The Norton's equivalent source in amperes as seen into the terminals X and Y is

1

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Solution

The correct option is A 1
Using source transformation theorem,

or we can simply the network,

Now from the circuit, we get

$I_{N} = \frac{5}{5} = 1 A$

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The Norton's equivalent source in amperes as seen into the terminals X and Y is 1

The correct option is A 1Using source transformation theorem, or we can simply the network, Now from the circuit, we get IN=55=1A

The Norton's equivalent source in amperes as seen into the terminals X and Y is

1

The correct option is A 1
Using source transformation theorem,

or we can simply the network,

Now from the circuit, we get

$I_{N} = \frac{5}{5} = 1 A$