The effective resistance between the points A and B of the circuit shown in the figure is:
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Whenever a Wheatstone bridge is not balanced, direct series-parallel reduction generally fails. In such cases, the Node Voltage Method is often the fastest and most reliable technique. Assume a voltage across the terminals, determine the current drawn, and then use
\[
R_{\text{eq}}=\frac{V}{I}.
\]
This approach works for every linear resistor network.
Step 1: Recognise the network. The circuit is a Wheatstone-bridge arrangement with arm resistors and a central bridge resistor. We label nodes A and B as the terminals and C, D as the two bridge junctions. Step 2: Test for balance. A bridge is balanced when the ratio of adjacent arms matches across the bridge. Here the ratios do not match, so the bridge is unbalanced and the central resistor carries current; pure series-parallel reduction will not work. Step 3: Choose the method. Assume a potential $V$ across A and B with $V_A = V$, $V_B = 0$, and unknown node potentials $V_C$ and $V_D$. We apply Kirchhoff's current law at C and D. Step 4: Apply KCL at the junctions. Writing current balance at node C and node D in terms of $V_C$, $V_D$ and $V$ gives two linear equations, which solve to fix $V_C$ and $V_D$ as fractions of $V$. Step 5: Get the total current and equivalent resistance. The current leaving A is the sum of the currents into the two arms from A; dividing $V$ by that total current gives $R_{AB} = \dfrac{V}{I}$. Carrying out the algebra for this network yields $R_{AB} = 14\,\Omega$. Step 6: Conclude. The effective resistance between A and B is $14\,\Omega$, which is option (1). \[ \boxed{R_{AB} = 14\,\Omega} \]
