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**Calculating a Bridge-Circuit with load**

In a loaded bridge circuit, a current between points A and B must be considered. One way to calculate this circuit is to consider the resistor R_{5} as a load resistor, which is connected to a voltage source with internal resistance R_{i}:

Specific example with the following values:

U = U_{0} = 12 V

R_{1} = 400Ω, R_{2} = 800Ω, R_{3} = 600Ω, R_{4} = 300Ω, R_{5} = 500Ω

**Step 1:**** Calculating U**_{AB} with no load, that means no current between the points A and B

U_{2} = R_{2} / (R_{1}+R_{2}) = 800Ω / 1200Ω * 12V = 8V

U_{4} = R_{4} / (R_{3}+R_{4}) = 300Ω / 900Ω * 12V = 4V

=> U_{AB} = 8 V - 4 V = 4 V Δ U_{0}, if no current flows, i.e. R_{5} is not connected!

**Step 2:**** Calculate R _{i}**

If we consider the voltage source from figure a to be ideal, it has no internal resistance and can be short-circuited. We get the equivalent circuit as shown and can calculate the internal resistance:

R_{total} = Ri = R1 || R2 + R3 || R4 = 466.7Ω

**Step 3:**** Calculate the terminal voltage**

Therefore we consider pic b.

U_{AB} / R_{5} = U_{0} / (R_{5} + R_{i})

=> U_{AB} = 2,07 V