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1. |
According to fuse law, the current carrying capacity varies as |
A. | diameter |
B. | (diameter)1.5 |
C. | \({\left( {{\rm{diameter}}} \right)^{\frac{1}{2}}}\) |
D. | \(\frac{1}{{diameter}}\) |
Answer» C. \({\left( {{\rm{diameter}}} \right)^{\frac{1}{2}}}\) | |