Animated Solution for Physics - Current Electricity: In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be

Welcome, future engineers and physicists! Today, we are going to dive into a classic and incredibly elegant problem from the realm of Current Electricity. We are dealing with a circuit that features a galvanometer showing zero deflection. This specific condition is the secret to unlocking the entire puzzle.

Let's take a close look at the circuit provided. We have a $12\text{ V}$ battery on the left, driving current into the system. There is a $500\Omega$ resistor on the top wire, an unknown resistor $R$ in the middle branch, and a rightmost branch containing a galvanometer $G$ in series with a $2\text{ V}$ battery.

The most crucial piece of information given is that the galvanometer shows zero deflection. But what does this actually mean in physical terms?

A galvanometer is a highly sensitive instrument used to detect small electric currents. When it shows zero deflection, it means that absolutely no current is flowing through the branch it is connected to.

Mathematically, we can state this as:

Because no current flows through the right branch, there is no voltage drop across the galvanometer itself, nor is there any voltage drop across the internal resistance of the $2\text{ V}$ battery. This implies that the potential difference across the entire right branch is solely determined by the electromotive force (EMF) of the battery in that branch.

To solve this elegantly, we will use nodal analysis. Let's assign a potential of $0\text{ V}$ to the entire bottom wire of the circuit.

Due to the $12\text{ V}$ battery on the left, the potential at the top-left node becomes exactly $12\text{ V}$.

Now, look at the right branch. Since it is connected between the top-middle node and the $0\text{ V}$ bottom wire, and it contains a $2\text{ V}$ battery with its positive terminal facing upwards, the potential at the top-middle node must be exactly $2\text{ V}$ to ensure no current flows. If the potential were anything else, a current would be forced through the galvanometer!

Now that we know the potentials at our key nodes, the rest is straightforward Ohm's Law. Let's look at the $500\Omega$ resistor. It is connected between the $12\text{ V}$ node and the $2\text{ V}$ node.

The potential difference across it is:

The current $I$ flowing through this resistor is:

Because no current splits off into the right branch (remember, $I_G=0$), this entire current of $\frac{1}{50}\text{ A}$ must flow straight down through the unknown resistor $R$.

The resistor $R$ is connected between the $2\text{ V}$ node and the $0\text{ V}$ ground. So, the voltage across $R$ is simply $2\text{ V}$.

Applying Ohm's Law one final time to find $R$:

And there we have it! The value of the unknown resistor $R$ is exactly $100\Omega$. This perfectly matches option (b).

Problems like this beautifully demonstrate how understanding the physical meaning of a "balanced" state can turn a seemingly complex circuit into a simple, logical sequence of steps.