Animated Solution for Physics - Current Electricity: A $3\text{ V}$ battery with negligible internal resistance is connected in a circuit as shown in the figure. The current $I$ in the circuit will be

Unraveling the Circuit: A Journey Through Series and Parallel Resistors

When you first look at a circuit diagram, it can sometimes feel like a maze of lines and squiggles. The key to solving any circuit problem is to break it down into manageable pieces.

In our given circuit, we have a $3\text{ V}$ battery supplying a total current $I$. This current travels along the wire until it hits a junction, which we can call node A. At this point, the current has a choice: it can either go straight down through the $3\Omega$ resistor on the left, or it can take the path to the right.

If we trace the path to the right, we see it goes through a $3\Omega$ resistor to reach node C, and then continues through another $3\Omega$ resistor to reach the bottom junction, node B.

Notice something special about the path on the right? The current that flows through the first resistor (from A to C) has absolutely nowhere else to go but through the second resistor (from C to B).

Because they share the exact same current path without any branching in between, these two resistors are connected in series.

To find the equivalent resistance of resistors in series, we simply add their values together:

We can now mentally (or physically, if you're redrawing the circuit) replace that entire right-hand triangle section with a single $6\Omega$ resistor connected directly between nodes A and B.

Now our circuit looks much simpler. We have the original $3\Omega$ resistor on the left, and our new equivalent $6\Omega$ resistor on the right.

Both of these resistors are connected directly across the same two points: node A at the top and node B at the bottom. This is the classic definition of a parallel connection.

For two resistors in parallel, the equivalent resistance is given by the product over sum formula:

Let's plug in our values:

So, the entire network of three resistors behaves exactly like a single $2\Omega$ resistor!

Ohm's Law

We've successfully reduced the complex circuit to a very simple one: a $3\text{ V}$ battery connected to a single $2\Omega$ equivalent resistor.

To find the total current $I$ drawn from the battery, we turn to the fundamental law of circuits, Ohm's Law:

We know the voltage $V$ is $3\text{ V}$, and our total equivalent resistance $R_{eq}$ is $2\Omega$. Substituting these values in:

The total current flowing in the circuit is $1.5\text{ A}$. By systematically breaking down the circuit into series and parallel parts, we turned a potentially confusing problem into a straightforward calculation.