In the given circuit, it is observed that the current I is independent of the value of the resistanc

Q: Why does $I$ being independent of $R_6$ mean no current flows through $R_6$?

If current flowed through $R_6$, changing its value would change the total equivalent resistance of the circuit. Since $I = V/R_{eq}$, changing $R_6$ would change $I$. The only way $I$ remains constant for *any* value of $R_6$ is if $R_6$ is effectively not part of the circuit, which happens when no current flows through it.

Q: What is a Wheatstone bridge?

A Wheatstone bridge is a specific circuit configuration used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. In this problem, $R_1, R_2, R_3$, and $R_4$ form the four arms of the bridge, and $R_6$ is the central galvanometer arm.

Q: How do we know the bridge is balanced?

The problem states that the total current is independent of $R_6$. As established, this means no current flows through $R_6$. The only physical condition that results in zero current through the central arm of a Wheatstone bridge is when the bridge is balanced, meaning the potential difference across $R_6$ is zero.

Q: Why is the condition for balance $\frac{R_1}{R_2} = \frac{R_3}{R_4}$?

For no current to flow through $R_6$, the potential at the junction of $R_1$ and $R_2$ must equal the potential at the junction of $R_3$ and $R_4$. This implies the voltage drop across $R_1$ must equal the voltage drop across $R_3$. Using Ohm's law and voltage divider principles, this leads directly to the ratio $\frac{R_1}{R_2} = \frac{R_3}{R_4}$.

Q: What if the battery was connected differently?

The balance condition depends strictly on the relative positions of the resistors. If the battery was connected across the nodes where $R_6$ is currently connected, and $R_6$ was moved to where the battery is, the new balance condition would be $\frac{R_1}{R_3} = \frac{R_2}{R_4}$, which mathematically is exactly the same!

Question

In the given circuit, it is observed that the current $I$ is independent of the value of the resistance $R_6$. Then, the resistance values must satisfy

Options:

(A) $R_1 R_2 R_5 = R_3 R_4 R_6$

(B) $\frac{1}{R_5} + \frac{1}{R_6} = \frac{1}{R_1 + R_2} + \frac{1}{R_3 + R_4}$

(C) $R_1 R_4 = R_2 R_3$

(D) $R_1 R_3 = R_2 R_4$

Sigma Percentile · Accepted Answer

### The Mystery of the Invisible Resistor: Mastering the Balanced Wheatstone Bridge

Imagine you are an electrical engineer tasked with designing a complex circuit. You place a resistor right in the middle of a network, but to your absolute astonishment, changing its value does absolutely nothing to the total current drawn from the battery. It is as if the resistor is completely invisible to the rest of the circuit!

This is not magic; it is the elegant physics of the **Wheatstone Bridge**. Let's dive deep into this fascinating problem and uncover the mathematical beauty that makes a resistor "disappear."

### Analyzing the Setup

Let's carefully trace the path of the current in our given circuit diagram. The battery acts as the pump, pushing a total current $I$ out of its positive terminal. This current first encounters the resistor $R_5$.

After passing through $R_5$, the current reaches a junction and splits into two parallel branches. The left branch consists of $R_1$ and $R_2$ in series, while the right branch consists of $R_3$ and $R_4$ in series. However, there is a twist! A central resistor, $R_6$, bridges the gap between the midpoint of the left branch and the midpoint of the right branch.

This specific diamond-like configuration of five resistors ($R_1, R_2, R_3, R_4$, and $R_6$) is universally known as a Wheatstone bridge. The total current $I$ supplied by the battery is governed by Ohm's Law:

$$I = \frac{V}{R_{eq}}$$

Here, $V$ is the voltage of the battery, and $R_{eq}$ is the total equivalent resistance of the entire circuit, which is simply $R_5$ plus the equivalent resistance of the bridge network.

### The Core Concept: Independence

The problem presents us with a crucial piece of information: the total current $I$ is **independent** of the value of $R_6$.

What does this physically mean? If we were to swap $R_6$ from a tiny $1 \ \Omega$ resistor to a massive $1,000,000 \ \Omega$ resistor, the current $I$ would not flinch. Since $V$ is constant, the only way $I$ remains constant is if the total equivalent resistance $R_{eq}$ remains completely unchanged.

Now, ask yourself: under what condition does a component in a circuit have absolutely zero effect on the total resistance?

The answer is profoundly simple: **when no current flows through it.**

If zero current passes through $R_6$, it acts exactly like an open switch. Whether it is there or not, the electrons simply ignore that path. Therefore, for $I$ to be independent of $R_6$, the current through $R_6$ must be strictly zero.

### The Master Equation: The Balance Condition

We have established that no current flows through the central arm $R_6$. But why would electrons choose to ignore a perfectly good conductive path?

Current only flows when there is a difference in electrical potential (voltage) between two points. Think of it like water; water only flows from a higher elevation to a lower elevation. If two pools are at the exact same height, no water flows between them through a connecting pipe.

Similarly, for zero current to flow through $R_6$, the electrical potential at the junction between $R_1$ and $R_2$ must be exactly equal to the electrical potential at the junction between $R_3$ and $R_4$.

When this magical state of equal potentials is achieved, we say the Wheatstone bridge is **balanced**.

Mathematically, this equal potential condition dictates that the voltage drop across $R_1$ must equal the voltage drop across $R_3$. Because the current splits and flows down the two parallel branches, this voltage equality forces the ratio of the resistances in the left arm to perfectly match the ratio of the resistances in the right arm.

This gives us our master equation:

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

### Final Calculation

We are now at the finish line. We have derived the fundamental condition for a balanced Wheatstone bridge based purely on the physical constraint provided in the problem.

To match our result with the given multiple-choice options, we simply perform a quick cross-multiplication on our master equation:

$$R_1 R_4 = R_2 R_3$$

Looking at the choices, this perfectly matches option (c).

By understanding the physical intuition behind a balanced bridge, we bypassed the need for complex Kirchhoff's loop equations or messy Star-Delta transformations. We saw right through the circuit's trick, proving once again that a strong conceptual foundation is the ultimate tool in physics!

Animated Solution for Physics - Current Electricity: In the given circuit, it is observed that the current $I$ is independent of the value of the resistance $R_{6}$ . Then, the resistance values must satisfy

Select Answer:

Visualized Solution

$R_{e q} = R_{5} + R_{b r i d g e}$

$I = \frac{V}{R _{e q}}$

$I_{R_{6}} = 0$

$\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}}$

$R_{1} R_{4} = R_{2} R_{3}$

$Unbalanced Bridge$

The Sigma Insight: Wheatstone Bridge

The Mystery of the Invisible Resistor

Analyzing the Setup

The Core Concept

The Master Equation

Final Calculation

Similar Questions

Which of the following statements is false?