A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistanc

Q: Why are the two paths considered to be in parallel?

Two circuit elements are in parallel if they are connected across the exact same two nodes. Here, both the short path EC and the long path EDABC start at node E and end at node C, meaning they share the same potential difference.

Q: How do we know the resistance of just a piece of the wire?

The problem states it's a uniform wire. For a uniform wire, resistance is directly proportional to length ($R = \rho L / A$). If the whole wire of length $4a$ has resistance $R$, a piece of length $L$ will have resistance $R \times (L / 4a)$.

Q: What if E was not the midpoint, but say, one-third of the way along CD?

The logic remains exactly the same! You would just calculate the new lengths. The short path would be $a/3$ (or $2a/3$), and the long path would be the remaining length. You'd find the new $R_1$ and $R_2$ and use the parallel formula again.

Q: Why does the denominator in the parallel formula just become R?

The denominator in the parallel formula is $R_1 + R_2$. Since $R_1$ and $R_2$ represent the two pieces that make up the entire square wire, their sum must naturally equal the total resistance of the wire, which is $R$.

Q: Can we use symmetry to solve this faster?

Symmetry is incredibly powerful, but it doesn't apply directly here because the two paths from E to C are not identical (one is much longer than the other). Symmetry works best when finding resistance across diagonally opposite corners (like A and C) or adjacent corners (like A and B).

Question

A wire of resistance $R$ is bent to form a square $ABCD$ as shown in the figure. The effective resistance between $E$ and $C$ is [$E$ is mid-point of arm $CD$]

Options:

(A) $\frac{7}{64} R$

(B) $\frac{3}{4} R$

(C) R

(D) $\frac{1}{16} R$

Sigma Percentile · Accepted Answer

This problem is a classic example of how geometric shapes can be translated into simple electrical circuits. Let's break down the thought process step-by-step to find the effective resistance between points $E$ and $C$.

### Analyzing the Setup
We are given a uniform wire of total resistance $R$ bent into a square $ABCD$. The point $E$ is the midpoint of the side $CD$. We need to find the equivalent resistance between $E$ and $C$.

Because the wire is uniform, its resistance is directly proportional to its length ($R \propto L$).

Let's assume the length of each side of the square is $a$. The total length of the wire is $4a$. Therefore, the resistance per unit length of the wire is:
$$\lambda = \frac{R}{4a}$$

### Identifying the Parallel Paths
If we connect a voltage source across points $E$ and $C$, the current entering at $E$ has two distinct paths to reach $C$:
1. **The Direct Path ($EC$):** This is the shortest route along the bottom edge.
2. **The Longer Path ($EDABC$):** This route goes around the rest of the square.

Since these two paths are connected across the same two points ($E$ and $C$), they are in **parallel**.

### Calculating Individual Resistances
Now, let's find the resistance of each path by multiplying their lengths by the resistance per unit length.

**Path 1 ($EC$):**
Since $E$ is the midpoint of $CD$, the length of $EC$ is $\frac{a}{2}$.
$$R_1 = \lambda 	imes L_{EC} = \left(\frac{R}{4a}ight) 	imes \frac{a}{2} = \frac{R}{8}$$

**Path 2 ($EDABC$):**
The length of this path includes the other half of side $CD$ (which is $ED = \frac{a}{2}$), plus the three full sides $DA$, $AB$, and $BC$.
$$L_{EDABC} = \frac{a}{2} + a + a + a = \frac{7a}{2}$$
$$R_2 = \lambda 	imes L_{EDABC} = \left(\frac{R}{4a}ight) 	imes \frac{7a}{2} = \frac{7R}{8}$$

Notice a quick sanity check: $R_1 + R_2 = \frac{R}{8} + \frac{7R}{8} = R$, which is the total resistance of the wire. Perfect!

### Final Calculation
Since $R_1$ and $R_2$ are in parallel, the equivalent resistance $R_{eq}$ is given by the standard parallel formula:
$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

Substitute the values we found:
$$R_{eq} = \frac{\left(\frac{R}{8}ight) \left(\frac{7R}{8}ight)}{\frac{R}{8} + \frac{7R}{8}}$$

The denominator simplifies beautifully to $R$:
$$R_{eq} = \frac{\frac{7R^2}{64}}{R} = \frac{7}{64} R$$

And there we have it! The effective resistance between $E$ and $C$ is $\frac{7}{64} R$.

Animated Solution for Physics - Current Electricity: A wire of resistance $R$ is bent to form a square $A B C D$ as shown in the figure. The effective resistance between $E$ and $C$ is [ $E$ is mid-point of arm $C D$ ]

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Resistance and Length}

\text{Identifying Parallel Paths}

\text{Resistance of Path EC } (R_1)

\text{Resistance of Path EDABC } (R_2)

\text{Equivalent Resistance Formula}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Combination of Resistors

Analyzing the Setup We are given a uniform wire of total resistance $R$ bent into a square $A B C D$

Identifying the Parallel Paths

Calculating Individual Resistances

Final Calculation

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