Question

a wheatstone bridge ABCD is arranged as follows : A B is =10 ohm , BC=30ohm, CD =15ohm and DA=20ohm A 2 volt battery of internal resistance 2 ohm is connected between A and C with A positive. A galvanometer of resistance 40 ohm is connected between B and D. Find the magnitude and direction of galvanometer current​

Answer 1

Answer:

A Wheatstone bridge ABCD is arranged as  A B is =10 ohm , BC=30ohm, CD =15ohm and DA=20ohm A 2 volt battery of internal resistance 2 ohm is connected between A and C with A positive. A galvanometer of resistance 40 ohm is connected between B and D. The magnitude of galvanometer current​ is $I_{g}=11.4mA$

Explanation:

Wheatstone bridge is used to measure an unknown resistance, and the galvanometer is used to detect an unbalanced condition in the Wheatstone bridge.

To measure the current through the galvanometer we can use Thevenin's theorem, which helps to replace the complicated network with a voltage source with one resistance in series.

The galvanometer is connected between B and D

Let the potential across the galvanometer be $V_{BD}=V_{B} -V_{D}$

where $V_{B}$ and $V_{D}$ are potentials at B and D.

From Thevenin's theorem $V_{BD}$ is the Thevenin's voltage that is,  $V_{th}=V_{BD}$

(because we need to find current along the path B and D where galvanometer is connected)

From the figure

$V_{BD}=V_{B} -V_{D}=I_{1} R_{1} -I_{2} R_{2}$

$I_{1} =\frac{V}{R_{1} +R_{3} } \\I_{2} =\frac{V}{R_{2} +R_{4} }$ where $V$ is the voltage source

using these values we get $V_{th}$ as

$V_{th}=V(\frac{R_{1} }{R_{1} +R_{2} }-\frac{R_{2} }{R_{2} +R_{4} } )$

substitute the values of resistors from the figure

$V_{th}=2(\frac{20}{35}-\frac{10}{40} )=0.64V$

next, we need to calculate Thevenin's resistance,

$R_{th}=R_{1} //R_{3}+R_{2}//R_{4}$

$R_{th}=\frac{R_1R_3}{R_1+R_3}+\frac{R_2R_4}{R_2+R_{4} }\c$$R_{th}=\frac{20\times 15}{35}+\frac{30\times 10}{40}=16.07\Omega$

Now, the current through the galvanometer is

$I_g=\frac{v_{th}}{R_{th}+R_g}$

From the question resistance of the galvanometer is

$R_{g}=40\Omega$

The galvanometer current is $I_{g}=11.4mA$