Q:     The fuse in the diagram is utilized to limit the current in the primary of the transformer. Supposing that the fuse restrict the value of I₁ to 1A, Tell the limit on the value of the secondary current?

SOLUTION:

The highest secondary current is found by using the limit on I₁ & the turn ratio like follows:

                                                I₂ =N₁/N₂ I₁

                                                    = (1/4)1A =250mA

If the secondary current tries to exceed limit of the 250mA, the primary current will exceed its limit & blow off the fuse.