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

Explanation

Figure (b) shows simplified circuit . The battery 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 . Now current through 45 `Omega` , resistor is 2A and 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 current i.e., 0.5 A