How much current will flow through the $1\Omega$ resistor in the circuit
Show Hint
Always check for the balanced Wheatstone bridge condition ($R_1/R_2 = R_3/R_4$) in complex-looking resistor networks. It often simplifies the problem by allowing you to remove the central branch.
Step 1: Understanding the Concept:
This problem typically refers to a Wheatstone Bridge circuit. A bridge is said to be balanced when the ratio of the resistances in the two arms is equal, resulting in no potential difference across the central galvanometer or resistor. Step 2: Detailed Explanation:
In a standard balanced Wheatstone bridge configuration where resistors $P, Q, R, S$ form a loop and a resistor (in this case $1\Omega$) is connected across the junctions:
If $\frac{P}{Q} = \frac{R}{S}$, then the bridge is balanced.
In a balanced state, the potential at the two ends of the central resistor is the same.
Potential Difference ($V$) = 0.
By Ohm's Law:
\[ I = \frac{V}{R} = \frac{0}{1} = 0\text{A} \]
Since most competitive questions of this format use balanced bridge values, the current through the central branch is zero. Step 3: Final Answer:
The current flowing through the 1Ω resistor is 0A.
