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# The Wheatstone Bridge

We can solve the problem using a potential divider or using loop currents.

Loop Currents:

We can assign loop currents to each loop as shown above. At balance the vector sum of the currents through the meter will be zero, we can therefore assign $I_2$to both the left and right loop, the currents will cancel through the meter.

Applying Kirchoff's voltage law to each loop we find that:

$\sum_{Voltages} Left Hand Loop = - (I_2 - I_1)R_1 - I_2 R_x = 0 $

$\sum_{Voltages}Right Hand Loop = -I_2 R_3 -(I_2 - I_1)R_2 = 0 $

We have two independent equations and two unknowns ($I_1$ and $I_2$).

From the left loop: $I_2 = \frac{R_1}{R_1 + R_x} I_1$

From the right loop: $I_2 = \frac{R_2}{R_3 + R_2} I_1$

Equating we see:

$R_1R_3 = R_2R_x$

Potential Divider:

At balance $V_b = V_d$

The potential at C is zero (ground). The potential at A is therefore divided between $R_x$ and $R_3$, in addition it is also divided between $R_1$and $R_2$ .

By potential divider:

$V_b = \frac{R_3}{R_x} V_a$

$V_d = \frac{R_2}{R_1}V_a$

Equating $V_b$ and $V_d$

$R_1 R_3 = R_2 R_x$

Extension:

If we replace:

$R_2 = Z_2$

We find $Z_2$ by combing the impedance of $C_2$ in parallel with $R_2$

$Z_2= \frac{R_2 \frac{1}{2 \pi f t C \bf i}} {R_2 + 2 \pi f t C \bf i }$

where $i = \sqrt{-1} = i$

$R_3 = Z_3 = $

We find $Z_3$ by combining the imperdance of $C_3$ in series with $R_3$

where $i = \sqrt{-1} = i$

From part 1 we know that:

$R_1Z_3 = Z_2R_x$

Substituting $Z_3$, $Z_4$ and equating real and imaginary terms we find that:

the real part tells us nothing about frequency (cancels)

The imaginary part tells us f = $\frac{1}{2 \pi} \sqrt{\frac{1}{C_3 C_2 R_3 R_2}} $ at balance

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Age 16 to 18

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We can solve the problem using a potential divider or using loop currents.

Loop Currents:

We can assign loop currents to each loop as shown above. At balance the vector sum of the currents through the meter will be zero, we can therefore assign $I_2$to both the left and right loop, the currents will cancel through the meter.

Applying Kirchoff's voltage law to each loop we find that:

$\sum_{Voltages} Left Hand Loop = - (I_2 - I_1)R_1 - I_2 R_x = 0 $

$\sum_{Voltages}Right Hand Loop = -I_2 R_3 -(I_2 - I_1)R_2 = 0 $

We have two independent equations and two unknowns ($I_1$ and $I_2$).

From the left loop: $I_2 = \frac{R_1}{R_1 + R_x} I_1$

From the right loop: $I_2 = \frac{R_2}{R_3 + R_2} I_1$

Equating we see:

$R_1R_3 = R_2R_x$

Potential Divider:

At balance $V_b = V_d$

The potential at C is zero (ground). The potential at A is therefore divided between $R_x$ and $R_3$, in addition it is also divided between $R_1$and $R_2$ .

By potential divider:

$V_b = \frac{R_3}{R_x} V_a$

$V_d = \frac{R_2}{R_1}V_a$

Equating $V_b$ and $V_d$

$R_1 R_3 = R_2 R_x$

Extension:

If we replace:

$R_2 = Z_2$

We find $Z_2$ by combing the impedance of $C_2$ in parallel with $R_2$

$Z_2= \frac{R_2 \frac{1}{2 \pi f t C \bf i}} {R_2 + 2 \pi f t C \bf i }$

where $i = \sqrt{-1} = i$

$R_3 = Z_3 = $

We find $Z_3$ by combining the imperdance of $C_3$ in series with $R_3$

where $i = \sqrt{-1} = i$

From part 1 we know that:

$R_1Z_3 = Z_2R_x$

Substituting $Z_3$, $Z_4$ and equating real and imaginary terms we find that:

the real part tells us nothing about frequency (cancels)

The imaginary part tells us f = $\frac{1}{2 \pi} \sqrt{\frac{1}{C_3 C_2 R_3 R_2}} $ at balance