Part A For the circuit shown in the figure(Figure 1) find the current through each resistor

here 4 ohm and 12 ohm resistance are in parellel

so the equilent resistance

= (4x12) / (4+12)

= 3 ohm

now 3 ohm and 5 ohm are in series

so 3 + 5 = 8 ohm

now 8 ohm and 24 ohm are in parallel

so the equivalent resistance = (8x24) / (8+24)

= 6 ohm

now apply KVL in loop

12 - 6I - 3 - 3I = 0

9I = 9

I = 1 Amp

so I(3 ohm) = I(6 ohm) = 1 Amp

so voltage V(3 ohm) = 1 x 3 = 3 volt

6 ohm is the resultant of parallel combination of 8 ohm and 24 ohm

so voltage across 6 ohm V(6 ohm) = 1 x 6 = 6 volt

voltage is same in parallel combination so

V(8 ohm) = V(24 ohm) = 6 volt

so I(8 ohm) = 6 / 8 = 0.75 Amp

and I(24 ohm) = 6 / 24 = 0.25 Amp

8 ohm is the resultant of series combination of 3 ohm and 5 ohm

current is same in series combination so

I(3 ohm) = I(5 ohm) = 0.75 Amp

so the voltage V(3 ohm) = 0.75 x 3 = 2.25 volt

and V(5 ohm) = 0.75 x 5 = 3.75 volt

3 ohm resistance is the resultant of parallel combination of 4 ohm and 12 ohm

so voltage is same in parallel so

V(4 ohm) = V(12 ohm) = V(3 ohm) = 2.25 volt

so current I(4 ohm) = 2.25 / 4 = 0.5625 Amp

and I(12 ohm) = 2.25 / 12 = 0.1875 Amp

(a) I(3 ohm) = 1 Amp I(24 ohm) = 0.25 Amp I(5 ohm) = 0.75 Amp I(4 ohm) = 0.5625 Amp

I(12 ohm) = 0.1875 Amp

(b) V(3 ohm) = 3 volt V(24 ohm) = 6 volt V(5 ohm) = 3.75 volt V(4 ohm) = 2.25 volt

V(12 ohm) = 2.25 volt