Wheatstone Bridge Simulation
This interactive simulation demonstrates Example 3.8, showing current distribution in a Wheatstone bridge circuit.
Bridge Circuit Simulation
Battery (10V)
Resistor
Galvanometer (15Ω)
100Ω
10Ω
5Ω
60Ω
Galvanometer Current:
Calculating...
Bridge Condition:
For balanced bridge (Ig = 0):
\[ \frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}} \]
Current bridge ratio: 100/10 ≠ 60/5 (Unbalanced)
Solution to Example 3.8
Applying Kirchhoff's Rules:
Mesh BADB:
\[100I_1 + 15I_g - 60I_2 = 0\]
Simplified: \(20I_1 + 3I_g - 12I_2 = 0\)
Mesh BCDB:
\[10(I_1 - I_g) - 15I_g - 5(I_2 + I_g) = 0\]
Simplified: \(2I_1 - 6I_g - I_2 = 0\)
Mesh ADCEA:
\[60I_2 + 5(I_2 + I_g) = 10\]
Simplified: \(13I_2 + I_g = 2\)
Solving the Equations:
From equations (b) and (a):
\[ I_2 = 31.5I_g \]
Substituting into equation (c):
\[ 13(31.5I_g) + I_g = 2 \]
\[ 410.5I_g = 2 \]
\[ I_g = 4.87 \, \text{mA} \]