The ratio of the equivalent resistance of the network (shown in figure) between the points a and b w

Q: Why are the top resistors in series when the switch is open?

When the switch is open, the path connecting the midpoints is broken. Therefore, any current that enters the top resistor $R$ has nowhere else to go but straight through the next resistor $2R$. Since the exact same current flows through both without splitting, they are in series.

Q: Why does closing the switch change the series combination to parallel?

Closing the switch connects the node between the top resistors to the node between the bottom resistors with a zero-resistance wire. This forces both nodes to be at the exact same electrical potential. Because the left resistors ($R$ and $2R$) now share the same starting potential (at $a$) and the same ending potential (at the switch), they are in parallel. The same logic applies to the right resistors.

Q: Can current actually flow through the closed switch?

Yes, absolutely! Because the bridge is unbalanced (the ratio of the left resistors is $1/2$, while the right is $2/1$), the natural potentials at the midpoints would be different if they weren't connected. The closed switch provides a path for current to flow between them to equalize that potential, effectively splitting the circuit into two parallel halves.

Q: What if the switch had a resistance of its own?

If the switch had resistance, the circuit would become a Wheatstone bridge. Because the ratio of the arms ($R/2R \neq 2R/R$) is not equal, it would be an unbalanced bridge. You could no longer use simple series/parallel formulas and would need to use Kirchhoff's laws or a Star-Delta transformation.

Q: Why do we use the product over sum rule for parallel resistors?

The product over sum rule ($R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$) is just a mathematically simplified version of the general parallel resistance formula $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$. It is incredibly handy and saves time when you are calculating the equivalent resistance of exactly two parallel resistors.

Question

The ratio of the equivalent resistance of the network (shown in figure) between the points $a$ and $b$ when switch is open and switch is closed is $x : 8$. The value of $x$ is

Sigma Percentile · Accepted Answer

Welcome to a classic puzzle of current electricity! This problem beautifully illustrates how a simple switch can completely alter the topology of a circuit, transforming series connections into parallel ones and vice versa. Let's break it down step by step.

### Analyzing the Setup

We are given a circuit with two main branches connected between terminals $a$ and $b$. The top branch consists of a resistor $R$ followed by a resistor $2R$. The bottom branch consists of a resistor $2R$ followed by a resistor $R$. Right in the middle, connecting the junction of the top resistors to the junction of the bottom resistors, is a switch $S$. We need to find the equivalent resistance in two distinct scenarios: when the switch is open, and when it is closed.

### Case 1: The Open Switch

When the switch $S$ is open, it acts like a broken bridge. No current can flow through that middle path. This forces the current entering at terminal $a$ to split into two independent paths: the top branch and the bottom branch.

Because there is no alternative route at the midpoints, the resistors in the top branch are strictly in series.

$$R_{top} = R + 2R = 3R$$

Similarly, the resistors in the bottom branch are also strictly in series.

$$R_{bottom} = 2R + R = 3R$$

Now, the entire circuit simplifies to two $3R$ resistors connected in parallel across terminals $a$ and $b$. Since they are identical, their equivalent resistance is simply half of one of them.

$$R_{eq} = \frac{3R \cdot 3R}{3R + 3R} = \frac{3R}{2}$$

### Case 2: The Closed Switch

Closing the switch changes the game entirely. The switch now acts as an ideal, zero-resistance wire. This creates a short circuit between the midpoint of the top branch and the midpoint of the bottom branch, forcing these two points to be at the exact same electrical potential.

Because these midpoints are tied together, the circuit is effectively split into two parallel sections that are in series with each other.

The first section (left side) consists of the top $R$ resistor and the bottom $2R$ resistor connected in parallel.

$$R_{p1} = \frac{R \cdot 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2R}{3}$$

The second section (right side) consists of the top $2R$ resistor and the bottom $R$ resistor connected in parallel.

$$R_{p2} = \frac{2R \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2R}{3}$$

Finally, these two parallel sections are in series with each other. We simply add them up to find the new equivalent resistance.

$$R'_{eq} = R_{p1} + R_{p2} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$$

### The Final Ratio

The problem asks for the ratio of the equivalent resistance when the switch is open to when it is closed. Let's divide our two results.

$$	ext{Ratio} = \frac{R_{eq}}{R'_{eq}} = \frac{\frac{3R}{2}}{\frac{4R}{3}}$$

$$	ext{Ratio} = \frac{3R}{2} 	imes \frac{3}{4R} = \frac{9}{8}$$

We are given that this ratio is equal to $x : 8$. Comparing the two, it is crystal clear that $x = 9$.

### The Way Forward: The Wheatstone Bridge Connection

**What if the switch wasn't just a wire, but a resistor itself?**

This is a fantastic thought experiment. If we replaced the switch with a resistor $R_s$, the circuit would form a classic Wheatstone bridge. To see if it's balanced, we would check the ratio of the adjacent arms: $\frac{R}{2R}$ versus $\frac{2R}{R}$.

Since $\frac{1}{2} 
eq 2$, the bridge would be **unbalanced**. This means current would flow through the middle resistor $R_s$, and we could no longer use simple series and parallel rules. We would have to deploy Kirchhoff's laws or a Star-Delta transformation to solve it. This highlights just how powerful that simple zero-resistance switch is in simplifying our circuit topology!

Animated Solution for Physics - Current Electricity: The ratio of the equivalent resistance of the network (shown in figure) between the points $a$ and $b$ when switch is open and switch is closed is $x : 8$ . The value of $x$ is

Enter Numerical Value:

Visualized Solution

$R_{e q}$ Analysis

Switch Open

Series Combination

Parallel Equivalent (Open)

Switch Closed

New Circuit Structure

Parallel Sections

Series Equivalent (Closed)

Final Ratio

Wheatstone Bridge Concept

The Sigma Insight: Combination of Resistors

Analyzing the Setup

Case 1

Case 2

The Final Ratio

The Way Forward

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