3.14 WHEATSTONE BRIDGE

3.25

Wheatstone Bridge

As an application of Kirchhoff’s rules consider the circuit shown in Fig. 3.25, which is called the Wheatstone bridge. The bridge has four resistors $R_{1}$ , $R_{2}$ , $R_{3}$ and $R_{4}$ . Across one pair of diagonally opposite points ( $A$ and $C$ in the figure) a source is connected. This (i.e., $A C$ ) is called the battery arm. Between the other two vertices, $B$ and $D$ , a galvanometer $G$ (which is a device to detect currents) is connected. This line, shown as $B D$ in the figure, is called the galvanometer arm.

For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current $I_{g}$ through $G$ . Of special interest, is the case of a balanced bridge where the resistors are such that $I_{g} = 0$ . We can easily get the balance condition, such that there is no current through $G$ . In this case, the Kirchhoff’s junction rule applied to junctions $D$ and $B$ (see the figure) immediately gives us the relations $I_{1} = I_{3}$ and $I_{2} = I_{4}$ . Next, we apply Kirchhoff’s loop rule to closed loops $A D B A$ and $C B D C$ . The first loop gives

$- I_{1} R_{1} + 0 + I_{2} R_{2} = 0 (I_{g} = 0)$ (3.81)

and the second loop gives, upon using $I_{3} = I_{1}$ , $I_{4} = I_{2}$

$I_{2} R_{4} + 0 - I_{1} R_{3} = 0$ (3.82)

From Eq. (3.81), we obtain, $\frac{I_{1}}{I_{2}} = \frac{R_{2}}{R_{1}}$ whereas from Eq. (3.82), we obtain, $\frac{I_{1}}{I_{2}} = \frac{R_{4}}{R_{3}}$ Hence, we obtain the condition

$\frac{R_{2}}{R_{1}} = \frac{R_{4}}{R_{3}}$ (3.83(a))

This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.

The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance. Let us suppose we have an unknown resistance, which we insert in the fourth arm; $R_{4}$ is thus not known. Keeping known resistances $R_{1}$ and $R_{2}$ in the first and second arm of the bridge, we go on varying $R_{3}$ till the galvanometer shows a null deflection. The bridge then is balanced, and from the balance condition the value of the unknown resistance $R_{4}$ is given by,

$R_{4} = R_{3} \frac{R_{2}}{R_{1}}$ (3.83(b))

A practical device using this principle is called the meter bridge. It will be discussed in the next section.

Example 3.8

The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances:
$A B = 100 Ω$ , $B C = 10 Ω$ , $C D = 5 Ω$ and $D A = 60 Ω$

3.26

A galvanometer of $15 Ω$ resistance is connected across $B D$ . Calculate the current through the galvanometer when a potential difference of $10 V$ is maintained across $A C$ .

VIEW SOLUTION

Considering the mesh $B A D B$ , we have

$\begin{array}{l} 100 I_{1} + 15 I_{g} - 6 O I_{2} = 0 \\ or 20 I_{1} + 3 I_{g} - 12 I_{2} = 0 \end{array}$ (3.84(a))

Considering the mesh $B C D B$ , we have

$\begin{array}{l} 10 (I_{1} - I_{g}) - 15 I_{g} - 5 (I_{2} + I_{g}) = 0 \\ 10 I_{1} - 30 I_{g} - 5 I_{2} = 0 \\ 2 I_{1} - 6 I_{a} - I_{2} = 0 \end{array}$ (3.84(b))

Considering the mesh $A D C E A$ ,

$\begin{array}{l} 60 I_{2} + 5 (I_{2} + I_{g}) = 10 \\ 65 I_{2} + 5 I_{g} = 10 \\ 13 I_{2} + I_{g} = 2 \end{array}$ (3.84(c))

Multiplying Eq. (3.84b) by $10$

$20 I_{1} - 60 I_{g} - 10 I_{2} = 0$ (3.84(d))

From Eqs. (3.84d) and (3.84a) we have

$\begin{array}{l} 63 I_{g} - 2 I_{2} = 0 \\ I_{2} = 31.5 I_{g} \end{array}$ (3.84(e))

Substituting the value of $I_{2}$ into Eq. [3.84(c)], we get $\begin{array}{l} 13 (31.5 I_{g}) + I_{g} = 2 \\ 410.5 I_{g} = 2 \\ I_{g} = 4.87 mA \end{array}$