Let there be $n$ resistors $R_1 \ldots R_n$ with $R_{max} = \max(R_1 \ldots R_n)$ and $R_{min} = \min\{R_1 \ldots R_n\}$. Show that when they are connected in parallel, the resultant resistance $R_P R_{max}$. Interpret the result physically.

Let there be n resistors R₁ … Rₙ with Rmax = max(R₁ … Rₙ) and Rmin = min\R₁ … Rₙ\. Show that when they are con

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Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

Four resistors are arranged in a diamond. A battery sits across the top and bottom points; a sensitive galvanometer bridges the left and right points. The question is: when does no current flow through the galvanometer? The answer is symmetry.

The Intuition

Treat the bridge as two parallel voltage dividers — two resistors in series on the left, two in series on the right — with the galvanometer joining their midpoints. If both dividers produce the same midpoint voltage, there is no potential difference across the galvanometer and no current flows: the bridge is balanced. This happens when the resistance ratio on the left arm equals the ratio on the right arm.

Note

Symmetry here means equal ratios, not equal resistances. $1\ \Omega$ with $2\ \Omega$ on one side balances $100\ \Omega$ with $200\ \Omega$ on the other.

The Precise Statement

Label the four resistors: $R_1$ top-left, $R_2$ bottom-left, $R_3$ top-right, $R_4$ bottom-right. The battery connects the top junction (between $R_1$ , $R_3$ ) to the bottom junction (between $R_2$ , $R_4$ ); the galvanometer connects the two midpoints. The bridge is balanced when

\frac{R_1}{R_2} = \frac{R_3}{R_4}

Important

The balance condition is independent of the battery voltage and of the galvanometer's resistance — it depends only on the four resistor values.

Why This Matters

Unknown resistance: put the unknown in place of $R_4$ , adjust the others until balanced, then $R_4 = R_3\,\dfrac{R_2}{R_1}$ .
Strain gauges: stretching slightly changes a resistance, unbalancing the bridge; the deflection measures the strain.
Temperature sensors: a temperature-dependent resistor unbalances the bridge in proportion to the change. …

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key result is:

When the bridge is balanced, no current flows through the galvanometer, and:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Let's understand why this is true — not just memorize it.

1. The Circuit Setup

A Wheatstone bridge has four resistors arranged in a diamond:

$R_1$ and $R_2$ in series on the left branch
$R_3$ and $R_4$ in series on the right branch
A galvanometer (sensitive current detector) connects the midpoints of the two branches
A battery connects across the top and bottom

        A
       / \
      /   \
     R1    R3
     |     |
     G-----|   (G = galvanometer)
     |     |
     R2    R4
      \   /
       \ /
        B

2. The Condition for Balance — "No Current Through G"

Balance means the galvanometer shows zero deflection — no current flows through it. This implies points C (between $R_1$ and $R_2$ ) and D (between $R_3$ and $R_4$ ) are at the same electric potential. If $V_C = V_D$ , no potential difference exists across the galvanometer, so no current flows.

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

With no current through G, the left branch carries a single current $I_1$ and the right branch a single current $I_2$ .

Step 2: Equal potentials at the midpoints

From A (common top) to the midpoints, since $V_C = V_D$ :

I_1 R_1 = I_2 R_3 \quad \text{(1)}

From the midpoints to B (common bottom), since $V_C = V_D$ :

I_1 R_2 = I_2 R_4 \quad \text{(2)}

Step 3: Divide (1) by (2)

\frac{I_1 R_1}{I_1 R_2} = \frac{I_2 R_3}{I_2 R_4}

The currents $I_1$ and $I_2$ cancel: …

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Because a parallel combination sums reciprocals, its reciprocal exceeds the largest single reciprocal $1/R_{\min}$ , giving $R_P < R_{\min}$ ; because a series combination sums resistances, its total exceeds the largest single term, giving $R_S > R_{\max}$ .

Parallel: $R_P < R_{\min}$

For $n$ resistors in parallel,

\frac{1}{R_P} = \sum_{i=1}^{n}\frac{1}{R_i} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}.

Among the reciprocals, the smallest resistance $R_{\min}$ gives the largest reciprocal $\dfrac{1}{R_{\min}}$ . The right-hand side contains this term plus the other reciprocals, all of which are positive. Hence the whole sum is strictly greater than $\dfrac{1}{R_{\min}}$ alone:

\frac{1}{R_P} = \frac{1}{R_{\min}} + \underbrace{\sum_{R_i\neq R_{\min}}\frac{1}{R_i}}_{>0} > \frac{1}{R_{\min}}.

Since both sides are positive, taking reciprocals reverses the inequality:

R_P < R_{\min}.

Watch out

Do not write $\tfrac{1}{R_i} > \tfrac{1}{R_{\min}}$ — that is false, because $R_{\min}$ gives the largest reciprocal. The correct argument keeps the $\tfrac{1}{R_{\min}}$ term and adds the other positive terms to it. (Check: $1\ \Omega \parallel 1000\ \Omega$ gives $R_P \approx 0.999\ \Omega < 1\ \Omega = R_{\min}$ , not $< R_{\min}/n$ .)

Series: $R_S > R_{\max}$

For $n$ resistors in series,

R_S = \sum_{i=1}^{n} R_i = R_1 + R_2 + \cdots + R_n.

This sum contains $R_{\max}$ together with the remaining resistances, each positive:

R_S = R_{\max} + \underbrace{\sum_{R_i\neq R_{\max}} R_i}_{>0} > R_{\max}.

Physical interpretation …

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Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

The Intuition

The Precise Statement

Why This Matters

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

1. The Circuit Setup

2. The Condition for Balance — "No Current Through G"

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

Step 2: Equal potentials at the midpoints

Step 3: Divide (1) by (2)

Parallel: $R_P < R_{\min}$

Series: $R_S > R_{\max}$

Physical interpretation …

Method: Proving Bounds on Series and Parallel Resistance Combinations

Steps

Parallel: RP<Rmin⁡R_P < R_{\min}RP​<Rmin​

Series: RS>Rmax⁡R_S > R_{\max}RS​>Rmax​

Physical interpretation …

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Parallel: $R_P < R_{\min}$

Series: $R_S > R_{\max}$

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