Animated Solution for Physics - Current Electricity: For the circuit shown in the figure

Circuit problems often look like an intimidating maze of wires and squiggly lines. But if you take a deep breath and trace the path of the current, they unravel into beautiful, logical puzzles. Today, we are going to dissect a classic JEE Advanced multiple-choice question that tests your mastery over equivalent resistance, Ohm's law, and power dissipation.

Imagine you are a tiny electron leaving the $24\text{ V}$ battery. Where do you go? Let's find out!

Let's start by carefully analyzing the circuit topology. We have a $24\text{ V}$ battery supplying a total current $I$. This current flows out of the battery and must pass entirely through the first resistor, $R_1$.

After passing through $R_1$, the path splits at a junction. The current divides, with some flowing through $R_2$ and the rest flowing through the load resistor, $R_L$. Because the current splits and then recombines, $R_2$ and $R_L$ are in parallel. Therefore, the entire circuit can be simplified as $R_1$ in series with the parallel combination of $R_2$ and $R_L$.

To find the total current, we first need the equivalent resistance of the entire circuit. Let's begin by calculating the equivalent resistance of the parallel branches, which we will call $R_p$.

The formula for two resistors in parallel is the product over the sum:

Now, let's substitute the given values into our formula. $R_2$ is $6\text{ k}\Omega$, and $R_L$ is $1.5\text{ k}\Omega$.

Let's do the math. Six times one point five gives us nine. And six plus one point five is seven point five. Dividing nine by seven point five, we get exactly $1.2\text{ k}\Omega$ for the parallel section.

Next, we add the series resistor $R_1$, which is $2\text{ k}\Omega$, to our parallel resistance. This gives a total equivalent resistance of $3.2\text{ k}\Omega$.

Using Ohm's law, the total current $I$ is the total voltage divided by the equivalent resistance.

This perfectly matches the first option. Option (a) is absolutely correct!

Let's check option (b), which talks about the potential difference across $R_L$. The voltage across the parallel combination is simply the total current multiplied by the parallel resistance $R_p$.

Since option (b) claims it is $18\text{ V}$, it is incorrect.

Moving on to option (c), we need the ratio of powers dissipated in $R_1$ and $R_2$. For $R_1$, we use the $I^{2}R$ formula because we know the total current flowing through it.

For $R_2$, it is easier to use the $V^2/R$ formula because we just found the voltage across the parallel branch.

Now, let's find their ratio.

Option (c) states the ratio is exactly 3, which is clearly not the case. So, option (c) is also incorrect.

Finally, let's evaluate option (d). Imagine we physically swap the resistors $R_1$ and $R_2$. Now, the series resistor $R_1$ becomes $6\text{ k}\Omega$, and the parallel branch resistor $R_2$ becomes $2\text{ k}\Omega$. Let's see how this impacts the load.

We must recalculate our resistances. The new parallel resistance, $R_p^{\prime }$, is:

Adding the new $R_1$, which is $6\text{ k}\Omega$, gives a new total resistance:

To find the new voltage across the load, $V_L^{\prime }$, we can use the voltage divider rule. It is the total voltage multiplied by the parallel resistance fraction.

The sevens cancel out, leaving twenty-four times six over forty-eight, which beautifully simplifies to exactly $3\text{ V}$.

Let's conclude. The power dissipated in $R_L$ is proportional to the square of the voltage across it ($P=V^2/R$).

The voltage dropped from $9\text{ V}$ to $3\text{ V}$, which is a factor of one-third.

Squaring this factor means the power decreases by a factor of 9. Thus, option (d) is absolutely correct!