I = current

R = resistor = 4.9 ohms

r = internal resistance = 0.2

E = I ( R + r)

I = E / R + r =

[tex] \frac{9}{4.9 + 0.2} = \frac{9}{5.1} = 1.8 \: amperes[/tex]

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