What is the voltage drop across a load with resistance 4.0 Ω and current 3.0 A?
Use Ohm’s law, $V = IR$. With $I = 3.0\,\text{A}$ and $R = 4.0\,\Omega$, the voltage drop is $V = (3.0)(4.0) = 12\,\text{V}$.
What you are being asked to find
You are given the resistance of a load and the current through it. The voltage drop across the load is the potential difference needed to push that current through that resistance.
Apply Ohm’s law
Ohm’s law relates voltage, current, and resistance:
$$V = IR$$
Substitute the given values:
$$V = (3.0\,\text{A})(4.0\,\Omega) = 12\,\text{V}$$
Quick reasonableness check
A $4\,\Omega$ load draws $1\,\text{A}$ for every $4\,\text{V}$ applied. At $3\,\text{A}$, you would expect $3 \times 4\,\text{V} = 12\,\text{V}$, which matches.
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