Animated Solution for Physics - Current Electricity: In the given circuit, the internal resistance of the 18 V cell is negligible. If $R_1=400\Omega$, $R_3=100\Omega$ and $R_4=500\Omega$ and the reading of an ideal voltmeter across $R_4$ is 5 V, then the value of $R_2$ will be

Let's carefully analyze the structure of this circuit. Notice how the current leaves the $18\text{ V}$ battery and first passes entirely through resistor $R_1$.

Then, it reaches a junction and splits. One path goes through $R_2$, while the other path goes through both $R_3$ and $R_4$. This means $R_1$ is in series with the parallel combination of the other branches.

We are given a crucial clue: the ideal voltmeter across $R_4$ reads exactly $5\text{ V}$. Since we know both the voltage and the resistance for $R_4$, we can use Ohm's Law to find the current flowing through this entire upper branch. Let's call this current $I_1$.

Let's plug in the values. The voltage is $5\text{ V}$, and the resistance is $500\Omega$. Dividing $5$ by $500$ gives us $0.01\text{ A}$. So, a current of $0.01\text{ A}$ is flowing through both $R_3$ and $R_4$.

Now, what is the total potential difference across this entire upper branch? Since $R_3$ and $R_4$ are in series, their equivalent resistance is simply their sum. Multiplying this total resistance by the current $I_1$ will give us the voltage across the parallel section, which we'll call $V_p$.

Substituting the values, we add $100\Omega$ and $500\Omega$ to get $600\Omega$. Multiplying $0.01\text{ A}$ by $600\Omega$ gives us exactly $6\text{ V}$. This means the voltage across the upper branch, and consequently across $R_2$, is $6\text{ V}$.

Let's zoom out and look at the big picture. The battery provides a total of $18\text{ V}$. According to Kirchhoff's Voltage Law, this total voltage must be the sum of the voltage drop across $R_1$ and the voltage drop across the parallel combination. So, the voltage across $R_1$ is simply the total voltage minus $V_p$.

Subtracting the $6\text{ V}$ we just found from the total $18\text{ V}$ gives us $12\text{ V}$. So, a significant $12\text{ V}$ is dropped across just the first resistor, $R_1$.

Knowing the voltage across $R_1$ and its resistance, we have everything we need to find the total current flowing out of the battery. We just apply Ohm's Law one more time to $R_1$.

We divide the $12\text{ V}$ by the $400\Omega$ of $R_1$. This calculates to $0.03\text{ A}$. This is the main current that will soon split at the junction.

Now, focus on the junction where the current splits. Kirchhoff's Current Law tells us that the total current entering the junction must equal the total current leaving it. So, the current through $R_2$, which we'll call $I_2$, is the total current minus the current $I_1$ that went to the upper branch.

Subtracting $0.01\text{ A}$ from $0.03\text{ A}$ leaves us with $0.02\text{ A}$. This is the exact current flowing through our unknown resistor, $R_2$.

Finally, the grand finale! We know the voltage across $R_2$ is $6\text{ V}$, and the current through it is $0.02\text{ A}$. Dividing $6$ by $0.02$ gives us $300\Omega$. And there we have it, the value of $R_2$ is $300\Omega$!

Before we wrap up, think about this: what if the voltmeter wasn't ideal? A real voltmeter has some finite resistance. This would create an extra parallel path across $R_4$, changing the equivalent resistance and altering all our current calculations. Always pay attention to whether instruments are ideal or real in these problems!