A current through 1 ohm resistance in the following circuit is
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When checking if a bridge with decimals is balanced, multiply the numbers out to clear the decimals mentally: $\frac{12.5}{2.5}$ is exactly the same as $\frac{125}{25} = 5$. Spotting a balanced bridge instantly eliminates the need for complex Kirchhoff's Loop Law matrix systems!
Step 1: Understanding the Question:
The problem presents a Wheatstone bridge network of four resistors with a total main line current of $2.1\text{ A}$ entering at junction $P$ and exiting at junction $R$.
We need to determine if the bridge is balanced to understand how the current distributes through the branches, specifically focusing on the $1\ \Omega$ resistor in arm $QR$.
Step 2: Key Formula or Approach:
For a standard Wheatstone bridge with arms arranged sequentially as $P$, $Q$, $R$, and $S$ corresponding to positions $PQ$, $QR$, $PS$, and $SR$, the balancing condition is given by:
$$\frac{R_{PQ}}{R_{QR}} = \frac{R_{PS}}{R_{SR}}$$
If this ratio holds true, the bridge is perfectly balanced, meaning the electric potential at junction $Q$ equals the potential at junction $S$ ($V_Q = V_S$). Consequently, no current flows through the central galvanometer branch ($\text{I}_G = 0$).
Step 3: Detailed Explanation:
Let's verify the balancing condition using the provided resistor values:
$$\text{Left ratio} = \frac{R_{PQ}}{R_{QR}} = \frac{5}{1} = 5$$
$$\text{Right ratio} = \frac{R_{PS}}{R_{SR}} = \frac{12.5}{2.5} = 5$$
Since $\frac{5}{1} = \frac{12.5}{2.5} = 5$, the Wheatstone bridge is perfectly balanced.
Because the bridge is balanced, the central galvanometer path carries zero current and can be treated as an open circuit (removed from the analysis).
This simplifies the network into two parallel branches connected across terminals $P$ and $R$:
1. The upper branch $PQR$ with an equivalent resistance of $R_{PQR} = 5\ \Omega + 1\ \Omega = 6\ \Omega$.
2. The lower branch $PSR$ with an equivalent resistance of $R_{PSR} = 12.5\ \Omega + 2.5\ \Omega = 15\ \Omega$.
Using the parallel current divider rule, the current flowing through the upper branch ($\text{I}_{PQR}$), which is the exact current passing through the $1\ \Omega$ resistor, is calculated as:
$$\text{I}_{1\Omega} = \text{I}_{\text{total}} \times \left( \frac{R_{PSR}}{R_{PQR} + R_{PSR}} \right)$$
Substitute the given numerical parameters into the expression:
$$\text{I}_{1\Omega} = 2.1\text{ A} \times \left( \frac{15}{6 + 15} \right)$$
$$\text{I}_{1\Omega} = 2.1 \times \frac{15}{21}$$
$$\text{I}_{1\Omega} = \frac{2.1}{21} \times 15 = 0.1 \times 15 = 1.8\text{ A}$$
Step 4: Final Answer:
The current through the $1\ \Omega$ resistor is $1.8\text{ A}$, which corresponds directly to option (A).
