Question

Two resistances 2 ohm and 3
ohm are connected across the two gaps of the metre bridge as shown in figure. Calculate
the current through the cell when the bridge is balanced and the specific resistance of the material of the metre bridge wire. Given the resistance of the bridge wire is 1.49 ohm and its diameter is 0.12 cm.​

Answer 1

Given :

Two resistances of 2 ohm and 3 ohm are connected across the two gaps of the metre bridge.

• Resistance of wire = 1.49 Ω
• Diameter = 0.12 cm

Potential difference = 12V

To Find :

Current flow in the circuit and specific resistance of the bridge wire.

Solution :

❒ Metre bridge is the important application of wheatstone bridge. Some other applications are Carey - Foster's bridge and Post office box.

• In balanced condition of metre bridge, no current flow through galvanometer so we can neglect resistance of galvanometer in calculation.

In order to find current flow, first we have to find equivalent resistance of the circuit.

★ Two resistors of 2Ω and 3Ω are connected in series. Therefore equivalent resistance of upper branch will be 2 + 3 = 5Ω

Finally 5Ω and 1.49Ω come in parallel.

$\sf:\implies\:\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{1.49}$

$\sf:\implies\:\dfrac{1}{R}=\dfrac{5+1.49}{5\times 1.49}$

$\sf:\implies\:\dfrac{1}{R}=\dfrac{6.49}{7.45}$

$\bf:\implies\:R=1.15\:{\Omega}$

As per ohm's law,

$\sf:\implies\:V=I\times R$

$\sf:\implies\:I=\dfrac{2}{1.15}$

$:\implies\:\underline{\boxed{\bf{\purple{I=1.74\:A}}}}$

★ Specific resistance of wire :

Radius of wire = 0.12/2 = 0.06 cm

• 0.06 cm = 0.06 × $\sf{10^{-2}}$

We know that,

$\sf:\implies\:R=\rho\times\dfrac{l}{A}$

$\sf:\implies\:\rho=\dfrac{RA}{l}$

$\sf:\implies\:\rho=\dfrac{R\times \pi r^2}{l}$

• Length of wire = 1 m

$\sf:\implies\:\rho=\dfrac{1.15\times3.14\times(0.06\times 10^{-2})^2}{1}$

$\sf:\implies\:\rho=3.61\times(6\times 10^{-4})^2$

$\sf:\implies\:\rho=3.61\times 36\times10^{-8}$

$\sf:\implies\:\rho=129.96\times10^{-8}$

$:\implies\:\underline{\boxed{\bf{\orange{\rho=1.29\times 10^{-6}\:{\Omega}\cdot m}}}}$

Answer 2

Given :

Two resistances of 2 ohm and 3 ohm are connected across the two gaps of the metre bridge.

Resistance of wire = 1.49 Ω

Diameter = 0.12 cm

Potential difference = 12V

To Find :

Current flow in the circuit and specific resistance of the bridge wire.

Solution :

❒ Metre bridge is the important application of wheatstone bridge. Some other applications are Carey - Foster's bridge and Post office box.

In balanced condition of metre bridge, no current flow through galvanometer so we can neglect resistance of galvanometer in calculation.

In order to find current flow, first we have to find equivalent resistance of the circuit.

★ Two resistors of 2Ω and 3Ω are connected in series. Therefore equivalent resistance of upper branch will be 2 + 3 = 5Ω

Finally 5Ω and 1.49Ω come in parallel.