Animated Solution for Physics - Current Electricity: In the figure given, the electric current flowing through the $5\text{ k}\Omega$ resistor is $x\text{ mA}$. The value of $x$ to the nearest Integer is ......

Imagine you are an electrical engineer tasked with analyzing a power distribution network. You are presented with a circuit powered by a $21\text{ V}$ battery. Your primary objective is to determine the exact amount of current flowing through a specific component: the $5\text{ k}\Omega$ resistor located at the top of the circuit.

At first glance, the circuit might look a bit tangled, but don't get intimidated. The key to solving any complex circuit is to break it down into smaller, manageable chunks. We need to look for obvious patterns, specifically resistors connected in series or parallel.

Focus your attention on the right side of the circuit diagram. You will notice three distinct branches, each containing a $3\text{ k}\Omega$ resistor. Because the top ends of all three resistors are connected to the same continuous top wire, and their bottom ends are all connected to the same bottom wire, they share the exact same potential difference. This is the classic definition of a parallel combination!

To simplify our circuit, we must find the equivalent resistance of this parallel block, which we will call $R_p$. The formula for parallel resistors is:

Let's substitute our known values into the equation. Since all three resistors are identical, the math becomes quite elegant:

Adding these fractions together gives us $\frac{3}{3}$, which simplifies perfectly to $1$. Therefore, the equivalent resistance of the entire parallel block is simply $1\text{ k}\Omega$.

Now, visualize replacing those three parallel resistors with our single, newly calculated $1\text{ k}\Omega$ equivalent resistor. Let's look at the structure of our new, simplified circuit.

The current leaves the battery, travels through the $5\text{ k}\Omega$ resistor, then passes through our $1\text{ k}\Omega$ equivalent resistor, and finally goes through the bottom $1\text{ k}\Omega$ resistor before returning to the battery. Because the current has only one path to follow and flows through each component sequentially, these three resistors are connected in series!

Finding the total resistance of a series circuit is straightforward; we simply add the individual resistances together:

We now know that the entire circuit behaves as if it were a single $7\text{ k}\Omega$ resistor connected to a $21\text{ V}$ battery. To find the total current ($I$) flowing out of the battery, we apply Ohm's Law:

Let's substitute our total voltage and total resistance into Ohm's Law:

Notice that we kept our resistance in kilo-ohms ($\text{k}\Omega$). When you divide volts by kilo-ohms, the resulting current is naturally in milliamperes ($\text{mA}$). This is a fantastic shortcut that saves you from writing out lots of zeros!

Finally, we must answer the original question: what is the current through the $5\text{ k}\Omega$ resistor? Because this resistor is located in the main series branch of the circuit, all of the total current must flow through it. Therefore, the current is exactly $3\text{ mA}$.

Comparing this to the problem statement, which asks for the value of $x$ in $x\text{ mA}$, we can confidently conclude that $x=3$.