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Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, only [tex]45 \times 10^{-6} \, \text{A}[/tex] is measured, but when a 3.0-V battery is used, [tex]25 \times 10^{-3} \, \text{A}[/tex] is measured.

Does the device obey Ohm's law?

Answer :

The device does not obey Ohm's law, as the value of resistance differs in both cases.

According to Ohm's law, the voltage put across a resistor is directly proportional to the current flowing through it.

Voltage = current × resistance,

V = IR,

First case: V = 1.5V.

I = 45×10⁻⁶ A

V = IR

Putting values,

R₁ = 1.5/45×10⁻⁶

R₁ = 33,333.33 Ω

Second case: V = 3 V

I = 25×10⁻³ A

V = IR

Putting values,

R₂ = 3/25×10⁻³

R₂ = 120 Ω

Since R₁ and R₂ are not equal. The device does not obey Ohm's law, because the resistance value differs in both cases.

To learn more about Ohm's law, here:

https://brainly.com/question/1247379

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