Potential Difference and Power in Resistor Circuit

[Gwozdzilla]

Homework Statement

I attached a screenshot of the problem(s) that my professor gave in a practice exam.

Homework Equations

∑I = 0 at each junction
∑V = 0 at each closed loop
P = I²R = V²/R
R_series = R₁ + R₂ +...

The Attempt at a Solution

Part A:
I think the current I in the center, broken line of current, is zero because that wire is broken. Then
I₁+I₂ = 0. So I₁ = I₂ = I

Going around the large loop, starting from the battery at the top:

V + V - IR - IR = 0
2V = 2IR
V=IR
12 = I(10)
I = 1.2

In the bottom loop...

V_a + V - IR + V = V_b
V_a - V_b = -2V + IR
V_a - V_b = -2(12) + (1.2)(10)
V_a - V_b = 12V in absolute value...
Is absolute value what I'm looking for? +12V is the answer in the solutions my professor provided.

Part B:
R_series = R₁ + R₂ +...
R_series = 10 + 10
R_series = 20

P = I²R = V²/R
P = (1.2)²(20) = 28.8W
P = (12)²/(20) = 7.2W

These aren't equal. I tried it with R in parallel as well (R_eq = 5), and it's still wrong. In the answers provided each of the power equations are multiplied by 2 and R = 10. Why is all of this?

Thanks!

 

[ehild]

Homework Statement

I attached a screenshot of the problem(s) that my professor gave in a practice exam.

Homework Equations

∑I = 0 at each junction
∑V = 0 at each closed loop
P = I²R = V²/R
R_series = R₁ + R₂ +...

The Attempt at a Solution

Part A:
I think the current I in the center, broken line of current, is zero because that wire is broken. Then
I₁+I₂ = 0. So I₁ = I₂ = I

Going around the large loop, starting from the battery at the top:

V + V - IR - IR = 0
2V = 2IR
V=IR
12 = I(10)
I = 1.2

In the bottom loop...

V_a + V - IR + V = V_b
V_a - V_b = -2V + IR
V_a - V_b = -2(12) + (1.2)(10)
V_a - V_b = 12V in absolute value...
Is absolute value what I'm looking for? +12V is the answer in the solutions my professor provided.

V_a-V_b=-12 V, The potential increases from a to b by 12 V. Look at your notes, what U_ab means, it might be V_b-V_a.

Part B:
R_series = R₁ + R₂ +...
R_series = 10 + 10
R_series = 20

P = I²R = V²/R
P = (1.2)²(20) = 28.8W
P = (12)²/(20) = 7.2W

The 28.8 V is correct. The voltage is 24 V across the two series resistors, so P=24²/20, the same as before.

ehild