Animated Solution for Physics - Current Electricity: Four resistances of $15\Omega$, $12\Omega$, $4\Omega$ and $10\Omega$ respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $10\Omega$ to balance the network is ......... $\Omega$.

Welcome to an exciting journey into the heart of circuit analysis! Today, we are going to unravel the elegance of the Wheatstone Bridge, a brilliant configuration that allows us to measure unknown resistances with incredible precision.

I know circuit diagrams can sometimes look like a tangled mess of wires, but let's take a breath and break this down step-by-step. Imagine you are an electrical detective, and your job is to perfectly balance a scale. But instead of weights, you are balancing electrical potentials.

The problem presents us with four resistances connected in a cyclic order to form a Wheatstone network. What does "cyclic order" mean? Imagine walking around the perimeter of the bridge from node A to B, then B to C, C to D, and finally D back to A. As you walk, you encounter the resistors in the exact sequence given: $15\Omega$, $12\Omega$, $4\Omega$, and $10\Omega$.

Let's formalize this: - Arm AB: $R_{AB}=15\Omega$ - Arm BC: $R_{BC}=12\Omega$ - Arm CD: $R_{CD}=4\Omega$ - Arm DA: $R_{DA}=10\Omega$

Our goal is to "balance" this network. But what does a balanced bridge actually look like?

A Wheatstone bridge is balanced when the potential difference across the central galvanometer branch (between nodes B and D) is exactly zero. When this happens, no current flows through the galvanometer. It's a state of perfect electrical equilibrium!

For this magic to happen, the ratio of the resistances in the adjacent arms must be equal. The master equation for a balanced bridge is:

This is our golden key. Let's substitute the known values into this equation to see what the resistance of arm AD should be to achieve balance.

Now, we execute a simple atomic compute. Let's simplify the fraction on the left side. Both 15 and 12 are divisible by 3, so $\frac{15}{12}$ simplifies to $\frac{5}{4}$.

By multiplying both sides by 4, the denominators cancel out beautifully, leaving us with:

This is a crucial revelation! For the bridge to be perfectly balanced, the effective resistance of arm AD must be exactly $5\Omega$.

Here is where the plot thickens. The current resistance of arm AD is $10\Omega$, but we need it to be $5\Omega$. How do we reduce the resistance of a branch?

If we add a resistor in series, the total resistance increases. But if we add a resistor in parallel, the total resistance decreases! The problem explicitly tells us to connect an unknown resistance, $X$, in parallel with the $10\Omega$ resistor.

The formula for the equivalent resistance of two resistors in parallel is the product of their resistances divided by their sum. We set this equal to our target resistance of $5\Omega$:

We are in the home stretch. Let's solve this algebraic equation for $X$. First, we multiply both sides by $(10+X)$ to clear the denominator:

Next, we distribute the 5 on the right side:

Now, we subtract $5X$ from both sides to isolate the terms with $X$:

Finally, dividing by 5 gives us our grand conclusion:

And there we have it! By connecting a $10\Omega$ resistor in parallel with the existing $10\Omega$ resistor in arm AD, we halve the resistance to $5\Omega$, perfectly balancing the Wheatstone bridge.

This problem is a beautiful demonstration of how abstract circuit principles translate into simple, logical algebraic steps. Keep practicing, and soon you'll be balancing complex networks in your sleep!