Electric Circuit using Kirchoff's rules

[Jakarto]

Homework Statement

We are given that R1 = 15.0 Ω, R2 = 10.0 Ω, and R3 = 8.00 Ω. What is the current and direction of the current through each resistor? Again, I have no idea how to do this, I got my answer from chegg, but I don't know if this is correct.

Homework Equations

V = IR[/B]

The Attempt at a Solution

Using Kirchoff's rules , for the loop ADCBA, starting from D, we have:
(-10ohm * i1) + (10V) - 15ohm(i1+i2) +(20V) = 0[/B]
(25ohm * i1) + (15ohm * i2) = 30V
(5 * i1) +(3 * i2) = 6 to the -(i) direction ?

Now, applying Kirchoff to DCFED, starting from D:
(-10ohm * i1) + (10V) + (8ohm * i2) = 0
(10ohm * i1) - (8ohm * i2) = 10V
(5 * i1) - (4 * i2) = 5V to the -(ii) direction ?
After solving, we get:
i1 = 1.14A for R2?
i2 = 0.142A for R3?
i3 = 1.256A for R1?

Are the currents and directions correct? If not, can you tell me what the direction is at least?

 

[axmls]

Have you tried reading your textbook or a textbook? This question should be one of the first things mentioned.

 

[CWatters]

Your equations look correct but after solving I get very slightly different answers...

I₁ = 1.114A
I₂ = 0.143A
I₃ = 1.257A

However...

Are the currents and directions correct? If not, can you tell me what the direction is at least?

Before you apply KVL/KCL it's essential to mark up the circuit with the currents I₁ and I₂ including arrows to define what you mean by +ve current flow. Then, after applying KVL/KCL and solving, if I₁ or I₂ turns out to be +ve or -ve you know which direction that is by referring to your drawing.

Note: You need BOTH the drawing and the calculated currents for a valid answer. That's because the direction you choose for +ve current is arbitrary. Your classmates may have chosen to define a different direction as +ve, in which case they would have got -ve answers for the currents. That would be an equally valid answer. The only time it's not an arbitrary choice is when the problem statement tells you to make a particular choice.

 

[Jakarto]

Okay, thanks again CWaters!