Q.Determine the current in each branch of the network shown in Fig. 3.20.

Check the balance condition first. With arms $AB = 10\ \Omega$, $BC = 5\ \Omega$, $AD = 5\ \Omega$, $DC = 10\ \Omega$:

These are unequal, so the bridge is unbalanced and current does flow through the $5\ \Omega$ bridge arm $BD$ — the 'no current in the middle' shortcut does not apply.

Solving with mesh/Kirchhoff equations (the $10\ \text{V}$ cell is in series with a $10\ \Omega$ resistor across $A$–$C$): a star–delta reduction of the bridge gives $7\ \Omega$ between $A$ and $C$, so the total current is …

The network. Nodes $A$ (left), $B$ (top), $C$ (right), $D$ (bottom) form a bridge: $AB = 10\ \Omega$, $BC = 5\ \Omega$, $AD = 5\ \Omega$, $DC = 10\ \Omega$, with a $5\ \Omega$ bridge arm $BD$. The supply is a $10\ \text{V}$ cell in series with a $10\ \Omega$ resistor across $A$ and $C$.

Is it balanced? $\dfrac{AB}{BC} = \dfrac{10}{5} = 2$ while $\dfrac{AD}{DC} = \dfrac{5}{10} = 0.5$; unequal, so the bridge is unbalanced and current flows in $BD$. We use mesh (loop) analysis.

Mesh equations. With clockwise mesh currents $i_1$ (loop $ABD$), $i_2$ (loop $BCD$), $i_3$ (loop $A$-$D$-$C$-battery):

Solving: $i_1 = \dfrac{4}{17}$, $i_2 = \dfrac{6}{17}$, $i_3 = \dfrac{10}{17}\ \text{A}$.

Branch currents.

Method: Mesh (Loop) Analysis for a Multi-Branch Resistor Network

Use this general technique to find every branch current in ANY resistor network with more than one loop — bridges, ladders, or arbitrary meshes — once you've confirmed no shortcut (symmetry, or a balanced-bridge condition) is available.

Steps

Step 1: Check for a shortcut first

Always check whether the network has exploitable symmetry, or is a balanced bridge (equal arm ratios, no source in the bridging branch). If neither applies, fall back to full mesh analysis.

Step 2: Identify independent meshes and assign loop currents

Redraw the network as a set of non-overlapping closed loops ("windows") that together cover the whole circuit. Assign each loop a single circulating current variable, drawn in a consistent sense (e.g. all clockwise) — a branch shared between two loops then carries the difference of the two adjacent loop currents.

Step 3: Apply Kirchhoff's voltage law to each loop

For each loop, sum the $IR$ drops around it — using the net current (own loop current minus any adjacent loop current, for a shared branch) — plus any EMF sources in that loop, and set the total to zero:

Step 4: Solve the resulting linear system …