Answer:
The current pass the [tex]R_2[/tex] is [tex]I = 0.25 A[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The voltage is [tex]V = 3V[/tex]
The first resistance is [tex]R_1 = 7 \Omega[/tex]
The second resistance is [tex]R_2 = 5 \Omega[/tex]
Since the resistors are connected in series their equivalent resistance is
[tex]R_{eq} = R_1 +R_2[/tex]
Substituting values
[tex]R_{eq} = 7 + 5[/tex]
[tex]R_{eq} = 12 \Omega[/tex]
Since the resistance are connected in serie the current passing through the circuit is the same current passing through [tex]R_2[/tex] which is mathematically evaluated as
[tex]I = \frac{V}{R_{eq}}[/tex]
Substituting values
[tex]I = \frac{3}{12}[/tex]
[tex]I = 0.25 A[/tex]
