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I am studying for a test, and I am stuck on a Linear Algebra problem that involves some knowledge of currents.

Here's the textbook problem:

The Wheatstone Bridge is used to determine an unknown resistance $R_3$ by adjusting a known resistance $R_1$, so that the measured current (in $r$) is zero. Determine the formula for $R_3$ using the linear system for the loop currents.

Here is the given image: WheatStone Bridge

I know that the point of balance (when $r = 0$) is $R_3/R_1 = R_4/R_2$.

Thus, $R_3 = R_4R_1/R_2$.

Using Kirchoff's Current Law, I started getting a linear system.

$$I_1 R_1 - I_rr - I_3R_3 = 0$$

$$I_2 R_2 - I_4R_4+I_rr = 0$$

Which would also represent the point of balance.

Though, I don't know how to rearrange the linear system so that it equals $R_3$. Am I correct so far?

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1 Answer 1

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The point of balance always occurs when $R_3 = R_4R_1/R_2$, no matter what the resistance $r$ is. In fact, $r$ plays essentially no role here. A derivation is given below. First let me label four points in your diagram A, B, C, and D.

circuit

Let us assume the resistance $R_1$ is adjusted so that we are at the balancing point, i.e., that $I_r = 0$. This means exactly that the voltage $V_B$ at $B$ has to be the same as the voltage $V_C$ at $C$. We know that

  1. $V_B - V_A = I_2R_2$.
  2. $V_C - V_A = -I_4R_4$ (the negative comes from the choice of direction of current in our diagram).
  3. $V_D - V_B = I_1R_1$.
  4. $V_D - V_C = -I_3R_3$.

Since $V_B = V_C$, equations (1) and (2) combine to give $I_2R_2 = -I_4R_4$, and equations (3) and (4) combine to give $I_1R_1 = -I_3R_3$. These are two equations in our system.

Another law we need to use is that the total current leaving any given point in the circuit is $0$. At point B, this law translates to $I_2 = I_1$, since by assumption $I_r = 0$. Similarly, at point C, this law gives $I_3 = I_4$. Putting this all together, we have four equations:

  1. $I_2R_2 = -I_4R_4$.
  2. $I_1R_1 = -I_3R_3$.
  3. $I_2 = I_1$.
  4. $I_3 = I_4$.

Plugging equations (3) and (4) into (2) gives $I_2R_1 = -I_4R_3$. On the other hand, using equation (1) we can solve for $I_4$ to get $I_4 = -I_2R_2/R_4$. Combining these gives $$I_2R_1 = -(-I_2R_2/R_4)R_3,$$ which, after rearranging, will give $R_3 = R_1R_4/R_2$.

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Okay, great. And the equations you found are considered a "linear system" for the loop currents? –  Grace C Oct 17 '12 at 3:34
    
This is a fantastic answer. –  Grace C Oct 17 '12 at 5:58

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