Q.The four arms of a Wheatstone bridge (Fig. 3.19) have the following resistances:
$AB = 100\ \Omega$, $BC = 10\ \Omega$, $CD = 5\ \Omega$, and $DA = 60\ \Omega$.

Concept: Wheatstone Bridge with Galvanometer Resistance – The bridge is unbalanced, so we cannot ignore the galvanometer branch. Use Kirchhoff’s laws to find $I_g$.

Step 1 – Assign currents.

Let $I_1$ flow from A to B, $I_2$ from A to D, and $I_g$ from B to D (downward). Then:

• B to C: $I_1 - I_g$
• D to C: $I_2 + I_g$

Step 2 – Apply Kirchhoff’s voltage law.

Loop ABDA:

$100 I_1 + 15 I_g - 60 I_2 = 0$

Loop BCDB:

$10 (I_1 - I_g) - 5 (I_2 + I_g) - 15 I_g = 0$

Loop ABCA (outer):

$100 I_1 + 10 (I_1 - I_g) = 10$

Step 3 – Solve the three equations.

From outer loop: $110 I_1 - 10 I_g = 10 \Rightarrow 11 I_1 - I_g = 1$ …(1)

From ABDA: $100 I_1 + 15 I_g = 60 I_2 \Rightarrow I_2 = \frac{100 I_1 + 15 I_g}{60}$ …(2)

From BCDB: $10 I_1 - 10 I_g - 5 I_2 - 5 I_g - 15 I_g = 0 \Rightarrow 10 I_1 - 30 I_g = 5 I_2$ …(3)

Substitute (2) into (3): …

Check for balance. For a Wheatstone bridge $\tfrac{AB}{BC}=\tfrac{DA}{CD}$ at balance. Here $\tfrac{100}{10}=10$ but $\tfrac{60}{5}=12$; the ratios differ, so the bridge is unbalanced and a current flows through the $15\ \Omega$ galvanometer across $BD$.

Assign currents. Let $I_1$ flow $A\to B$ and $I_2$ flow $A\to D$, with $I_g$ flowing $B\to D$ through the galvanometer. By the junction rule the current in $BC$ is $I_1-I_g$ and in $DC$ is $I_2+I_g$.

Kirchhoff's voltage law (three independent loops):

Loop $ABDA$:

Loop $BCDB$:

Loop $ABC$ with the $10\ \text{V}$ source across $AC$:

Solve. From (3): $I_g=11I_1-1$. Substituting into (2) gives $I_2=6-64I_1$. Putting both into (1): …

Method: Finding the Galvanometer Current in an Unbalanced Wheatstone Bridge

Use this when a Wheatstone bridge is unbalanced and you must find the actual current through the galvanometer (not just confirm it's zero) — this needs Kirchhoff's laws, since the ratio shortcut only tells you IF the bridge is balanced, not what happens when it isn't.

Steps

Step 1: Check the balance condition first

Before reaching for Kirchhoff's laws, always check whether $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$ across the four arms. If it holds, the galvanometer current is exactly zero and no further work is needed. Only proceed to full loop analysis once you've confirmed the ratios genuinely differ.

Step 2: Assign independent branch currents

Label a current in each independent branch (e.g. $I_1$ through one arm from the entry node, $I_2$ through the adjacent arm, and $I_g$ through the galvanometer branch itself). Use Kirchhoff's junction rule to express every other branch current in terms of these, keeping the number of unknowns to a minimum.

Step 3: Write Kirchhoff's voltage law around independent loops

Choose enough closed loops to cover every branch of the bridge (typically: one loop through the galvanometer and two arms, a second loop through the galvanometer and the other two arms, and a third outer loop through the driving source). Around each loop, sum the $IR$ drops (and any EMF) to zero, using a consistent sign convention for the direction of traversal: …