Question
What is Wheatstone bridge? Explain its principle.
✓
Answer
$\rightarrow $ The circuit shown in the figure is called the wheatstone bridge. It uses four resistors $R_1, R_2$, $R _3$ and $R _4$ out of them three resistors are known and one is unknown, wheatstone bridge is used to find the value of unknown resistance.
$\rightarrow$ As shown in the figure, across one pair of diagonally opposite points $(A$ and $C$ in the figure$)$ a source is connected hence $AC$ is called the battery arm.
$\rightarrow$ Between the other two vertices $, B$ and $D,$ a galvanomenter $G$ is connected hence $BD$ is called the galvanometer arm.
$\rightarrow$ When battery is connected, the currents flowing through the resistors $R_1, R_2, R_3$ and $R_4$ are $I_1$, $I _2, I _3$ and $I _4$ respectively.
$\rightarrow$ Here, there resistors are chosen in such a way that current flowing through galvanometer is zero $\left(I_g=0\right.$ ).
$\rightarrow$ When the current flowing through the galvanometer becomes zero, the bridge is said to be in balanced condition.
$\rightarrow$ From the figure, in balanced condition
$I _1= I _3 $ and $I _2= I _4$
$\rightarrow $ Applying Kirchhoff's loop rule to closed loop
$A - D - B - A$
$- I _1 R _1+0+ I _2 R _2=0$
$\therefore I _1 R _1= I _2 R _2$
$\rightarrow$ Applying similarly, for closed loop $C - B - D - C$
$I _4 R _4+0- I _3 R _3=0$
$\therefore I _3 R _3= I _4 R _4$
$\rightarrow$ Taking ratio of equation $(1)$ and $(2)$
$\therefore \frac{ I _1 R _1}{ I _3 R _3}=\frac{ I _2 R _2}{ I _4 R _4}$
But $I _1= I _3$ and $I _2= I _4$
$\therefore \frac{ R _1}{ R _3}=\frac{ R _2}{ R _4}$ OR $\frac{ R _1}{ R _2}=\frac{ R _3}{ R _4}$
$\rightarrow$ which is a condition for the whetstone bridge to be in balanced condition.
$\rightarrow$ If three resistors $R_1, R_2$ and $R _3$ are known then unknown resistence of $R_4$ is given by
$R _4= R _3 \cdot \frac{ R _2}{ R _1}$
$\rightarrow A$ practical device using this principle is called the meter bridge.
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