Current Electricity • NEET Revision

Example 3.37

Network Reduction (Node Merging)

Question

Find the current drawn from a cell of emf 2 V connected to the network of 1 Ω resistors shown below. (The figure shows the cell as 2 V, 1 Ω — the figure value r = 1 Ω is used, which gives the standard answer.)

Given

EMF \(E = 2\ \text{V}\), internal resistance \(r = 1\ \Omega\)
Six resistors of \(1\ \Omega\) each: top \(AB\), bottom \(DC\), and two on each diagonal (\(A{-}C\) and \(D{-}B\))
\(A\) and \(D\) joined by a plain wire; \(B\) and \(C\) joined by a plain wire (cell leads tap these wires)

Asked

Current \(I\) drawn from the cell.

Formula

Points joined by a resistance-free wire are the same node.

Parallel combination: \(\dfrac{1}{R_{eq}} = \sum \dfrac{1}{R_i}\)

\(I = \dfrac{E}{R_{eq} + r}\)

Solution

Since \(A\) and \(D\) are connected by a wire, treat them as one node \(P\). Similarly \(B\) and \(C\) form one node \(Q\).
List every path from \(P\) to \(Q\): (i) \(A \to B\): \(1\ \Omega\); (ii) \(D \to C\): \(1\ \Omega\); (iii) diagonal \(A \to C\): \(1 + 1 = 2\ \Omega\); (iv) diagonal \(D \to B\): \(1 + 1 = 2\ \Omega\).
All four paths are in parallel between \(P\) and \(Q\): \(\dfrac{1}{R_{eq}} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{2} = 3\ \Rightarrow\ R_{eq} = \dfrac{1}{3}\ \Omega\).
Current from the cell: \(I = \dfrac{E}{R_{eq} + r} = \dfrac{2}{\frac{1}{3} + 1} = \dfrac{2}{\frac{4}{3}} = 1.5\ \text{A}\).

\(I = 1.5\ \text{A}\)

Short Cut

Whenever a scary-looking network has plain wires joining corners, merge those nodes first and redraw. The 'crossed square' instantly collapses into four parallel branches: \(1, 1, 2, 2\ \Omega\). Adding conductances \(1 + 1 + 0.5 + 0.5 = 3\ \text{S} \Rightarrow R = \tfrac{1}{3}\ \Omega\).