The value of current i_1 flowing from A to C in the circuit diagram is
(a) 1 A
(b) 5 A
(c) 2 A
(d) 4

Q: Why can we just remove the 5 ohm resistors? Isn't that changing the circuit?

We aren't changing the physics of the circuit; we are just simplifying our model of it. Because the bridge is perfectly balanced, the nodes connected by the $5\,\Omega$ resistors are at the exact same potential. Since current requires a potential difference to flow, zero current passes through them. A resistor with no current has no effect on the rest of the circuit, so we can safely ignore it.

Q: What if the battery was connected between nodes B and D instead of A and C?

That would completely change the problem! The symmetry would be entirely different, and the $5\,\Omega$ resistors would absolutely carry current. You would likely have to use Kirchhoff's laws or Star-Delta transformations to solve it.

Q: Do I need to calculate the equivalent resistance of the entire circuit to find the total current first?

Not at all! Because the three branches are connected in parallel directly across the battery, the voltage across each individual branch is independent of the others. You only need to look at the middle branch to find $i_1$.

Q: How do I know the voltage across the middle branch is exactly 8V?

Look at the nodes. The battery is connected directly to node A and node C. The middle branch also starts exactly at node A and ends exactly at node C. Therefore, it must experience the full $8\text{ V}$ potential difference.

Q: What if the middle resistors were 4 ohms and 8 ohms instead?

Then the symmetry would be broken for that specific branch. While the top and bottom branches might still have a $1:1$ ratio, the center node would have a different potential than nodes B and D. Consequently, current would flow through the $5\,\Omega$ resistors, and you could no longer ignore them.

Question

The value of current $i_1$ flowing from $A$ to $C$ in the circuit diagram is

Options:

(A) 1 A

(B) 5 A

(C) 2 A

(D) 4 A

Sigma Percentile · Accepted Answer

### Demystifying the Complex Bridge: A Lesson in Symmetry

At first glance, the circuit diagram presented in this problem looks like an absolute nightmare. With multiple interconnected branches, vertical cross-resistors, and a battery spanning the entire structure, it seems like a prime candidate for a long, tedious application of Kirchhoff's Voltage and Current Laws.

However, in the world of competitive physics, whenever you see a highly structured, grid-like circuit, your first instinct should always be to look for **symmetry**.

### The Power of Symmetry

Let's break down the circuit by analyzing the three horizontal paths connecting node $A$ to node $C$:
1.  **The Upper Branch:** Contains two resistors of $2\,\Omega$ each.
2.  **The Middle Branch:** Contains two resistors of $4\,\Omega$ each.
3.  **The Lower Branch:** Contains two resistors of $2\,\Omega$ each.

Notice a pattern? In every single branch, the ratio of the left-side resistance to the right-side resistance is exactly $1:1$.

Because these ratios are identical across all parallel paths, this circuit is essentially a multi-layered, perfectly balanced **Wheatstone Bridge**.

### The Dummy Resistors

What happens in a balanced Wheatstone bridge? The intermediate nodes—in this case, node $B$ on the top, the center node $O$ in the middle, and node $D$ on the bottom—will all share the exact same electrical potential.

$$V_B = V_O = V_D$$

Current only flows when there is a potential difference. Since there is zero potential difference between these nodes, absolutely no current will flow through the two $5\,\Omega$ vertical resistors connecting them. Electrically speaking, these resistors are "invisible" or "dummy" resistors. They exist purely to intimidate you.

### The Final Calculation

Once we confidently remove the $5\,\Omega$ resistors from our mental model, the circuit simplifies drastically. We are left with three independent parallel branches connected directly across the $8	ext{ V}$ battery.

The question specifically asks for the current $i_1$ flowing through the middle branch. Because the branches are in parallel, the middle branch experiences the full $8	ext{ V}$ potential difference provided by the battery.

The total resistance of this middle branch is simply the sum of its two series resistors:

$$R_{	ext{middle}} = 4\,\Omega + 4\,\Omega = 8\,\Omega$$

Finally, we apply Ohm's Law to find the current $i_1$:

$$i_1 = \frac{V}{R_{	ext{middle}}} = \frac{8	ext{ V}}{8\,\Omega} = 1	ext{ A}$$

By recognizing the symmetry, a problem that looked like a 10-minute mathematical grind was solved conceptually in under a minute!

Animated Solution for Physics - Current Electricity: The value of current $i_{1}$ flowing from $A$ to $C$ in the circuit diagram is

Select Answer:

Visualized Solution

$V = 8 V$

$\frac{R _{left}}{R _{right}} = \frac{2}{2} = \frac{4}{4} = 1$

$V_{B} = V_{O} = V_{D}$

$Δ V = 0 ⟹ I = 0$

$V_{branch} = 8 V$

$R_{middle} = 4 + 4 = 8 Ω$

$i_{1} = \frac{V}{R} = \frac{8}{8} = 1 A$

The Sigma Insight: Kirchhoff's Laws

Demystifying the Complex Bridge

The Power of Symmetry

The Dummy Resistors

The Final Calculation

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Animated Solution for Physics - Current Electricity: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of current i1​ flowing from A to C in the circuit diagram is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=8 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rright​Rleft​​=22​=44​=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VB​=VO​=VD​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔV=0⟹I=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vbranch​=8 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rmiddle​=4+4=8Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i1​=RV​=88​=1 A