Question
Explain with a neat circuit diagram. How you will determine the unknown resistances using a meter bridge.
✓
Answer
Construction:
i. Metrebridge consists of a one-metre long wire of uniform cross-section, stretched on a metre scale which is fixed on a wooden table.
ii. The ends of the wire are fixed below two $L$ shaped metallic strips. A single metallic stripe separates the two L-shaped strips leaving two gaps, left gap and right gap.
iii. iii. Usually, an unknown resistance $X$ is connected in the left gap and a resistance box is connected in the other gap.
iv. One terminal of a galvanometer is connected to point $C$ on the central strip, while the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire $A B$ can be established with the help of the jockey.
v. A cell of emf E along with a key and a rheostat is connected between points $A$ and $B$.
Working:
i. A suitable resistance $R$ is selected from the resistance box.
ii. The jockey is brought in contact with $A B$ at various points on the wire $A B$ and the balance point (null point), $D$ is obtained. The galvanometer shows no deflection when the jockey is at the balance point (point $D$ ).
iii. Let the respective lengths of the wire between $A$ and $D$, and that between $D$ and $C$ be $I_x$ and $I_R$.
iv. Then using the balancing conditions,
$\frac{ X }{ R }=\frac{ R _{ AD }}{ R _{ DB }} \text {. }$
where $R_{A D}$ and $R_{D B}$ are a resistance of the parts $A D$ and $D B$ of the wire respectively.
$v$. If $I$ is the length of the wire, $\rho$ is its specific resistance, and $A$ is its area of cross-section then
$ R _{ AD }=\frac{\rho l_{ x }}{ A }$
$R _{ DB }=\frac{\rho l_{ R }}{ A } $
From equations (1), (2) and (3),
$ \frac{ X }{ R }=\frac{ R _{ AD }}{ R _{ DC }}=\frac{\rho l_{ x } / A }{\rho l_{ R } / A }$
$\therefore \frac{ X }{ R }=\frac{l_{ x }}{l_{ R }}$
$\therefore X =\frac{l_{ x }}{l_{ R }} R $
Thus, knowing $R, I_x$ and $I_R$, the value of the unknown resistance can be determined.
i. Metrebridge consists of a one-metre long wire of uniform cross-section, stretched on a metre scale which is fixed on a wooden table.
ii. The ends of the wire are fixed below two $L$ shaped metallic strips. A single metallic stripe separates the two L-shaped strips leaving two gaps, left gap and right gap.
iii. iii. Usually, an unknown resistance $X$ is connected in the left gap and a resistance box is connected in the other gap.
iv. One terminal of a galvanometer is connected to point $C$ on the central strip, while the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire $A B$ can be established with the help of the jockey.
v. A cell of emf E along with a key and a rheostat is connected between points $A$ and $B$.
Working:
i. A suitable resistance $R$ is selected from the resistance box.
ii. The jockey is brought in contact with $A B$ at various points on the wire $A B$ and the balance point (null point), $D$ is obtained. The galvanometer shows no deflection when the jockey is at the balance point (point $D$ ).
iii. Let the respective lengths of the wire between $A$ and $D$, and that between $D$ and $C$ be $I_x$ and $I_R$.
iv. Then using the balancing conditions,
$\frac{ X }{ R }=\frac{ R _{ AD }}{ R _{ DB }} \text {. }$
where $R_{A D}$ and $R_{D B}$ are a resistance of the parts $A D$ and $D B$ of the wire respectively.
$v$. If $I$ is the length of the wire, $\rho$ is its specific resistance, and $A$ is its area of cross-section then
$ R _{ AD }=\frac{\rho l_{ x }}{ A }$
$R _{ DB }=\frac{\rho l_{ R }}{ A } $
From equations (1), (2) and (3),
$ \frac{ X }{ R }=\frac{ R _{ AD }}{ R _{ DC }}=\frac{\rho l_{ x } / A }{\rho l_{ R } / A }$
$\therefore \frac{ X }{ R }=\frac{l_{ x }}{l_{ R }}$
$\therefore X =\frac{l_{ x }}{l_{ R }} R $
Thus, knowing $R, I_x$ and $I_R$, the value of the unknown resistance can be determined.
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