Q.Determine the current in each branch of the network shown in Fig. 3.17.

Concept: Kirchhoff's laws applied to an unbalanced bridge network

A $5\,\text{V}$ cell sits in the diagonal $BD$, so this bridge is not balanced — the shortcut $R_1/R_2=R_3/R_4$ does not apply, and the network must be solved directly with Kirchhoff's rules. …

The network

The four nodes $A,B,C,D$ form a bridge: arms $AB=AD=4\,\Omega$, arms $BC=CD=2\,\Omega$, diagonal $AC$ carries a $1\,\Omega$ resistor in series with a $10\,\text{V}$ cell, and diagonal $BD$ carries an (ideal, resistance-free) $5\,\text{V}$ cell. Because an emf sits directly in a diagonal, the balanced-bridge condition $R_1/R_2=R_3/R_4$ (which only applies when the diagonal carries no source and the bridge is at null deflection) is irrelevant here — this must be solved as a general Kirchhoff's-laws problem.

Solving by node potentials

Take $D$ as the reference node, $V_D=0$. The diagonal $BD$ is an ideal $5\,\text{V}$ cell directly connecting $B$ and $D$, so it fixes $B$'s potential immediately:

For the branch $AC$ (a $1\,\Omega$ resistor in series with a $10\,\text{V}$ cell), the current flowing from $C$ to $A$, $I_{AC}$, obeys

Kirchhoff's junction rule at $A$: current entering from $C$ equals current leaving through $AB$ and $AD$:

Substituting $V_B=5,\ V_D=0$ and simplifying:

Kirchhoff's junction rule at $C$: current entering from $B$ equals current leaving through $AC$ and $CD$:

Substituting and simplifying:

Setting (i) = (ii): $\dfrac{4V_C+45}{6} = 2V_C+7.5 \Rightarrow 4V_C+45 = 12V_C+45 \Rightarrow V_C = 0$, and then $V_A = 7.5\,\text{V}$.

Reading off the branch currents

With $V_A=7.5\,\text{V}$, $V_B=5\,\text{V}$, $V_C=0\,\text{V}$, $V_D=0\,\text{V}$:

Method: Solving an Unbalanced Bridge by Node (Nodal) Potentials

Use this whenever a resistor network — a bridge or any multi-loop circuit — contains a source (battery/cell) placed directly in an interior branch, so the simple balance-ratio shortcut ($R_1/R_2 = R_3/R_4$) does not apply and the full network must be solved.

Steps

Step 1: Check whether the balance condition applies

Compute the two arm ratios of the bridge (e.g. $AB/BC$ vs $AD/DC$). If they're unequal, OR if a real EMF source (not just a galvanometer) sits in the bridging diagonal, the network is unbalanced — you cannot assume zero current in that branch and must solve the full circuit.

Step 2: Choose a reference node and assign potentials

Pick one node as the reference (set its potential to $0$). If an ideal EMF source with no internal resistance directly connects two nodes, it immediately fixes their potential difference — use this to eliminate one unknown right away:

Step 3: Apply Kirchhoff's junction rule at every unfixed node …