In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are

Q: Why doesn't the voltmeter affect the currents $I_1$ and $I_2$?

We assume the voltmeter is ideal, meaning it has infinite resistance. Therefore, it draws zero current, allowing us to treat the left and right branches as simple series circuits.

Q: Why did we calculate the potential from node Q instead of node P?

You can calculate it from either node! If you start from P, $V_S = 50 - I_1 R_1$. Both methods yield the exact same potential for S and T. Starting from Q (assuming it's 0V) just makes the multiplication slightly more direct.

Q: Does the temperature change affect the other resistors?

The problem specifically states that only the temperature of $R_3$ is increased. In physics problems, we only apply changes to the components explicitly mentioned.

Q: What if the battery had internal resistance?

If the battery had internal resistance, the voltage across nodes P and Q would be less than 50V. You would first need to find the equivalent resistance of the entire bridge to find the main current and the actual voltage drop across PQ.

Q: Why is the bridge unbalanced after the temperature increase?

A Wheatstone bridge is balanced only when the ratio of resistances in the adjacent arms is equal ($R_1/R_3 = R_2/R_4$). Since $R_3$ increased while the others remained constant, the ratio changed, breaking the balanced condition.

Question

In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance $R_3$ has temperature coefficient $0.0004\ ^\circ	ext{C}^{-1}$. If the temperature of $R_3$ is increased by $100\ ^\circ	ext{C}$, the voltage developed between S and T will be ______ volt.

Sigma Percentile · Accepted Answer

### The Beauty of Unbalanced Bridges

Welcome to a thrilling problem that beautifully marries the concepts of **Current Electricity** with the **Thermal Properties of Matter**. At first glance, we are presented with a classic Wheatstone bridge. But as we will soon discover, a simple change in temperature can disrupt the delicate balance of the circuit, forcing us to dive deep into parallel branch analysis.

### Analyzing the Initial Setup

Before we jump into any complex calculations, it is always wise to check the initial state of the circuit. Is the Wheatstone bridge balanced?

Let's look at the ratio of the resistances in the adjacent arms. On the left side, we have $R_1 = 60\ \Omega$ and $R_3 = 300\ \Omega$. Their ratio is:

$$\frac{R_1}{R_3} = \frac{60}{300} = \frac{1}{5}$$

On the right side, we have $R_2 = 100\ \Omega$ and $R_4 = 500\ \Omega$. Their ratio is:

$$\frac{R_2}{R_4} = \frac{100}{500} = \frac{1}{5}$$

Since the ratios are perfectly equal, the bridge is initially in a **balanced condition**. This means that the potential at node S is exactly equal to the potential at node T ($V_S = V_T$), and an ideal voltmeter connected between them would read exactly zero volts.

### The Thermal Twist

But physics problems rarely let us rest on a balanced bridge! The problem introduces a twist: the temperature of resistor $R_3$ is increased by $100\ ^\circ	ext{C}$. We know that the resistance of a conductor increases with temperature according to the relation:

$$R'_3 = R_3(1 + \alpha \Delta T)$$

Let's substitute the given values to find the new resistance. The original resistance $R_3$ is $300\ \Omega$, the temperature coefficient $\alpha$ is $0.0004\ ^\circ	ext{C}^{-1}$, and the change in temperature $\Delta T$ is $100\ ^\circ	ext{C}$.

$$R'_3 = 300 (1 + 0.0004 	imes 100)$$

$$R'_3 = 300 (1 + 0.04) = 300 	imes 1.04 = 312\ \Omega$$

Our new resistance $R'_3$ is $312\ \Omega$. Because this resistance has changed while the others remained constant, the ratio $\frac{R_1}{R'_3}$ is no longer equal to $\frac{R_2}{R_4}$. The bridge is now **unbalanced**, and a potential difference will develop between nodes S and T.

### Unbalanced Bridge Dynamics

To find this new voltage difference, we must analyze the circuit as two independent parallel branches connected across the $50\ 	ext{V}$ battery. The battery is connected directly across nodes P and Q, meaning the entire $50\ 	ext{V}$ drops across both the left branch (S-side) and the right branch (T-side).

Let's calculate the current flowing through each branch using Ohm's Law ($I = \frac{V}{R_{eq}}$).

For the left branch, the total series resistance is $R_1 + R'_3 = 60 + 312 = 372\ \Omega$. The current $I_1$ is:

$$I_1 = \frac{50}{372}\ 	ext{A}$$

For the right branch, the total series resistance is $R_2 + R_4 = 100 + 500 = 600\ \Omega$. The current $I_2$ is:

$$I_2 = \frac{50}{600}\ 	ext{A}$$

### The Master Equation

Now that we have the branch currents, we can determine the absolute electrical potential at nodes S and T. To make our calculations straightforward, let's define the potential at the negative terminal of the battery (node Q) as $0\ 	ext{V}$. Consequently, the potential at the positive terminal (node P) is $50\ 	ext{V}$.

The potential at node S ($V_S$) is simply the voltage drop across the new resistor $R'_3$ as we move from Q to S:

$$V_S = I_1 R'_3 = \left(\frac{50}{372}ight) 	imes 312$$

Similarly, the potential at node T ($V_T$) is the voltage drop across resistor $R_4$:

$$V_T = I_2 R_4 = \left(\frac{50}{600}ight) 	imes 500$$

### Final Calculation

The voltage developed between S and T is the difference $V_S - V_T$. Let's compute the exact values carefully to avoid any silly mistakes.

$$V_S = 50 	imes \frac{312}{372} \approx 41.935\ 	ext{V}$$

$$V_T = 50 	imes \frac{500}{600} = 50 	imes \frac{5}{6} \approx 41.667\ 	ext{V}$$

Subtracting these potentials gives us the reading on the voltmeter:

$$V_S - V_T = 41.935 - 41.667 = 0.268\ 	ext{V}$$

Rounding to two decimal places, we arrive at our final answer of **$0.27\ 	ext{V}$**. This problem is a fantastic reminder of how interconnected physical phenomena are—a slight change in thermal energy can completely alter the electrical equilibrium of a system!

Enter Numerical Value:

Visualized Solution

$V_{S} = V_{T}$

$R_{3}^{'} = R_{3} (1 + α Δ T)$

$R_{3}^{'} = 300 (1 + 0.0004 \times 100)$

$R_{3}^{'} = 312 Ω$

$V_{P Q} = 50 V$

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

$I_{1} = \frac{50}{372} A, I_{2} = \frac{50}{600} A$

$V_{S} = I_{1} R_{3}^{'}, V_{T} = I_{2} R_{4}$

$V_{S} - V_{T} = I_{1} R_{3}^{'} - I_{2} R_{4}$

$V_{S} - V_{T} = 0.268 V$

$V_{S} - V_{T} \approx 0.27 V$

The Sigma Insight: Electrical Instruments

The Beauty of Unbalanced Bridges

Analyzing the Initial Setup

The Thermal Twist

Unbalanced Bridge Dynamics

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VS​=VT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3′​=R3​(1+αΔT)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3′​=300(1+0.0004×100)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R3′​=312 Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VPQ​=50 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=Req​V​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​=37250​ A,I2​=60050​ A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VS​=I1​R3′​,VT​=I2​R4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VS​−VT​=I1​R3′​−I2​R4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VS​−VT​=0.268 V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VS​−VT​≈0.27 V