# Current through combination of parallel and series resistors

I have created a simple example in order to understand what is going on with the current through combination of resistors (parallel and in series) in a electrical circuit. I ask my self to calculate the current through resistor R3.

R1 is in parallel with R2 (R1||R2) and R2 in series with R3.So $$R_{eq} = \frac{5*4}{5+4}+R3 = \frac{20}{9}+8=\frac{92}{9} \approx 10.222$$.

Now the current that flows through R1 is (I use the current division here):

$$i_{1} = \frac{i}{R_1}\left(\frac{R1*R2}{R1+R2}\right) = i\left(\frac{R2}{R1+R2}\right) =\frac{16}{3}=5.333$$

for i_2

$$i_{2} = \frac{i}{R_2}\left(\frac{R1*R2}{R1+R2}\right) = i\left(\frac{R1}{R1+R2}\right) =\frac{60}{9}=6.666$$

Both currents i_1 and i_2 add up to the current that the source is giving to the circuit.And now because R3 is in series with R2 the current is the same for R3. Am I correct here? Or I am missing something?

Edit

• All 3 are in parallel. Label your junctions and all 3 resistors are connected to both junctions. Commented May 14, 2023 at 18:35
• @StainlessSteelRat I made an edit in the picture in OR but explain me why R3 is in parallel? I would appreciate it. Commented May 14, 2023 at 18:45
• @HomerJaySimpson They are still all in parallel. Rotate and move R3 next to R2 and you'll see. All three resistors connect one terminal to one node, and other terminal to the orher node. There are only two nodes. DEFGH are same node and ABC are same node. If you think R2 is in series with R3, you should explain why you think they are, because they aren't. Commented May 14, 2023 at 18:47
• If a wire connects the junctions, they are the same junction. A junction has 3 or more wires. A, B and C are the same junction. Commented May 14, 2023 at 18:48
• The current's going from A to H. It can go through A, B, R1, G, H. And A, B, C, R2, F, G, H. And A, B, C, R3, D, E, F, G, H. It never goes through two resistors, so none of them can be in series. Commented May 14, 2023 at 18:51

R1 is in parallel with R2 (R1||R2) and R2 in series with R3

No, that is incorrect; all three are in parallel.

Am I correct here? Or I am missing something?

The latter.

Learn to redraw schematics:

simulate this circuit – Schematic created using CircuitLab

That's all it is that you have.

First, get the least common multiple of those numbers, which is 40. Use that and replace all those resistors with their whole-number conductances: $$\S_1=\frac{40}5=8\$$, $$\S_2=\frac{40}4=10\$$, and $$\S_3=\frac{40}8=5\$$. Sum those up to get $$\S=23\$$. Then the proportion of $$\I_1\$$ that each gets is $$\\frac{S_1}{S}=\frac8{23}\$$, $$\\frac{S_2}{S}=\frac{10}{23}\$$, and $$\\frac{S_3}{S}=\frac5{23}\$$.

You could use nodal and solve for the unknown voltage of that top wire and then compute the current in $$\R_3\$$, too.

(The Rank-Nullity Theorem explains exactly why there's no need to connect the other side of $$\I_1\$$ to anything and why there's no need to separately do KCL for the ground node. In fact, the mathematics won't work if you do write KCL for ground. But that's another topic.)