The Wheatstone bridge shown in figure here, gets balanced when the carbon resistor is used as R1 has

Q: Why is the current through the galvanometer zero?

In a balanced Wheatstone bridge, the ratio of resistances creates a situation where the electrical potential at the top node is exactly equal to the potential at the bottom node. Since current only flows when there is a potential difference, no current flows through the galvanometer.

Q: How do I easily remember the resistor color code?

You can use the classic mnemonic: "**B**B **R**OY of **G**reat **B**ritain had a **V**ery **G**ood **W**ife". The bold letters stand for Black(0), Brown(1), Red(2), Orange(3), Yellow(4), Green(5), Blue(6), Violet(7), Gray(8), and White(9).

Q: What if the battery and galvanometer are swapped in the circuit?

This is a fascinating property of the Wheatstone bridge! If you swap the positions of the battery and the galvanometer, the bridge remains balanced, and the condition $\frac{R_1}{R_2} = \frac{R_3}{R_4}$ still holds perfectly true.

Q: Why is the multiplier for the brown band $10^1$?

In the standard 4-band resistor color code, the first two bands represent the significant digits, and the third band represents the decimal multiplier, which is $10^n$. Since Brown corresponds to the digit 1, its multiplier is $10^1$.

Q: Can we write the balance condition as $\frac{R_1}{R_3} = \frac{R_2}{R_4}$ instead?

Yes, absolutely! Mathematically, $\frac{R_1}{R_2} = \frac{R_3}{R_4}$ is exactly the same as $\frac{R_1}{R_3} = \frac{R_2}{R_4}$. You can use whichever form is easier for your specific calculation.

Question

The Wheatstone bridge shown in figure here, gets balanced when the carbon resistor is used as $R_1$ has the color code (orange, red, brown). The resistors $R_2$ and $R_4$ are $80\ \Omega$ and $40\ \Omega$, respectively.
Assuming that the color code for the carbon resistors gives their accurate values, the color code for the carbon resistor is used as $R_3$ would be

Options:

(A) brown, blue, black

(B) brown, blue, brown

(C) grey, black, brown

(D) red, green, brown

Sigma Percentile · Accepted Answer

### The Balanced Bridge

Imagine you are an electrical detective, and your first clue is the word **balanced**.

When a Wheatstone bridge is balanced, it means the bridge has reached a state of perfect electrical equilibrium. The potential at the top node is exactly equal to the potential at the bottom node.

Because there is no potential difference, the galvanometer sitting between these nodes experiences absolutely zero current. It shows no deflection.

This beautiful symmetry gives us our master equation. The ratio of the resistances in the adjacent arms must be perfectly equal:

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

### Decoding the Colors

Before we can use our master equation, we need to decode the hidden value of $R_1$.

The problem gives us a color code: **Orange, Red, Brown**.

If you recall the classic mnemonic for resistor color codes, you know that Orange corresponds to the digit $3$, and Red corresponds to the digit $2$.

The third band, Brown, represents the decimal multiplier. Brown corresponds to $1$, which means our multiplier is $10^1$.

Putting it all together, the resistance $R_1$ is:

$$ R_1 = 32 	imes 10^1 = 320 \ \Omega $$

### The Master Equation

Now that we have all our pieces, let's plug them into the balance condition.

We know $R_1 = 320 \ \Omega$, $R_2 = 80 \ \Omega$, and $R_4 = 40 \ \Omega$.

Substituting these into our master equation, we get:

$$ \frac{320}{80} = \frac{R_3}{40} $$

This is a straightforward linear equation. Let's isolate $R_3$:

$$ R_3 = 40 	imes \frac{320}{80} $$

Since $320$ divided by $80$ is exactly $4$, the calculation simplifies beautifully:

$$ R_3 = 40 	imes 4 = 160 \ \Omega $$

### The Final Color Code

We have found the numerical value of $R_3$, but the question asks for its color code.

We need to reverse-engineer the value $160 \ \Omega$ back into colored bands.

First, let's write $160$ in the standard two-digit format with a multiplier:

$$ 160 = 16 	imes 10^1 \ \Omega $$

Now, we map each part back to its corresponding color.

The first digit is $1$, which corresponds to **Brown**.

The second digit is $6$, which corresponds to **Blue**.

The multiplier is $10^1$, and the power $1$ corresponds to **Brown**.

Therefore, the final color code for $R_3$ is **Brown, Blue, Brown**.

Select Answer:

Visualized Solution

$Wheatstone Bridge$

$Balance Condition$

$Decoding R_{1}$

$Substitution$

$Solving for R_{3}$

$Encoding R_{3}$

$Symmetry of the Bridge$

The Sigma Insight: Electrical Instruments

The Balanced Bridge

Decoding the Colors

The Master Equation

The Final Color Code

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Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for R3​

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

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