Circuit with two voltage sources (Kirchoff Rules)

[XianForce]

Homework Statement

The circuit shown in the figure below is connected for 3.10 min. (Assume R1 = 7.90 Ω, R2 = 2.60 Ω, and V = 16.0 V.)

The circuit shown in the figure below is connected for 3.10 min. (Assume R1 = 7.90 Ω, R2 = 2.60 Ω, and V = 16.0 V.)

Homework Equations

Kirchoff's Rules
V = IR

The Attempt at a Solution

So naming the current in the left segment I1 downwards, the current in the middle section I2 downwards, and the current in the right most section I3 upwards, I find from the junction rule that I1 + I2 = I3. Then using the current rule, I find that -16 +7.9 * I1 + 2.6 * I3 = 0. Substituting in I3 as I1 + I2, and then solving for I1 I find that I1 = (16 - 2.6I2) / 10.5 . Using the current rule again, I find that -4 + 7.9 * I1 - 6.0 * I2 = 0 and so I substitute for I1 and then solve for I2 to find that I2 = 1.0103. Then use that value back in other equations to find that I1 = 1.2736 Amperes downward, I2 = 1.0103 Amperes downward, and I3 = 2.2839 Amperes Upward.

Web Assign keeps saying these values are wrong and I have reworked these problems several times and keep getting the same answers.

 

[ehild]

Homework Statement

The circuit shown in the figure below is connected for 3.10 min. (Assume R1 = 7.90 Ω, R2 = 2.60 Ω, and V = 16.0 V.)

The circuit shown in the figure below is connected for 3.10 min. (Assume R1 = 7.90 Ω, R2 = 2.60 Ω, and V = 16.0 V.)

Homework Equations

Kirchoff's Rules
V = IR

The Attempt at a Solution

So naming the current in the left segment I1 downwards, the current in the middle section I2 downwards, and the current in the right most section I3 upwards, I find from the junction rule that I1 + I2 = I3. Then using the current rule, I find that -16 +7.9 * I1 + 2.6 * I3 = 0.

R2 is connected to a 3 ohm resistor in series.

ehild

 

[XianForce]

I cannot believe I missed that. Thank you!

 

[ehild]

I cannot believe I missed that. Thank you!

You are welcome.

ehild