A Wheatstone bridge ABCD (labeled in clockwise direction) with a galvanometer G joined to the junctions B and D. A cell E is connected across AC. Resistances of 15, 10, 4 and X ohms are inserted into the arms AB, BC, CD and DA respectively. The value of X is _________ ohms.
R1=15 ohms R2= 10 ohms R3= 4 ohms First, Kirchhoff's first rule is used to find the currents in junctions B and D:
B) I1 - I2 - Ig = 0 ( Ig - current through the galvanometer) D) Ix - I3 - Ig = 0 ( Ix - current trough Rx)
Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD: ABD) I1*R1 + Ix*Rx + Ig*Rg = 0 BCD) I2*R2 + I3*R3+ Ig*Rg = 0
When the bridge is balanced, then Ig = 0, so the second set of equations can be rewritten as: I1*R1 = - Ix*Rx I2*R2 = - I3*R3 and I1=I2 Ix=I3 From the first rule and the second rule we get: