1.

Find the value of R in fig. so that there is no current in the 15 Omega resistor

Answer»

Solution :This is the WHEATSTONE bridge with the galvanometer replaced by `15 OMEGA` RESISTOR. The bridge is balanced because there is no CURRENT is 15 `Omega` resistor, hence,
`20//10 = 40//R`
`THEREFORE R= (40 xx 10 Omega)/(20) = 20 Omega`


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