Animated Solution for Physics - Current Electricity: A $16\Omega$ wire is bend to form a square loop. A $9\text{ V}$ supply having internal resistance of $1\Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is ...... $\times 10^{-1}\text{ V}$.

Imagine a uniform wire with a total resistance of $16\Omega$. When this wire is bent into a perfect square, its length is divided into four equal segments. Since resistance is directly proportional to length, the resistance is also divided equally among the four sides.

Therefore, each side of the square loop has a resistance of $\frac{16}{4}=4\Omega$.

Now, a $9\text{ V}$ battery with an internal resistance of $1\Omega$ is connected across one of these sides. This creates an interesting parallel circuit. The current from the battery reaches a corner of the square and splits into two paths.

One path goes directly through the side connected to the battery, which has a resistance of $4\Omega$. The other path must travel through the remaining three sides of the square to reach the other terminal. Since these three sides are connected end-to-end, they are in series, giving a combined resistance of $4\Omega +4\Omega +4\Omega =12\Omega$.

We now have a parallel combination of a $4\Omega$ resistor and a $12\Omega$ resistor. Let's find the equivalent resistance of this square loop.

Using the formula for two resistors in parallel, we get:

But we must not forget the internal resistance of the battery! The total resistance of the entire circuit is the sum of the loop's equivalent resistance and the battery's internal resistance.

With the total resistance known, we can easily find the total current flowing out of the battery using Ohm's Law.

The total current of $\frac{9}{4}\text{ A}$ splits at the junction. To find the potential drop across the diagonal, we need to know the current flowing through the upper branch (the $12\Omega$ path).

Using the Current Divider Rule, the current in the $12\Omega$ branch is:

The diagonal of the square connects two opposite corners. If we trace the path along the square's perimeter between these two corners, it spans exactly two sides. The resistance of these two sides is $4\Omega +4\Omega =8\Omega$.

The potential drop across this diagonal is simply the current flowing through it multiplied by its resistance.

The question asks for the answer in the format of $\cdots \times 10^{-1}\text{ V}$. We can rewrite $4.5\text{ V}$ as $45\times 10^{-1}\text{ V}$.

Thus, the required integer answer is 45.