Two resistances R1 = X \Omega and R2 = 1 \Omega are connected to a wire AB of uniform resistivity, a

Q: Why can't we just use the average radius for each half to calculate the resistance?

Because resistance depends on the inverse square of the radius ($1/r^2$). The average of $1/r^2$ is mathematically not equal to $1/r_{avg}^2$. You must integrate to account for the continuous non-linear change in cross-sectional area.

Q: How did we get the equation for $r(x)$?

We used the two-point form of a straight line. We know the radius at $x=0$ is $0.2$ and at $x=1$ is $1.0$. The slope is $(1.0 - 0.2) / 1 = 0.8$. So the equation is $r(x) = 0.2 + 0.8x$.

Q: Why did we change the variable of integration from $dx$ to $dr$?

It makes the integration much simpler. Since the denominator has $r^2$, integrating with respect to $r$ directly gives us a standard power rule integral $\int r^{-2} dr = -r^{-1}$, avoiding messy chain rules.

Q: Does the value of resistivity $\rho$ matter in this problem?

No, it doesn't! Because we are finding the ratio of the resistances of the two halves, the uniform resistivity $\rho$ (along with $\pi$ and the slope factor $0.8$) completely cancels out.

Q: What if the galvanometer was not connected at the exact center?

If it were connected at a distance $L_1$ from A, we would simply change the limits of integration. The first integral would be from $x=0$ to $x=L_1$, and the second from $x=L_1$ to $x=L$.

Question

Two resistances $R_1 = X \ \Omega$ and $R_2 = 1 \ \Omega$ are connected to a wire AB of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from $0.2 	ext{ mm}$ at A to $1 	ext{ mm}$ at B. A galvanometer (G) connected to the center of the wire, $50 	ext{ cm}$ from each end along its axis, shows zero deflection when A and B are connected to a battery. The value of X is _____.

Sigma Percentile · Accepted Answer

### The Setup: A Disguised Bridge

When you first look at this circuit, it might seem like a strange hybrid of a standard resistor network and a continuous wire. But the key lies in the galvanometer.

The problem states that the galvanometer shows **zero deflection**. This is the universal signature of a **balanced Wheatstone bridge**.

In a balanced bridge, the ratio of the resistances in the upper arms must perfectly match the ratio of the resistances in the lower arms.

$$ \frac{R_1}{R_2} = \frac{R_{A,mid}}{R_{mid,B}} $$

Here, the lower arms are simply the two halves of the wire AB, split exactly at the 50 cm mark. Our mission is to find the resistance of these two halves.

### Taming the Tapered Wire

If the wire were uniform, both halves would have the same resistance, and the ratio would be 1. But this wire is tapered! Its radius increases linearly from $0.2 	ext{ mm}$ at end A to $1.0 	ext{ mm}$ at end B.

Let's set up a coordinate system. Let $x$ be the distance from end A in meters. The total length is $L = 1 	ext{ m}$. We can write the radius $r(x)$ as a linear equation:

$$ r(x) = r_A + \left(\frac{r_B - r_A}{L}ight)x $$

Substituting our values, we get:

$$ r(x) = 0.2 + \left(\frac{1.0 - 0.2}{1}ight)x = 0.2 + 0.8x $$

This equation tells us exactly how thick the wire is at any point $x$.

### The Magic of Calculus

To find the total resistance of a varying shape, we must slice it into infinitesimally thin disks of thickness $dx$. The resistance of one such disk is:

$$ dR = \frac{ho dx}{A} = \frac{ho dx}{\pi r(x)^2} $$

Integrating this directly with respect to $x$ can be slightly messy. Let's use a brilliant substitution. Since $r = 0.2 + 0.8x$, we can differentiate both sides to get $dr = 0.8 dx$, which means $dx = \frac{dr}{0.8}$.

Now, let's calculate the resistance of the first half ($x = 0$ to $x = 0.5 	ext{ m}$). At $x = 0$, $r = 0.2 	ext{ mm}$. At $x = 0.5$, $r = 0.2 + 0.8(0.5) = 0.6 	ext{ mm}$.

$$ R_{A,mid} = \int_{0.2}^{0.6} \frac{ho}{\pi r^2} \frac{dr}{0.8} = \frac{ho}{0.8\pi} \left[ -\frac{1}{r} ight]_{0.2}^{0.6} $$

Evaluating the limits:

$$ R_{A,mid} = \frac{ho}{0.8\pi} \left( \frac{1}{0.2} - \frac{1}{0.6} ight) = \frac{ho}{0.8\pi} \left( 5 - \frac{5}{3} ight) = \frac{ho}{0.8\pi} \left( \frac{10}{3} ight) $$

Now, for the second half ($x = 0.5$ to $x = 1.0 	ext{ m}$). The radius goes from $0.6 	ext{ mm}$ to $1.0 	ext{ mm}$.

$$ R_{mid,B} = \int_{0.6}^{1.0} \frac{ho}{\pi r^2} \frac{dr}{0.8} = \frac{ho}{0.8\pi} \left[ -\frac{1}{r} ight]_{0.6}^{1.0} $$

Evaluating these limits:

$$ R_{mid,B} = \frac{ho}{0.8\pi} \left( \frac{1}{0.6} - \frac{1}{1.0} ight) = \frac{ho}{0.8\pi} \left( \frac{5}{3} - 1 ight) = \frac{ho}{0.8\pi} \left( \frac{2}{3} ight) $$

### Bringing It All Together

We have the resistances of both halves! Let's find their ratio:

$$ \frac{R_{A,mid}}{R_{mid,B}} = \frac{\frac{10}{3}}{\frac{2}{3}} = 5 $$

The first half has exactly 5 times the resistance of the second half.

Returning to our master Wheatstone bridge equation:

$$ \frac{R_1}{R_2} = 5 $$

We are given that $R_1 = X$ and $R_2 = 1 \ \Omega$. Therefore:

$$ \frac{X}{1} = 5 \implies X = 5 $$

The beauty of this problem lies in how the messy resistivity $ho$ and the constants perfectly cancel out, leaving us with a clean, elegant integer.

Enter Numerical Value:

Visualized Solution

\text{Understanding the Circuit}

\text{Balanced Wheatstone Bridge Condition}

\text{Analyzing the Tapered Wire}

\text{Radius as a Function of } x

\text{Resistance of an Elemental Ring}

\text{Resistance of the First Half } (R_{A,mid})

\text{Evaluating } R_{A,mid}

\text{Resistance of the Second Half } (R_{mid,B})

\text{Evaluating } R_{mid,B}

\text{Finding the Ratio}

\text{Calculating } X

\text{What if the profile was different?}

The Sigma Insight: Electrical Instruments

The Setup

Taming the Tapered Wire

The Magic of Calculus

Bringing It All Together

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Analyzing the Tapered Wire} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Radius as a Function of } x .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Resistance of an Elemental Ring} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Resistance of the First Half } (R_{A,mid}) .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Evaluating } R_{A,mid} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Resistance of the Second Half } (R_{mid,B}) .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Evaluating } R_{mid,B} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{What if the profile was different?} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }