Kirchoff's Loop Rule Question

Question

https://imgur.com/a/wtE03TQ Diagram in link

Answer 1

Part (a)

Using Kirchhoff's Current Law, we know that the sum of all currents flowing into a node equal the sum of currents flowing out of the node. Looking at the top-center node, I can see this relationship:

I₂ + I₃ = I₁

I₁ = 2.20A (given)

I₂ = ?

I₃ = ?

To find I₂, we're told that R₂ has a potential difference of 5.10V and its resistance is 5.00 Ohms. Using Ohm's law we can find the current to be:

I₂ = V_R2 / R₂ = 5.10V / 5.00 Ohms = 1.02A

To find I₃, rearrange the equation:

I₃ = I₁ - I₂ = 2.20A - 1.02A = 1.18A

Part (b)

Always start a Kirchhoff loop at the positive terminal of your battery, then make its way around that same point. A few things to note when looping:

1. If you cross a power source with the negative terminal first, the voltage will be positive. If you encounter the positive terminal first, it will be negative.
2. If you cross R in the direction of the current, voltage will be negative. If you cross R opposite the direction of the current, it will be positive.
3. Kirchhoff's Voltage Law: The change in voltage in a closed loop is always 0.

Starting from positive terminal of E₁:

-I₁R₁ - I₁r₁ - I₃R₃ + E₃ + E₁ = 0

E₃ = I₁R₁ + I₁r₁ + I₃R₃ - E₁

E₃ = (2.20A)(3.00 Ohms) + (2.20A)(1.50 Ohms) + (1.18A)(3.50 Ohms) - 5.00V

E₃ = 9.03V

Part (c)

Starting from positive terminal of E₂:

-I₂R₂ - E₃ + I₃R₃ - I₂r₂ + E₂ = 0

r₂ = (-I₂R₂ - E₃ + I₃R₃ + E₂) / I₂

r₂ = [-(1.02A)(5.00 Ohms) - 9.03V + (1.18A)(3.50 Ohms) + 12.00V] / 1.02A

r₂ = 1.96 Ohms

Answer 2