Wheatstone Bridge

A wheatstone bridge circuit can be drawn as shown below.

To find $R_{4}$ as a function of the other values, we first find $V_{g}$ as a function of the other values using KVL on the loop consisting of the positive terminal of the battery, $R_{1}$ , the voltmeter, $R_{3}$ , and the negative end of the battery and the loop consisting of the positive end of the battery, $R_{1}$ , $R_{4}$ , and the negative end of the battery.

{V_{b} - V_{1} - V_{g} - V_{3} = 0 V_{b} - V_{1} - V_{4} = 0

Substituting, we get

V_{g} = V_{4} - V_{3}

Now we want to find $V_{4}$ and $V_{3}$ as a function of the resistances and $V_{b}$ .

⎩ ⎨ ⎧ V_{b} = I_{1} (R_{1} + R_{4}) V_{4} = I_{1} R_{4} V_{b} = I_{2} (R_{2} + R_{3}) V_{3} = I_{2} R_{3}

Substituting, we get

V_{4} V_{3} = V_{b} \cdot \frac{R _{4}}{R _{1} + R _{4}} = V_{b} \cdot \frac{R _{3}}{R _{2} + R _{3}}

Our final formula for the reading of the voltmeter is

V_{g} = V_{b} (\frac{R _{4}}{R _{1} + R _{4}} - \frac{R _{3}}{R _{2} + R _{3}})

Now, solving for $R_{4}$ , the “unknown” resistance, we get

\frac{V _{g}}{V _{b}} \frac{R _{4}}{R _{1} + R _{4}} R_{4} R_{4} (1 - \frac{V _{g}}{V _{b}} - \frac{R _{3}}{R _{2} + R _{3}}) R_{4} = \frac{R _{4}}{R _{1} + R _{4}} - \frac{R _{3}}{R _{2} + R _{3}} = \frac{V _{g}}{V _{b}} + \frac{R _{3}}{R _{2} + R _{3}} = (R_{1} + R_{4}) (\frac{V _{g}}{V _{b}} + \frac{R _{3}}{R _{2} + R _{3}}) = R_{1} (\frac{V _{g}}{V _{b}} + \frac{R _{3}}{R _{2} + R _{3}}) = \frac{R _{1} ( \frac{V _{g}}{V _{b}} + \frac{R _{3}}{R _{2} + R _{3}} )}{1 - \frac{V _{g}}{V _{b}} - \frac{R _{3}}{R _{2} + R _{3}}} = \frac{R _{3} V _{b} - R _{2} V - R _{3} V}{R _{2} V _{b} + R _{2} V + R _{3} V}

🪴 m.y. notes

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Wheatstone Bridge

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