The relationship given in the problem states that the current \( I \) varies directly as the voltage \( V \) and inversely as the resistance \( R \). This can be represented mathematically as:
\[ I = k \frac{V}{R} \]
where \( k \) is a constant of proportionality.
Step 1: Find the constant \( k \)
Using the initial conditions: \( I = 16 , A \), \( V = 75 , V \), and \( R = 15 , \Omega \):
\[ 16 = k \frac{75}{15} \]
Calculating the fraction:
\[ \frac{75}{15} = 5 \]
Substituting back into the equation:
\[ 16 = k \cdot 5 \]
Now solve for \( k \):
\[ k = \frac{16}{5} = 3.2 \]
Step 2: Find the current for the new conditions
Now we can use this value of \( k \) to find the current when \( V = 170 , V \) and \( R = 13 , \Omega \):
\[ I = 3.2 \frac{170}{13} \]
Calculating the fraction \( \frac{170}{13} \):
\[ \frac{170}{13} \approx 13.07692307 \]
Now substituting this back:
\[ I = 3.2 \times 13.07692307 \]
Calculating the final result:
\[ I \approx 41.15384615 \]
Rounding to a reasonable number of significant figures, the current is approximately:
\[ \boxed{41.15 , A} \]
Thus, the current when the voltage is 170 V and the resistance is 13 Ohms is approximately 41.15 amperes.