Animated Solution for Physics - Current Electricity: An electrical circuit is shown in figure. Calculate the potential difference across the resistor of $400\Omega$ as will be measured by the voltmeter $V$ of resistance $400\Omega$ either by applying Kirchhoff's rules or otherwise.

At first glance, this circuit looks like a tangled web of resistors that might require a tedious application of Kirchhoff's loop and junction rules. The problem asks us to find the potential difference across a $400\Omega$ resistor, which is being measured by a voltmeter.

However, there is a crucial detail hidden in the problem statement: the voltmeter is not ideal. An ideal voltmeter has infinite resistance and draws zero current, acting like an open circuit. But here, the voltmeter has a finite resistance of $R_V=400\Omega$. This means we cannot just ignore it; it actively participates in the circuit as a parallel resistor.

The voltmeter is connected directly across the $400\Omega$ resistor. Because they are in parallel, we can replace this entire section with a single equivalent resistance. Let's call this equivalent resistance $Q$.

By replacing the bulky voltmeter and its companion resistor with a clean $200\Omega$ equivalent, the true geometry of the circuit begins to reveal itself.

Wheatstone's Elegance

If we redraw the circuit by carefully tracing the nodes, a beautiful structure emerges. The circuit is actually a classic Wheatstone bridge! Let's map out the arms of this bridge:

- The top-left arm is $P=100\Omega$. - The top-right arm is our newly calculated $Q=200\Omega$. - The bottom-left arm is $R=100\Omega$. - The bottom-right arm is $S=200\Omega$. - Connecting the junctions between these arms is a middle resistor of $100\Omega$.

The magic of the Wheatstone bridge lies in its balance condition. Let's check the ratios of the resistances on the left and right sides:

Since $\frac{P}{R}=\frac{Q}{S}$, the bridge is perfectly balanced!

What does a balanced bridge mean physically? It means that the electrical potential at the top junction (between $P$ and $Q$) is exactly equal to the electrical potential at the bottom junction (between $R$ and $S$).

Because there is zero potential difference across the middle $100\Omega$ resistor, Ohm's law ($I=\frac{V}{R}$) dictates that zero current will flow through it. Electrically speaking, this middle resistor is a ghost. It does absolutely nothing, and we can completely remove it from our circuit diagram to simplify our lives.

With the middle resistor gone, the circuit breaks down into two simple, independent parallel branches connected directly to the $10\text{ V}$ battery. The top branch consists of $P$ and $Q$ in series, and the bottom branch consists of $R$ and $S$ in series.

We need to find the reading on the voltmeter, which is simply the voltage across our equivalent resistor $Q$. Since $P$ and $Q$ are in series across the $10\text{ V}$ source, we can elegantly bypass calculating the total current by using the voltage divider rule:

Substituting our values:

And there we have it! By recognizing the non-ideal nature of the voltmeter and spotting the hidden Wheatstone bridge, we bypassed pages of tedious Kirchhoff equations to arrive at a clean, elegant solution.