## Balanced Bridge

### Task number: 2169

A bridge containing resistors *R*_{1}, *R*_{2}, *R*_{3} and a resistor *R*_{4} whose resistance is temperature dependent, is balanced. Calculate the temperature *t* of resistor *R*_{4}.

At 0 °C (so-called reference temperature) resistor *R*_{4} has resistance of 4.5 kΩ, temperature coefficient of resistance equals 0.004 °C^{-1} and the values of the other resistors are
*R*_{1} = 2.0 kΩ;
*R*_{2} = 2.3 kΩ;
*R*_{3} = 4.0 kΩ.

#### Hint 1

Recall or look up the meaning of “balanced bridge”.

#### Hint 2

How does one calculate the resistance of a temperature-dependent resistor?

#### Hint 3

What voltage is there between points

**A**and**B**if no current runs through the galvanometer? What does it imply about the electric potential of these points?#### Analysis

If the bridge is balanced, then there is no current running through galvanometer

**G**. These thus apply:-
Current running through resistor
*R*_{1}will be equal to the current running through resistor*R*_{2}(the same applies to resistors*R*_{3}and*R*_{4}). -
The voltage between points
**A**and**B**is zero. Voltage between two points is equal to the difference of their potential, thus nodes**A**and**B**have the same electric potential. Therefore, voltages on resistors*R*_{1}and*R*_{3}are equal (same goes for*R*_{2}and*R*_{4}).

Using Ohm’s law and these thoughts, we get a rule for a balanced bridge:

**the quotient of resistances of resistors**.*R*_{1}and*R*_{2}is equal to the quotient of resistances of resistors*R*_{3}and*R*_{4}The resistance of resistor

\[R_4\,=\,R_0\left(1\,+\,\alpha \Delta t\right),\]*R*_{4}is temperature-dependent. The resistance can be calculated using this relation:where

*R*is the resistor’s resistance at temperature*t*,*R*_{0}is the resistor’s resistance at initial (so-called reference) temperature*t*_{0},*α*is temperature coefficient of electric resistance and Δ*t*is the temperature difference Δ*t*=*t*_{0}.-
Current running through resistor
#### Solution

If the bridge is balanced, then no current runs through galvanometer

**G**. Thus nodes**A**and**B**will have the same electric potential and:**1.**Current running through resistor*R*_{1}is equal to current running through resistor*R*_{2}. Current running through resistor*R*_{3}is equal to current running through resistor*R*_{4}. Thus:**2.**Voltages on resistors*R*_{1}and*R*_{3}are the same because nodes**A**and**B**have the same electric potential. The same applies to resistors*R*_{2}and*R*_{4}.Ohm’s law says:

\[U_x\,=\,R_xI_x,\]where

*U*_{x}is the loss of voltage at an appliance,*R*_{x}is the resistance of the appliance and*I*_{x}is the current that runs through the appliance.Now we will formulate equations for resistors:

\[R_1I_1\,=\,R_3I_3\] \[R_2I_2\,=\,R_4I_4.\]We divide both equations:

\[\frac{R_1I_1}{R_2I_1}\,=\,\frac{R_3I_3}{R_4I_3}.\]Currents

\[\frac{R_1}{R_2}\,=\,\frac{R_3}{R_4}.\]*I*_{1}and*I*_{3}will be reduced in the fractions and we will receive the condition for a balanced bridge:Now we express resistance

\[R_4\,=\,\frac{R_2R_3}{R_1}.\tag{*}\]*R*_{4}from the equation:The resistance of resistor

\[R_4\,=\,R_0\left(1\,+\,\alpha \Delta t\right),\]*R*_{4}is temperature-dependent. This dependence is described by relationwhere

*R*is the resistor’s resistance at temperature*t*,*R*_{0}is the resistor’s resistance at initial (so-called reference) temperature*t*_{0},*α*is temperature coefficient of electric resistance and Δ*t*is the temperature difference Δ*t*=*t*_{0}.We will formulate from this equation the temperature difference Δ

\[1\,+\,\alpha\Delta t\,=\,\frac{R_4}{R_0}\] \[\alpha\Delta t\,=\,\frac{R_4}{R_0}-1\,=\,\frac{R_4-R_0}{R_0}\] \[\Delta t\,=\,t-t_0\,=\,\frac{\frac{R_4-R_0}{R_0}}{\alpha}\,=\,\frac{R_4-R_0}{\alpha R_0}.\]*t*:For temperature

\[t\,=\,\frac{R_4-R_0}{\alpha R_0}\,+\,t_0.\]*t*this applies:Now we will substitute for

\[t\,=\,\frac{\frac{R_2R_3}{R_1}-R_0}{\alpha R_0}\,+\,t_0,\]*R*_{4}using relation (*):which will give us the relation for temperature

\[t\,=\,\frac{R_2R_3\,-\,R_0R_1}{\alpha R_0R_1}\,+\,t_0.\]*t*of resistor*R*_{4}:#### List of known information and substitution

We know from the assignment:

*R*_{1}= 2,0 kΩresistances of resistors *R*_{1},*R*_{2},*R*_{3}of the bridge*R*_{2}= 2,3 kΩ*R*_{3}= 4,0 kΩ*R*_{0}= 4,5 k Ωresistance of resistor *R*_{4}at temperature 0 °Cα = 0,004 °C ^{-1}temperature coefficient of electric resistance of resistor *R*_{4}t = ? (°C) temperature of resistor R _{4}

**Note:**All resistances are given in kiloohms and the unit ohm cancels out, so we do not have to convert them.#### Answer

If the bridge is balanced, then the temperature of resistor

*R*_{4}is 5.5 °C.#### Other ways of solving this problem

Another possible, though more complicated, way of solving this problem is the usage of Kirchhoff’s laws. You can formulate the current running through the galvanometer under general conditions and then calculate when the current will be equal to zero.

Kirchhoff’s laws and a “cookbook“ of their usage can be found in Using Kirchhoff’s laws to solve circiut with two power supplies

#### Comment

Measuring bridges are used for measuring of electric capacity, resistance and inductance. Measuring with bridges is based on reaching balance between branches of the bridge. This balance is determined by galvanometer which is connected to one diagonal of the bridge. A current source is connected to the other, dc or ac depending on the measured quantity (ac for capacity and inductance, both ac and dc for resistance, dc being the better one).

Measuring with bridges is very precise but can be used only in a small range (depends on the bridge’s parameters).

Specially constructed devices are based on balanced bridges and are used for measuring electric resistances. They are called ohmmeters.