Is This a Series or Parallel Circuit?

  • #1
bmarc92
8
0
Thread moved from the technical forums to the schoolwork forums
Homework Statement
If the potential from each voltage source is 10V and the resistance of each resistor is 5Ohms, what is the current exiting point A into R3?
Relevant Equations
Loop rule: \oint {\vec{E}} \cdot \left{\vec{dl}} = 0

Junction rule: \sum{I_{in}} = \sum{I_{out}}
Consider the problem below

1716149859420.png


I used Kirchoff's loop/junction rules to correctly arrive at answer C (1.3A). Before using Kirchoff's rules I tried to get ##V_3## by treating this as a Fermi problem using the following lines of logic, but there is a flaw somewhere, can someone identify the flaw?


(a) If ##V_1 = V_2##, we can assume no current from either power source branches at junction A and everything exits through ##R_3##.

(b) If there is no current branching, we can treat each loop independently as a series circuit.

(c) If I can treat each loop independently as a series circuit,

$$I_I = \frac{V_1=10}{(R_1=5) + (R_3=5)}$$
$$I_2 = \frac{V_2=10}{(R_2=5) + (R_3=5)}$$

(d) ##I_3 = I_1 + I_2##

(e) ##I_3 = 2##


(c) & (d) & (e) follows from (b), implying each loop cannot be treated independently. But why must this circuit be treated as parallel if there is no current branching?
 
Last edited:
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  • #2
The question asks for the current, not the voltage.

If the two voltage sources are identical, and share the same common node, then they can be joined in parallel, and one eliminated.
Then R1 and R2 are in parallel, and in series with R3.
The resistance of the circuit is then known, (5/2 + 5) and so is the current.
 
  • #3
bmarc92 said:
(b) If there is no current branching, we can treat each loop independently as a series circuit.
No, because the conductive capacity of ##R_3## is shared in some way between those two loops. Given the symmetry, it is shared equally, so in each loop it provides resistance ##2R_3##.
 
  • #4
If the correct answer is; (c). 1.3 C/s (33%)
What does 1.3 C/s mean?
What does (33%) mean?
 
  • #5
bmarc92 said:
But why must this circuit be treated as parallel if there is no current branching?
Why is branching the criterion you're using? The currents "join" at point A. Isn't that just branching in reverse?
 
  • #6
Baluncore said:
What does (33%) mean?
The fraction of responders who chose that answer?
 
  • #7
Mister T said:
The fraction of responders who chose that answer?
Then 2% did not tick any box.
 
  • #8
Baluncore said:
What does 1.3 C/s mean?
Coulombs/sec?
 
  • #9
haruspex said:
Coulombs/sec?
Coulombs/sec = amp.
That seems likely, but the answer is still a little bit wrong.
R = 5/2 + 5 = 15/2 ohms.
I = 10 * 2/15 = 20/15 = 4/3 amp ≠ 1.3 amp
 
  • #10
Baluncore said:
Then 2% did not tick any box.
Not surprising. I guess "responder" means student that started the assignment.
 
  • #11
Baluncore said:
Coulombs/sec = amp.
That seems likely, but the answer is still a little bit wrong.
R = 5/2 + 5 = 15/2 ohms.
I = 10 * 2/15 = 20/15 = 4/3 amp ≠ 1.3 amp
I thought you were asking the OP so you could verify that he understood the units are not volts.

C/s is the correct combination of symbols according to the SI.

##4/3 \approx 1.3##, correct to 2 sig figs.

Edit: Or rounded to one decimal place.
 
  • #12
Mister T said:
C/s is the correct combination of symbols according to the SI.
The ampere was a base unit in the SI.
The coulomb was a derived unit. C =A⋅s
C/s = A⋅s / s = A
When did things change?
 
  • #14
Baluncore said:
The ampere was a base unit in the SI.
The coulomb was a derived unit. C =A⋅s
C/s = A⋅s / s = A
When did things change?
They didn't. I should have C/s is a correct combination of SI symbols. The symbol A is also correct.
 

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