Question:
The effective resistance between \(A\) and \(B\) is _ _ _, where \(R=3\,\Omega\).
The effective resistance between \(A\) and \(B\) is _ _ _, where \(R=3\,\Omega\).
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Before solving a resistor network, first merge all points connected by ideal wires into single nodes. Then identify series and parallel combinations between those nodes.
Updated On: Jul 9, 2026
- \(\dfrac{5}{3}\,\Omega\)
- \(5\,\Omega\)
- \(3\,\Omega\)
- \(\dfrac{3}{5}\,\Omega\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
First identify nodes that are directly connected by ideal wires. Resistors lying between the same two nodes are in parallel.
Step 1: Identify the common nodes. The left vertical wire connects the point \(A\) directly to the junction before the lower \(2R\) resistor. The right vertical wire connects the junction after the upper \(3R\) resistor directly to \(B\). Hence all three branches are connected between the same two nodes \(A\) and \(B\). Upper branch: \[ R+2R+3R=6R. \] Middle branch: \[ R+2R=3R. \] Lower branch: \[ 2R+3R+5R=10R. \] Thus the network reduces to three resistances in parallel: \[ 6R,\quad 3R,\quad 10R. \]
Step 2: Calculate the equivalent resistance. \[ \frac{1}{R_{\text{eq}}} = \frac{1}{6R} + \frac{1}{3R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{1+2}{6R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{1}{2R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{5+1}{10R} = \frac{6}{10R} = \frac{3}{5R}. \] Therefore, \[ R_{\text{eq}} = \frac{5R}{3}. \]
Step 3: Substitute \(R=3\,\Omega\). \[ R_{\text{eq}} = \frac{5(3)}{3}. \] \[ R_{\text{eq}} = 5\,\Omega. \]
Step 4: Write the final answer. \[ \boxed{R_{\text{eq}}=5\,\Omega} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Identify the common nodes. The left vertical wire connects the point \(A\) directly to the junction before the lower \(2R\) resistor. The right vertical wire connects the junction after the upper \(3R\) resistor directly to \(B\). Hence all three branches are connected between the same two nodes \(A\) and \(B\). Upper branch: \[ R+2R+3R=6R. \] Middle branch: \[ R+2R=3R. \] Lower branch: \[ 2R+3R+5R=10R. \] Thus the network reduces to three resistances in parallel: \[ 6R,\quad 3R,\quad 10R. \]
Step 2: Calculate the equivalent resistance. \[ \frac{1}{R_{\text{eq}}} = \frac{1}{6R} + \frac{1}{3R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{1+2}{6R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{1}{2R} + \frac{1}{10R}. \] \[ \frac{1}{R_{\text{eq}}} = \frac{5+1}{10R} = \frac{6}{10R} = \frac{3}{5R}. \] Therefore, \[ R_{\text{eq}} = \frac{5R}{3}. \]
Step 3: Substitute \(R=3\,\Omega\). \[ R_{\text{eq}} = \frac{5(3)}{3}. \] \[ R_{\text{eq}} = 5\,\Omega. \]
Step 4: Write the final answer. \[ \boxed{R_{\text{eq}}=5\,\Omega} \] \[ \boxed{\text{Answer = (B)}} \]
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