Animated Solution for Physics - Current Electricity: The resistive network shown below is connected to a DC source of $16\text{ V}$. The power consumed by the network is $4\text{ W}$. The value of $R$ is

Welcome to a fascinating journey into the world of current electricity! At first glance, a circuit diagram with multiple branches and resistors can look like an intimidating maze. But the true beauty of physics lies in breaking down the complex into the simple.

In this problem, we are presented with a resistive network connected to a $16\text{ V}$ DC source. We are told that the entire network consumes a total power of $4\text{ W}$. Our mission is to unravel this network and find the value of the unknown resistance, $R$.

Let's take a deep breath and dive into the structure of this circuit.

Divide and Conquer

The key to conquering any complex circuit is to look for recognizable patterns—specifically, series and parallel combinations. Instead of trying to tackle the whole network at once, let's divide it into smaller, manageable blocks.

Notice the first block on the left side of the circuit. Let's call this Part A. It consists of two resistors, each with a value of $4R$, connected in parallel.

We know that for two resistors in parallel, the equivalent resistance can be quickly found using the "product over sum" rule.

Let's calculate the equivalent resistance for Part A, which we'll call $R_A$:

This is a beautiful simplification! We can now mentally (and visually) replace that entire parallel block with a single resistor of value $2R$.

Now, let's shift our focus to the right side of the circuit. Here, we spot another parallel block. Let's call this Part B. This block contains a $6R$ resistor and a $12R$ resistor connected in parallel.

Once again, we apply our trusty product over sum rule to find its equivalent resistance, $R_B$:

Perfect! We can replace this second block with a single resistor of value $4R$.

Bringing It All Together

Look at what we have achieved through these two simple steps. By simplifying the parallel blocks, our initially complex network has transformed into a straightforward series circuit.

We now have four resistors connected end-to-end in series: $2R$, $R$, $4R$, and $R$.

Because they are in series, finding the total equivalent resistance of the entire network, $R_{eq}$, is as simple as adding them up:

Now we have a single expression for the total resistance of the circuit. It's time to connect this to the physical quantities given in the problem: power and voltage.

We know the fundamental relationship that links power ($P$), voltage ($V$), and resistance ($R$):

Unveiling the Unknown

We are given that the total power consumed by the network is $P=4\text{ W}$, and the source voltage is $V=16\text{ V}$. We also just determined that our total equivalent resistance is $R_{eq}=8R$.

Let's substitute all these known values into our master power equation:

Now, it's just a matter of simple algebra. Let's square the voltage:

We can simplify the right side by dividing $256$ by $8$, which gives us $32$:

Finally, to isolate our unknown variable $R$, we rearrange the equation:

And there we have it! By systematically breaking down the circuit and applying fundamental electrical principles, we have found that the value of the unknown resistance $R$ is $8\Omega$.

This matches option (b), completing our journey through this elegant problem.