between the right side and left side since the ratio of currents in a parallel combination
is the conductance ratio. Therefore, current to the right is 1/3 of 24 mA8 mA and
current to the left is 2/3 of 24 mA 16 mA as shown in Fig. 3.1-10(b).
The 8 mA to the right side flows through 12.5 k and reaches the node-B where
it gets divided into 10 k and 30 k. The conductance ratio here is 3:1 and hence 75%
of 8 mA 6 mA goes into 10 k and 25% of 8 mA2 mA goes into 30 k as shown in
Fig. 3. 1-10(c).
Similarly, the 16 mA flowing to the left side from current source node will divide
into 8 mA each in the 20 k resistors on the left ...
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