Five resistors in Wheatstone bridge — find current through galvanometer

Question

In a Wheatstone bridge, $P = 10\ \Omega$ , $Q = 15\ \Omega$ , $R = 6\ \Omega$ , and $S = 10\ \Omega$ . A galvanometer of resistance $G = 20\ \Omega$ is connected across BD. A battery of emf 6 V is connected across AC. Find the current through the galvanometer.

Solution — Step by Step

The Wheatstone bridge is balanced when:

\frac{P}{Q} = \frac{R}{S}

Check: $\frac{P}{Q} = \frac{10}{15} = \frac{2}{3}$ and $\frac{R}{S} = \frac{6}{10} = \frac{3}{5}$

Since $\frac{2}{3} \neq \frac{3}{5}$ , the bridge is unbalanced. Current flows through the galvanometer.

We must solve using Kirchhoff’s laws.

The Wheatstone bridge circuit:

Battery connected between A (positive) and C (negative)
P is between A and B; Q is between B and C
R is between A and D; S is between D and C
Galvanometer G is between B and D

Assign currents:

$I_1$ through P (A→B)
$I_2$ through R (A→D)
$I_g$ through G (B→D, assuming B is at higher potential)
$I_1 - I_g$ through Q (B→C)
$I_2 + I_g$ through S (D→C)

Total current: $I = I_1 + I_2$ (from battery)

Loop ABD (A → B via P, then B → D via G):

I_1 P + I_g G - I_2 R = 0

10I_1 + 20I_g - 6I_2 = 0 \quad \ldots (1)

Loop BDC (B → D via G, then D → C via S, and B → C via Q):

(I_1 - I_g)Q - (I_2 + I_g)S - I_g G = 0

Wait — let me re-apply KVL carefully for loop BDCB (going around the inner loop):

Starting at B: go B→D via G: drop $I_g \cdot G$ ; go D→C via S: drop $(I_2 + I_g)S$ ; go C→B via Q (reversed — going against current): gain $(I_1 - I_g)Q$ .

-I_g G - (I_2 + I_g)S + (I_1 - I_g)Q = 0

-20I_g - 10(I_2 + I_g) + 15(I_1 - I_g) = 0

15I_1 - 10I_2 - 45I_g = 0

3I_1 - 2I_2 - 9I_g = 0 \quad \ldots (2)

Outer loop (A→B via P → B→C via Q → C to A via battery):

\varepsilon = I_1 P + (I_1 - I_g) Q

6 = 10I_1 + 15(I_1 - I_g)

6 = 25I_1 - 15I_g \quad \ldots (3)

From equations (1), (2), (3):

From (1): $10I_1 - 6I_2 = -20I_g \implies 5I_1 - 3I_2 = -10I_g \quad (1')$

From (2): $3I_1 - 2I_2 = 9I_g \quad (2)$

Multiply (2) by 1.5: $4.5I_1 - 3I_2 = 13.5I_g \quad (2')$

Subtract (1’) from (2’): $(4.5 - 5)I_1 = 13.5I_g + 10I_g = 23.5I_g$

$-0.5I_1 = 23.5I_g \implies I_1 = -47I_g$

Hmm, let me redo this more carefully. I’ll use $I_1$ and $I_2$ from equation (3) and substitute.

From (3): $25I_1 - 15I_g = 6 \implies I_1 = \frac{6 + 15I_g}{25} \quad (3')$

We need another relation. Use the outer loop through R and S:

\varepsilon = I_2 R + (I_2 + I_g)S = 6I_2 + 10(I_2 + I_g) = 16I_2 + 10I_g

6 = 16I_2 + 10I_g \implies I_2 = \frac{6 - 10I_g}{16} \quad (4)

Substitute (3’) and (4) into (1):

10 \cdot \frac{6 + 15I_g}{25} + 20I_g - 6 \cdot \frac{6 - 10I_g}{16} = 0

\frac{2(6 + 15I_g)}{5} + 20I_g - \frac{3(6 - 10I_g)}{8} = 0

Multiply through by 40 (LCM of 5 and 8):

16(6 + 15I_g) + 800I_g - 15(6 - 10I_g) = 0

96 + 240I_g + 800I_g - 90 + 150I_g = 0

6 + 1190I_g = 0

I_g = -\frac{6}{1190} = -\frac{3}{595} \approx -0.00504 \text{ A}

The negative sign means current flows from D→B (opposite to assumed direction). Magnitude:

|I_g| = \frac{3}{595} \approx 5.04 \times 10^{-3} \text{ A} \approx 5 \text{ mA}

Why This Works

Kirchhoff’s laws encode the two fundamental conservation laws:

KCL (junction rule): Conservation of charge — current in = current out at every node
KVL (loop rule): Conservation of energy — sum of potential differences around any loop = 0

For a Wheatstone bridge, the balanced condition ( $P/Q = R/S$ ) is a special case where $I_g = 0$ — derived by setting the potential at B and D equal. When this ratio doesn’t hold, we must use the full Kirchhoff analysis.

Alternative Method — Thevenin’s Theorem

For an unbalanced Wheatstone bridge, Thevenin’s theorem is often faster:

Remove the galvanometer
Find the open-circuit voltage $V_{BD}$ (Thevenin voltage)
Find the equivalent resistance $R_{th}$ between B and D (with battery replaced by wire)
Current: $I_g = \frac{V_{BD}}{R_{th} + G}$

$V_{BD} = V_B - V_D = I_1 P - I_2 R$ (with galvanometer removed, $I_1 = \frac{6}{P+Q} = \frac{6}{25} = 0.24$ A, $I_2 = \frac{6}{R+S} = \frac{6}{16} = 0.375$ A)

$V_{BD} = 0.24 \times 10 - 0.375 \times 6 = 2.4 - 2.25 = 0.15$ V

$R_{th} = \frac{PQ}{P+Q} + \frac{RS}{R+S} = \frac{10 \times 15}{25} + \frac{6 \times 10}{16} = 6 + 3.75 = 9.75\ \Omega$

$I_g = \frac{0.15}{9.75 + 20} = \frac{0.15}{29.75} \approx 5.04 \times 10^{-3}$ A ✓

Thevenin’s theorem is significantly faster for Wheatstone bridge problems. JEE Main prefers Kirchhoff’s for shown working, but Thevenin gives the same answer in fewer steps. Learn both — Thevenin for quick MCQ checks, Kirchhoff for long-answer proofs.

Common Mistake

Students often assume a Wheatstone bridge is balanced and set $I_g = 0$ without checking the balance condition first. Always check $P/Q = R/S$ before concluding the bridge is balanced. Here, $10/15 \neq 6/10$ , so $I_g \neq 0$ . Treating an unbalanced bridge as balanced gives completely wrong answers.

Question

Solution — Step by Step

Why This Works

Alternative Method — Thevenin’s Theorem

Common Mistake

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