According to the information given, the current \( I \) varies inversely with the resistance \( R \). This relationship can be expressed mathematically as:
\[ I \times R = k \]
where \( k \) is a constant.
From the problem, when \( I = 5 \) A, \( R = 30 \) ohms. We can calculate the constant \( k \):
\[ k = I \times R = 5 , \text{A} \times 30 , \text{ohms} = 150 , \text{A} \cdot \text{ohms} \]
Now, we need to find the resistance \( R \) when the current \( I \) is 6 A. Using the constant \( k \):
\[ k = I \times R \]
Substituting the known values:
\[ 150 = 6 , \text{A} \times R \]
Now, solving for \( R \):
\[ R = \frac{150}{6} = 25 , \text{ohms} \]
Thus, the resistance when the current is 6 A is \( \boxed{25} \) ohms.