1.

In the circuit shown in figure . (a) find the current flowing through the 100Omega resistor connecting points U and S.

Answer»

Solution :Figure (b) shows simplified CIRCUIT . Thebattery is directly attached to resistor `90Omega` HENCE current in it is 2A, SEE figure (c ), The total resistance of SECOND branch is also `90Omega` , hence current divides equally .Nowcurrent through 45 `Omega` , resistor is 2Aand it is a combination of two EQUAL `90Omega` resistors.. Once again current divides equally . `90Omega` resistor is a series combination of `40Omega` and `50Omega` , hence current through them is equal , i.e .,


1 A . As `50Omega` resistor is a parallel combination of two equal `100Omega` resistors , they must have the same currenti.e., 0.5 A


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