Let's calculate the equivalent resistances on both circuits.
On Diagram A we have a series connection of the resistors. The equivalent resistance will be the sum of all resistances:
[tex]R_{eq}=1+1+1\\\\\boxed{R_{eq}=3\Omega}[/tex]
On diagram B we have a parallel connection of the resistors. The reciprocal of the equivalent resistance will be the sum of the reciprocals of all the resistances:
[tex]\frac{1}{R_{eq}} = \frac{1}{1} +\frac{1}{1} +\frac{1}{1} \\\frac{1}{R_{eq}}=3\\\\\boxed{R_{eq}=\frac{1}{3}}[/tex]
Therefore, the larger resistance occurs on diagram A.
For the current, recall
[tex]V=IR[/tex]
Where [tex]I[/tex] stands for current [tex]R[/tex] is the resistance and [tex]V[/tex] is the voltage. Rearranging the equation we have
[tex]I = \frac{V}{R}[/tex]
We can see that the larger the resistance, the smaller the current gets. So the larger current must happen in the diagram with smaller resistance. Therefore, the larger current occurs on diagram B.
Glad to help, wish you great studies ;)
Mark brainliest if you deem the answer worthy
