If a closed loop contains a source of $12\,\text{V}$ and two voltage drops of $5\,\text{V}$ and $3\,\text{V}$, what must the remaining voltage drop be for Kirchhoff's Voltage Law to hold?

For a closed loop, total voltage rise equals total voltage drop. The source provides $12\,\text{V}$, and known drops are $5\,\text{V}$ and $3\,\text{V}$, totaling $8\,\text{V}$. The remaining drop must therefore be $12-8=4\,\text{V}$.