In the previous question, what is the current crossing resistor R1 if the current leaving the battery is 5.0A?

the current crossing resistor R1 if the current leaving the battery is 5.0A is 2.0 Ω

Explanation:

In the previous question;

R1 = 15 Ω, R2 = 3.0 Ω, R3 = 7.0 Ω just as shown in the image uploaded along this  answer; R2 and R3 are in series

so Rq1 = R2 + R3 = 3 + 7 = 10 Ω

now Rq1 and r1 are in pararell

so,

1/Rq = 1/Rq1 + 1/R1 = 1/10Ω  + 1/15Ω  

Rq = 6 Ω

Now In the previous question, what is the current crossing resistor R1 if the current leaving the battery is 5.0A?

Rq1 = 3 + 7 = 10 Ω

R1 = 15 Ω

current in R1 = I₁

therefore

I₁ = I(Rq1 / ( R1 + Rq1))

so we substitute

I₁ = 5 × ( 10 / ( 15 + 10))

I₁ = 5 × 10/25

I₁ = 50 / 25

I₁ = 2.0 Ω

so the current crossing resistor R1 if the current leaving the battery is 5.0A is 2.0 Ω

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )

the force on the thing and the distance the force moves the thing