Animated Solution for Physics - Current Electricity: Four resistances $40\Omega ,60\Omega ,90\Omega$ and $110\Omega$ make the arms of a quadrilateral $ABCD$. Across $AC$ is a battery of emf $40\text{ V}$ and internal resistance negligible. The potential difference across $BD$ (in volt) is ......... .

Circuit analysis can sometimes look like a tangled web of wires, but the moment you identify the core structure, everything falls into place. In this problem, we are presented with a quadrilateral $ABCD$ made of four resistors, and a battery connected across the diagonal $AC$. At first glance, it might look like a Wheatstone bridge, but there is no resistor or galvanometer connected across $BD$. This makes our job significantly easier!

Let's break down the circuit. The battery is connected directly to nodes $A$ and $C$. This means the entire circuit is simply two parallel branches connected across the $40\text{ V}$ source.

The upper branch consists of the path $A\to B\to C$, which contains the $40\Omega$ and $60\Omega$ resistors in series.

The lower branch consists of the path $A\to D\to C$, which contains the $90\Omega$ and $110\Omega$ resistors in series.

Because these two branches are in parallel, the full $40\text{ V}$ potential difference is applied across both of them independently. This is the crucial insight that unlocks the problem.

Finding the Currents

To find the potential at any point in the circuit, we first need to know how much current is flowing through each branch. We can use Ohm's Law, $I=\frac{V}{R}$, for this.

For the upper branch ($ABC$): The total series resistance is $R_{ABC}=40\Omega +60\Omega =100\Omega$. The current flowing through this branch, let's call it $i_1$, is:

For the lower branch ($ADC$): The total series resistance is $R_{ADC}=90\Omega +110\Omega =200\Omega$. The current flowing through this branch, let's call it $i_2$, is:

Now, we need to find the potential difference between points $B$ and $D$. The most robust way to do this is to assign a reference potential (ground) to one of the nodes.

Let's assume the potential at node $C$ is zero, so $V_C=0\text{ V}$. Since the battery provides a $40\text{ V}$ boost from $C$ to $A$, the potential at node $A$ must be $V_A=40\text{ V}$.

Now, let's trace the path from $A$ to $B$. As current $i_1$ flows through the $40\Omega$ resistor, there is a voltage drop. The potential at $B$ will be the potential at $A$ minus this drop:

Similarly, let's trace the path from $A$ to $D$. As current $i_2$ flows through the $90\Omega$ resistor, the potential drops:

We now have the absolute potentials at both points of interest. The potential difference across $BD$ is simply the difference between these two values:

By systematically breaking the circuit into parallel branches, calculating the currents, and tracking the voltage drops from a known reference point, we turned a potentially confusing quadrilateral into a straightforward application of Ohm's Law. Always look for these simple parallel and series structures hidden within complex diagrams!