A 12-V battery is connected to a device and 24 mA (24 x 10' Amps) of current flows
through it. If the device obeys Ohm's law, how much current will flow when a 24-V battery
is used?

Correct  Ic₂ = 48 mA = 48 · 10⁻³ A

Explanation:

Given:

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

God is with you!!!

any one of the statements could be an accurate comparison. but the statement that would define the correct choice and seal the deal is not even on the list:

"the visibilities of vector-x and vector-y in the drawing provided with the question are identical and nonexistent."

electrical power