In a Wheatstone's bridge, three resistances P, Q and R are connected in the three arms and the fourt

Q: Why does the galvanometer show zero deflection when balanced?

When balanced, the potential at the two nodes connected to the galvanometer is exactly the same. With zero potential difference, no current flows through it!

Q: Does interchanging the battery and galvanometer affect the balance point?

No! The Wheatstone bridge is symmetric in that regard. Interchanging the source and the galvanometer will not disturb the balance condition.

Q: What if $S_1$ and $S_2$ were in series instead of parallel?

If they were in series, the equivalent resistance of the fourth arm would simply be $S_1 + S_2$, making the balance condition $P/Q = R/(S_1 + S_2)$.

Q: Why do we use a Wheatstone bridge instead of a simple ohmmeter?

A Wheatstone bridge is a null-deflection method. It doesn't depend on the calibration of the galvanometer, making it highly accurate for measuring unknown resistances.

Q: Can the bridge be balanced if the resistances are complex impedances (AC circuit)?

Yes! The same principle applies to AC circuits (like the Maxwell bridge), but the balance condition must satisfy both magnitude and phase (real and imaginary parts).

Question

In a Wheatstone's bridge, three resistances $P, Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_1$ and $S_2$ connected in parallel. The condition for the bridge to be balanced will be

Options:

(A) $\frac{P}{Q} = \frac{2R}{S_1 + S_2}$

(B) $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$

(C) $\frac{P}{Q} = \frac{R(S_1 + S_2)}{2S_1 S_2}$

(D) $\frac{P}{Q} = \frac{R}{S_1 + S_2}$

Sigma Percentile · Accepted Answer

The Wheatstone bridge is one of the most elegant and frequently tested concepts in circuit analysis. It provides a beautiful way to measure unknown resistances with high precision. In this problem, we are given a slight twist on the classic setup. Let's break it down step-by-step!

### Analyzing the Setup

Imagine the standard Wheatstone bridge. It consists of four arms forming a closed loop, with a galvanometer bridging the opposite junctions.

In our specific problem, the first three arms are straightforward: they contain the resistances $P$, $Q$, and $R$.

However, the fourth arm is where the trick lies. Instead of a single resistor, it contains two resistors, $S_1$ and $S_2$, connected in **parallel**.

### The Master Equation

The core principle of a Wheatstone bridge is its balanced state. When the bridge is balanced, the potential difference across the galvanometer is zero, meaning no current flows through it.

Mathematically, this balanced condition is expressed as the ratio of the resistances in the adjacent arms being equal.

We write this as:
$$ \frac{P}{Q} = \frac{R}{S_{eq}} $$

Here, $S_{eq}$ represents the total equivalent resistance of the entire fourth arm.

### Solving the Parallel Arm

To use our master equation, we first need to find $S_{eq}$. Since $S_1$ and $S_2$ are in parallel, we use the standard parallel resistance formula:
$$ \frac{1}{S_{eq}} = \frac{1}{S_1} + \frac{1}{S_2} $$

Taking the common denominator and inverting the expression, we get the product-over-sum rule:
$$ S_{eq} = \frac{S_1 S_2}{S_1 + S_2} $$

### Final Calculation

Now, we simply substitute this equivalent resistance back into our balance condition.

Plugging it in gives:
$$ \frac{P}{Q} = \frac{R}{\left( \frac{S_1 S_2}{S_1 + S_2} \right)} $$

To simplify this complex fraction, the denominator of the bottom fraction ($S_1 + S_2$) flips up to multiply with the numerator $R$.

This yields our final, elegant condition:
$$ \frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2} $$

This perfectly matches option (b). Always remember to carefully evaluate any hidden series or parallel combinations within the arms of a bridge before applying the balance formula!

Animated Solution for Physics - Current Electricity: In a Wheatstone's bridge, three resistances $P, Q$ and $R$ are connected in the three arms and the fourth arm is formed by two resistances $S_{1}$ and $S_{2}$ connected in parallel. The condition for the bridge to be balanced will be

Select Answer:

Visualized Solution

$Wheatstone Bridge Setup$

$Balanced Bridge Condition$

$Equivalent Resistance$

$Substituting S_{e q}$

$Final Expression$

$Practical Applications$

The Sigma Insight: Electrical Instruments

Analyzing the Setup

The Master Equation

Solving the Parallel Arm

Final Calculation

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