How much current will flow through the $1\Omega$ resistor in the circuit
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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.
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Approach Solution -2
This circuit is a classic balanced Wheatstone bridge, where a resistor (here \(1\ \Omega\)) bridges the midpoints of two parallel resistor arms. Rather than computing currents branch by branch, it helps to reason about what "balanced" means physically and test that against each option.
15A: A current this large through the bridge resistor would require a significant potential difference across its two ends. In a balanced bridge configuration, the two midpoints are held at the exact same potential by design, so no such large current can exist here.
10A: The same reasoning applies: any nonzero current through the bridge arm implies the two junctions it connects are at different potentials. A balanced bridge specifically rules this out regardless of the bridge resistor's own value.
5A: Even a moderate current still requires a potential difference across the \(1\ \Omega\) resistor. Since the arms of the bridge are proportioned so that both paths bring the current to the same potential at the midpoints, this value is also inconsistent with the circuit's balance.
0A: When a bridge is balanced, both midpoints sit at identical potential no matter what resistor connects them, so Ohm's law gives zero current through that connecting resistor regardless of whether it were \(1\ \Omega\), \(10\ \Omega\), or any other value.
Since a balanced bridge forces equal potential at both ends of the connecting resistor by construction, only a current of zero is consistent with the circuit's design, independent of the resistor's own value.
