In the experimental set up of meter bridge shown in the figure, the null point is obtained at a dist

Q: Why does the null point shift to the right when resistance is added in series to the left gap?

The balancing condition is $R_{left}/R_{right} = l_{left}/l_{right}$. If $R_{left}$ increases while $R_{right}$ stays constant, the ratio $R_{left}/R_{right}$ increases. To maintain balance, the ratio $l_{left}/l_{right}$ must also increase, which means $l_{left}$ must get larger, shifting the null point to the right.

Q: How do we know the new balancing length is 50 cm and not 30 cm?

The problem states the null point 'shifts by 10 cm'. Because we added resistance to the left gap ($R_1$), the balancing length on the left must increase to compensate. Therefore, it shifts from 40 cm to $40 + 10 = 50\text{ cm}$.

Q: Why must the equivalent resistance return exactly to $R_1$ for the null point to return to 40 cm?

The right gap resistance $R_2$ hasn't changed. Since the target balancing length is 40 cm (the same as the initial state), the ratio of the gap resistances must be exactly the same as it was initially. Therefore, the left gap's total resistance must equal its initial value, $R_1$.

Q: What happens if we connect the resistance $x$ in parallel to $R_2$ instead?

Connecting a resistor in parallel to $R_2$ would decrease the resistance in the right gap. This would cause the null point to shift even further to the right, away from the 40 cm mark, rather than returning to it.

Q: Can we solve this without finding the exact values of $R_1$ and $R_2$?

Yes, you can solve it algebraically. You know $R_1/R_2 = 2/3$ and $(R_1+10)/R_2 = 1$. This gives $R_1+10 = (3/2)R_1$, so $R_1/2 = 10$, meaning $R_1 = 20$. While you don't strictly need $R_2$ for the final step, finding $R_1$ is necessary to evaluate the parallel combination.

Question

In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of $40	ext{ cm}$ from $A$. If a $10\,\Omega$ resistor is connected in series with $R_1$, the null point shifts by $10	ext{ cm}$. The resistance that should be connected in parallel with $(R_1 + 10)\,\Omega$ such that the null point shifts back to its initial position is

Options:

(A) $60\,\Omega$

(B) $20\,\Omega$

(C) $30\,\Omega$

(D) $40\,\Omega$

Sigma Percentile · Accepted Answer

Welcome to a fascinating exploration of the Meter Bridge! This problem is a classic example of how changing resistances in the gaps affects the balancing length. Let's break it down step-by-step and uncover the logic behind the shifts.

### Analyzing the Initial Setup

Initially, we have a standard meter bridge setup with unknown resistances $R_1$ and $R_2$ in the left and right gaps, respectively. The problem states that the null point is obtained at a distance of $40	ext{ cm}$ from end A.

For a balanced meter bridge, the fundamental principle is that the ratio of the resistances in the gaps equals the ratio of the corresponding balancing lengths. Mathematically, this is expressed as:

$$\frac{R_1}{R_2} = \frac{l}{100 - l}$$

Plugging in our initial balancing length $l = 40	ext{ cm}$, we get:

$$\frac{R_1}{R_2} = \frac{40}{100 - 40} = \frac{40}{60}$$

Simplifying this fraction gives us our first crucial relationship:

$$\frac{R_1}{R_2} = \frac{2}{3} \quad \dots (i)$$

### The First Shift: Adding Series Resistance

Next, the problem introduces a change: a $10\,\Omega$ resistor is connected in series with $R_1$. This increases the total resistance in the left gap to $(R_1 + 10)\,\Omega$.

Because the resistance on the left has increased, the null point must shift to the right to maintain the balance. The problem tells us it shifts by $10	ext{ cm}$, making the new balancing length $l' = 40 + 10 = 50	ext{ cm}$.

Applying the meter bridge principle to this new configuration:

$$\frac{R_1 + 10}{R_2} = \frac{50}{100 - 50}$$

Since $50 / 50 = 1$, this simplifies beautifully to:

$$R_1 + 10 = R_2 \quad \dots (ii)$$

Now we have a system of two equations. Let's substitute the expression for $R_2$ from equation (ii) into equation (i):

$$\frac{R_1}{R_1 + 10} = \frac{2}{3}$$

Cross-multiplying yields:

$$3R_1 = 2(R_1 + 10)$$
$$3R_1 = 2R_1 + 20$$

Solving for $R_1$, we find **$R_1 = 20\,\Omega$**. Consequently, $R_2 = 20 + 10 = 30\,\Omega$.

### The Second Shift: Restoring the Null Point

Finally, we are asked to find a resistance, let's call it $x$, that must be connected in parallel with the entire left gap combination $(R_1 + 10)$ to shift the null point back to its initial position of $40	ext{ cm}$.

For the null point to return to $40	ext{ cm}$, the equivalent resistance of the left gap must return to its original value, which is $R_1$ (or $20\,\Omega$).

The left gap now consists of $x$ in parallel with $(R_1 + 10)$. Since $R_1 = 20\,\Omega$, the series combination is $20 + 10 = 30\,\Omega$. The equivalent resistance of $x$ and $30\,\Omega$ in parallel must equal $20\,\Omega$:

$$\frac{x \cdot 30}{x + 30} = 20$$

Let's solve this for $x$:

$$30x = 20(x + 30)$$
$$30x = 20x + 600$$
$$10x = 600$$

**$$x = 60\,\Omega$$**

Therefore, a $60\,\Omega$ resistor must be connected in parallel to restore the original balance point. This makes option (a) the correct answer. Mastering these sequential shifts is key to conquering complex meter bridge problems!

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Visualized Solution

$I ni t ia l S e t u p$

$M e t er B r i d g e P r in c i pl e$

$F i r s t E q u a t i o n$

$S im pl i f y in g R a t i o$

$A dd in g S er i es R es i s t an ce$

$S eco n d E q u a t i o n$

$S im pl i f y in g S eco n d R a t i o$

$S o l v in g f or R_{1}$

$C a l c u l a t in g R es i s t an ces$

$A dd in g P a r a l l e l R es i s t an ce$

$C o n d i t i o n f or I ni t ia l N u l l P o in t$

$P a r a l l e l R es i s t an ce E q u a t i o n$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Electrical Instruments

Analyzing the Initial Setup

The First Shift

The Second Shift

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Meter Bridge Principle

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adding Series Resistance

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Second Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for R1​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adding Parallel Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Initial Null Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Resistance Equation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward