Animated Solution for Physics - Current Electricity: In the circuit shown $P\ne R$, the reading of galvanometer is same with switch $S$ open or closed. Then

Imagine you are an electrical detective, tasked with uncovering the hidden secrets of a circuit. The Wheatstone bridge is your magnifying glass. It is a beautifully symmetrical arrangement of four resistors, designed to measure unknown resistances with extreme precision. In our specific problem, we are presented with a classic Wheatstone bridge setup consisting of four arms: resistors $P$, $Q$, $R$, and a galvanometer $G$. A switch $S$ connects the top node $B$ and the bottom node $D$.

The problem presents us with a fascinating clue: the galvanometer reading remains completely unchanged whether the switch $S$ is open or closed. What does this mean physically?

When the switch $S$ is open, the circuit is simply two parallel branches connected across the battery. The upper branch consists of $P$ and $Q$ in series, while the lower branch consists of $R$ and $G$ in series. Because $R$ and $G$ are in series, the current flowing through them must be identical. Mathematically, we can state that $I_R=I_G$.

Now, what happens when we close the switch? Suddenly, a new path is available between nodes $B$ and $D$. In a typical circuit, current would rush through this new path, altering the equivalent resistance of the entire network and drastically changing the currents in all the branches. However, the problem explicitly states that the galvanometer current $I_G$ does not change.

If closing a switch between two nodes does not alter the currents in the surrounding branches, it implies a profound truth: absolutely no current flows through the switch itself!

Why would no current flow through a perfectly good conducting path? Current only flows when there is a potential difference. If $I_S=0$, it must mean that the electrical potential at node $B$ is exactly equal to the electrical potential at node $D$. In mathematical terms, $V_B=V_D$. This is the hallmark of a balanced Wheatstone bridge.

Because the bridge is balanced and no current flows through the middle branch, the resistors in the upper arm ($P$ and $Q$) remain effectively in series, and the resistors in the lower arm ($R$ and $G$) also remain effectively in series.

Therefore, the current that enters resistor $R$ has nowhere else to go but straight through the galvanometer $G$. Even with the switch closed, the relationship holds true: the current through $R$ is equal to the current through $G$.

This perfectly matches option (a).

Why $P eq R$?

In physics problems, every piece of information is a breadcrumb. Why did the author explicitly state that $P eq R$?

Let's entertain a thought experiment. What if $P$ was equal to $R$? If $P=R$, the bridge would be perfectly symmetrical horizontally. The current splitting at the left node would divide equally between the upper and lower branches, meaning $I_P=I_R$. Since $I_P=I_Q$ and $I_R=I_G$, this would also mean that $I_Q=I_R$.

If $I_Q=I_R$ were true, then option (d) would also be a correct answer! By explicitly stating that $P eq R$, the problem setter ensures that the currents in the upper and lower branches are different ($I_Q eq I_R$), leaving option (a) as the single, unambiguous correct answer. Always pay attention to these subtle constraints; they are the keys to mastering JEE physics!