(a) In a meter bridge , balance point is found to be a 39.5 cm from end A , when resistor Y is of 12

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3.Current Electricity

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$(a)$ In a meter bridge , the balance point is found to be a $39.5\; cm$ from the end $A$, when the resistor $Y$ is of $12.5\; \Omega$ Determine the resistance of $X .$ Why are the connections between resistors in a Wheatstone or meter bridge made of thick coppestrips?

$(b)$ Determine the balance point of the bridge above if $X$ and $Y$ arinterchanged.

$(c)$ What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer shot any current?

Option A

Option B

Option C

Option D

Solution

A metre bridge with resistors $X$ and $Y$ is represented in the given figure.

$(a)$ Balance point from end $A, l_{1}=39.5\, cm$

Resistance of the resistor $Y=12.5\, \Omega$

Condition for the balance is given as,

$\frac{X}{Y}=\frac{100-l_{1}}{l_{1}}$

$X=\frac{100-39.5}{39.5} \times 12.5=8.2 \,\Omega$

Therefore, the resistance of resistor $X$ is $8.2\, \Omega .$

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

$(b)$ If $X$ and $Y$ are interchanged, then $l_{1}$ and $100-l_{1}$ get interchanged.

The balance point of the bridge will be $100-l_{1}$ from $A.$

$100-l_{1}=100-39.5=60.5 \,cm$

Therefore, the balance point is $60.5\, cm$ from $A.$

$(c)$ When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

Standard 12

Physics

Wheatstone bridge principle is used to measure the specific resistance $\left(S_1\right)$ of given wire, having length $L$, radius $r$. If $X$ is the resistance of wire, then specific resistance is: $S_1=X\left(\frac{\pi r^2}{L}\right)$. If the length of the wire gets doubled then the value of specific resistance will be :

hard

(JEE MAIN-2024)

In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced

medium

Fig. shows rough sketch of meter bridge. $(G)$ deflects zero at length $\ell \, cm$. Now $R_1$ and $R_2$ are interchanged then balancing length increases by $25\, cm$. Find $R_1/R_2$

medium

The figure shows a meter-bridge circuit, with $AB = 100\, cm$,$ X = 12\,\Omega$ and $R = 18\,\Omega$ , and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$ , through what distance will $J$ have to be moved to obtain balance? ………….. $cm$

hard

In the circuit shown $P \ne R$, the reading of the galvanometer is same with switch $S$ open or closed. Then

easy

(IIT-1999)

Solution

Similar Questions

In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced

Fig. shows rough sketch of meter bridge. $(G)$ deflects zero at length $\ell \, cm$. Now $R_1$ and $R_2$ are interchanged then balancing length increases by $25\, cm$. Find $R_1/R_2$

The figure shows a meter-bridge circuit, with $AB = 100\, cm$,$ X = 12\,\Omega$ and $R = 18\,\Omega$ , and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$ , through what distance will $J$ have to be moved to obtain balance? ………….. $cm$

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