A 12 V battery is connected to a device and 24 mA of current is measured. If the device obeys Ohm's law, how much current is present when a(n) 108 V battery is used?

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Answer:

Correct answer: Ic₂ = 48 mA = 48 · 10⁻³ A

Explanation:

U₁ = 12 V DC first battery voltage

Ic₁ = 24 mA = 24 · 10⁻³ A Intensity of current with the first battery

U₂ = 24 V DC second battery voltage

Ic₂ = ? Intensity of current with the second battery

The formula that applies to a simple electric circuit under the Ohm's law is:

R = U / Ic

where R is the total resistance in the electrical circuit and it is constant.

R = U₁ / Ic₁ = U₂ / Ic₂ ⇒ U₁ / Ic₁ = U₂ / Ic₂ ⇒ Ic₂ = (U₂ · Ic₁) / U₁

Ic₂ = (24 · 24) / 12 = 48 mA

Ic₂ = 48 mA = 48 · 10⁻³ A

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