The four arms ABCD of a Wheatstones network have the following resistances: AB = 2 Ω, BC=4 Ω, CD= 4 - Physics - Current Electricity

Question

The four arms ABCD of a Wheatstones network have the following
resistances:
AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω A
galvanometer of resistance 10 Ω is
connected between B and D Find the current through the
galvanometer, when
the potential difference between A and C is 5 V

Abhay Kumar · Accepted Answer

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Applying Kirchoff's current law at node B we get;
(V1-5)/2 + (V2-V1)/10 -V1/4 = 0

Applying Kirchoff's current law at node D we get;
(V2-5)/8 + (V1-V2)/10 -V2/4 = 0

from here we get the values of V2 and V1 as 14 28 volt , 3 57
volt
Current comes out to be: (â€‹14 28 volt - 3 57
volt)/10 =1 071 ampere

Sorry, for late reply!

The four arms ABCD of a Wheatstones network have the following resistances: AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω. A galvanometer of resistance 10 Ω is connected between B and D. Find the current through the galvanometer, when the potential difference between A and C is 5 V.

The four arms ABCD of a Wheatstones network have the following resistances:

AB = 2 Ω, BC=4 Ω, CD= 4 Ω and DA= 8 Ω. A galvanometer of resistance 10 Ω is

connected between B and D. Find the current through the galvanometer, when

the potential difference between A and C is 5 V.