To solve the problem, we can use the formula for current in a wire, which is given by Ohm's law:
\[ I = \frac{V}{R} \]
However, since the current varies directly as the voltage and inversely as the resistance, we can write this relationship as:
\[ I = k \cdot \frac{V}{R} \]
where \( I \) is the current in amperes, \( V \) is the voltage in volts, \( R \) is the resistance in ohms, and \( k \) is a constant of proportionality.
From the problem, we know the following values:
- \( I = 15 \) A when \( V = 80 \) V and \( R = 8 \) ohms.
We can first calculate the constant \( k \):
\[ 15 = k \cdot \frac{80}{8} \]
Simplifying the fraction:
\[ \frac{80}{8} = 10 \]
So we have:
\[ 15 = k \cdot 10 \]
Now, solving for \( k \):
\[ k = \frac{15}{10} = 1.5 \]
Now we have established that the relationship between current, voltage, and resistance can be expressed as:
\[ I = 1.5 \cdot \frac{V}{R} \]
Next, we need to find the current \( I \) when \( V = 111 \) V and \( R = 12 \) ohms:
\[ I = 1.5 \cdot \frac{111}{12} \]
Calculating \( \frac{111}{12} \):
\[ \frac{111}{12} = 9.25 \]
Now substituting back into the formula for \( I \):
\[ I = 1.5 \cdot 9.25 \]
Calculating \( 1.5 \cdot 9.25 \):
\[ I = 13.875 \]
Therefore, the current when the voltage is 111 V and the resistance is 12 ohms is
\[ \boxed{13.875 \text{ A}} \]