Animated Solution for Physics - Current Electricity: First, a set of $n$ equal resistors of $10\Omega$ each are connected in series to a battery of emf $20\text{V}$ and internal resistance $10\Omega$. A current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased $20$ times, then the value of $n$ is ......... .

Series vs Parallel

Imagine you are an electrical engineer tasked with designing a circuit. You have a handful of identical resistors, each with a resistance of $10\Omega$, and a $20\text{V}$ battery. But this battery isn't perfect; it has an internal resistance of $10\Omega$.

Our journey begins by connecting all $n$ resistors in a single, continuous chain—a series combination.

Resistors in Series

In a series circuit, the equivalent resistance is simply the sum of all individual resistances. Since we have $n$ resistors of $10\Omega$ each, the equivalent external resistance is $10n$.

However, we must not forget the battery's internal resistance, $r=10\Omega$, which is always in series with the external circuit. Using Ohm's law, the total current $I_s$ flowing through this setup is the total EMF divided by the total resistance:

By factoring out the $10$ in the denominator, we can simplify this expression to:

Resistors in Parallel

Now, let's completely rewire the system. We take those exact same $n$ resistors and connect them in parallel across the same battery.

When identical resistors are connected in parallel, the equivalent resistance drops drastically. It becomes the resistance of one resistor divided by the number of resistors. So, our new external equivalent resistance is $\frac{10}{n}$.

Applying Ohm's law once again, the new total current $I_p$ is:

To clean up this complex fraction, we take the least common multiple in the denominator:

Factoring out the $10$ again, we get a beautifully symmetric equation:

The problem provides us with a thrilling constraint: the current in the parallel circuit is exactly $20$ times the current in the series circuit. This gives us our master equation:

Let's substitute our simplified expressions for $I_p$ and $I_s$ into the master equation:

Notice how the $(n+1)$ term appears in the denominator on both sides. Since $n$ represents a physical count of resistors, it must be a positive integer. Therefore, $(n+1)$ is never zero, and we can safely multiply both sides by $(n+1)$ to cancel it out.

This leaves us with a wonderfully simple linear equation:

Dividing by $2$, we arrive at our final answer:

We used exactly $20$ resistors!

A Moment of Physics Intuition: Take a step back and think about the internal resistance. If the battery had been ideal ($r=0$), the ratio of parallel current to series current would have been $\frac{I_p}{I_s}=n^2=400$. The current would have skyrocketed! But because of the $10\Omega$ internal resistance acting as a bottleneck, the current only increased by a factor of $n=20$. This perfectly illustrates how internal resistance protects real-world circuits from drawing infinite current when shorted or heavily loaded in parallel.