The current flowing through the $$1\Omega$$ resistor is $$\frac{n}{10}$$ A. The value of $$n$$ is ________

step 1: identify key nodes

Let node C be reference (0 V).

We want current through the 1 Ω resistor (between left junction and C).

step 2: simplify structure

Notice:

• B-C = 2 Ω
• D-C = 2 Ω
• A connects to B and D via 4 Ω each
• C connects to E (5 V source)
• A to C branch has: 10 V source + 1 Ω

step 3: use symmetry

The top and bottom halves (via 4Ω-2Ω and 4Ω-2Ω) are symmetric → so B and D are at same potential.

Thus currents through those two paths are equal.

So equivalent of those two paths from A to C:

Each path: 4 + 2 = 6 Ω

Two identical 6 Ω in parallel → equivalent = 3 Ω

step 4: reduced circuit

Now between A and C we have:

• one branch: 10 V source in series with 1 Ω
• another branch: 3 Ω

And C to E is just a wire with 5 V source setting potential.

step 5: set node voltages

Let C = 0 V
Then E = 5 V

Let $$A=V_A$$

step 6: currents from A to C

Branch 1 (via 3Ω):
$$I₁=V_A/3$$

Branch 2 (via 1Ω + 10V source):

Careful with polarity: from diagram, going A → C, the 10 V source gives a drop.

So:
$$I₂=(V_A−10)/1=V_A−10$$

step 7: KCL at node A

Total current leaving A = 0:

$$V_A/3+(V_A−10)=0$$

Multiply by 3:

$$V_A+3V_A−30=0$$
$$4V_A=30$$
$$V_A=7.5V$$

step 8: current through 1Ω

$$I=V_A−10=7.5−10=−2.5A$$

Magnitude = 2.5 A

step 9: match form

I = n/10

2.5 = n/10
n = 25