Electrical circuit with resistors and current flow values.

Looking at the circuit:

We have a 12V battery connected to three resistors: R1 = 1Ω, R2 = 2Ω, and R3 = 2Ω.

Currents are labeled:
- 6A flows from the battery into the top node (before splitting).
- At that node, current splits: some goes through R2, some through R3.
- The current going down through R3 is labeled as 3A.
- The red “X” is placed on the wire after R2 — this likely indicates we need to find the current at point X.

Let’s use Kirchhoff’s Current Law (KCL) at the junction where the 6A enters.

At that junction:
Total incoming current = 6A
This splits into two branches: one through R2, one through R3.

We’re told the current through R3 is 3A.

So, by KCL:
Current through R2 + Current through R3 = Total current entering
→ I_R2 + 3A = 6A
→ I_R2 = 3A

Now, look at point X — it’s on the wire *after* R2, meaning all the current that went through R2 continues through X.

Therefore, current at X = current through R2 = 3A.

Wait — let’s double-check with voltage drops to make sure everything adds up.

Voltage across R3: V = I × R = 3A × 2Ω = 6V
Voltage across R2: V = 3A × 2Ω = 6V
Since R2 and R3 are in parallel, they should have same voltage drop — yes, both 6V. Good.

Then, the remaining voltage must be dropped across R1.

Total voltage from battery = 12V
Voltage used in parallel branch (R2 or R3) = 6V
So voltage across R1 = 12V - 6V = 6V

Current through R1: I = V/R = 6V / 1Ω = 6A — which matches the total current coming out of the battery. Perfect.

Also, note: the 6A from the battery goes through R1 first? Wait — looking again at the diagram...

Actually, re-examining the diagram carefully:

The battery positive terminal connects to R1 first? No — wait, the diagram shows:

Battery (+) → then splits? Actually, no — let's trace:

From battery (+), wire goes right to R1, then from R1 it goes down to join the other side of R2 and R3? That doesn’t match the labels.

Wait — actually, looking again:

The 6A arrow is pointing downward on the left vertical wire — so that means 6A is flowing down the left side.

That suggests the battery is on top, and current leaves the positive terminal, goes left, then down the left wire (labeled 6A), then splits at the bottom-left corner? But there’s a resistor R2 horizontally in the middle.

Actually, standard interpretation:

The battery is at the top. Positive terminal on left, negative on right? Or vice versa? The symbol shows long line on left — that’s positive. So current flows out of left (positive), goes down the left wire (6A labeled), then at the junction below, it splits: part goes right through R2, part goes down through R3.

Then, after R2 and R3, they rejoin, go up through R1? But R1 is on the right side, vertically.

Actually, tracing the loop:

Start at battery (+):
→ Left wire down (6A)
→ At bottom-left junction: splits to R2 (right) and R3 (down then right)
→ After R2 and R3, they meet at bottom-right
→ Then up through R1? But R1 is drawn vertically on the right, connecting back to battery (-)?

Wait — the battery has two terminals: left is +, right is -.

Wire from + goes left, then down (6A label).

At the bottom-left node: splits to R2 (going right) and R3 (going down then right).

After R2: reaches a node (where X is marked) — then continues to connect to the bottom of R1? And R3 also connects to that same node.

Then from that bottom-right node, current goes up through R1 to the battery’s negative terminal.

So R2 and R3 are in parallel, and their combination is in series with R1.

Yes — that makes sense.

So total current from battery: 6A — this goes through R1? No — if R1 is on the return path, then yes.

Actually, current path:

Battery (+) → down left wire (6A) → splits at bottom-left:
- Branch 1: through R2 (2Ω) → to bottom-right node
- Branch 2: through R3 (2Ω) → to bottom-right node
→ Both branches combine at bottom-right node → then up through R1 (1Ω) → back to battery (-)

So R1 is in series with the parallel combination of R2 and R3.

Total current from battery is 6A — this same 6A must flow through R1, because it’s in series with the whole parallel block.

But earlier I calculated voltage across R1 as 6V, giving 6A — consistent.

And since R2 and R3 are identical (both 2Ω), and in parallel, current splits equally: 3A each.

Given that, the current through R2 is 3A, and point X is on the wire immediately after R2 — so current at X is still 3A (no branching after R2 until it meets R3’s output).

Thus, answer is 3A.

Final Answer:
3A