Question

If the current through the middle part of the loop is I1 = 4.75 amps, what is the current through the top loop, I2, in amps?

Answer 1

Answer:

3.5 A counterclockwise

Explanation:

According to Kirchhoff's voltage law, the sum of the voltage drops and voltage gains around a loop is 0.

-I₁R₁ − I₂R₂ + ε₁ − I₂r₁ = 0

I₂ (r₁ + R₂) = ε₁ − I₁R₁

I₂ (0.5 Ω + 2.5 Ω) = 18 V − (4.75 A) (6.0 Ω)

I₂ = -3.5 A

The current through the top loop is 3.5 A. (The negative sign means it is in the opposite direction of what is drawn.)