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# In  a wheatstone bridge the 4 resistance arms of the bridge are ab=2 ohm bc=3ohm cd= 4ohm and da =1 ohm .a cell of emf 2volt and negligible internal resistance is connected across ac.and a galvanometer of resistance 20 ohm is connected between b and d.find the current through the galvanometer.

Asked by raghubrnmurthy 29th May 2021, 12:04 PM
Figure shows the wheatstone bridge in which resistances are connected as per the question.
Current distribution in the netwrok is as shown in figure.

Let us apply Kirchoff's voltage law to the closed loop adcfa , we get

i2 + 4 ( i2 + i3 ) = 2    or   5 i2 + 4 i3 =  2  ...........................(1)

Let us apply Kirchoff's voltage law to the closed loop abcfe , we get

2 i1 + 3 i4 = 2   ........................... (2)

at node-b , by Kirchoff's current law ,  we have , i1 = i3 + i4  ..........................(3)

Let us apply Kirchoff's voltage law to the closed loop bcdb , then we get

3 i4 - 4 ( i2 + i3 ) - 20 i3 = 0    or    i4 = ( 4/3 ) i2 + 8 i3 .......................(4)

Let us substitute for i1 using eqn.(3), and rewrite eqn.(2) as

2 i3 + 5 i4 = 2  .................................(5)

Let us substitute i4 using eqn.(4) and rewrite eqn.(5) as ,

42 i3 + (20/3) i2  = 2  or      126 i3 + 20 i2 =  6  .............................(6)

By solving equations (1) and (6) , we get i2 = 0.415 A  and i3 = -0.0182 A

Hence current through Galvanometer = 18.2 mA and current direction is from the node-d to node-a

Answered by Expert 29th May 2021, 4:18 PM
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