21.For the network shown in Figure below, determine the current flowing in each branch.
R1=R2=R3=1ΩR_1=R_2=R_3=1\OmegaR1=R2=R3=1Ω
R2,R3R_2 ,R_3R2,R3
Parallel combination
R3=R2R3R2+R3=1×11+1=0.5ΩR_3=\frac{R_2R_3}{R_2+R_3}=\frac{1\times1}{1+1}=0.5\OmegaR3=R2+R3R2R3=1+11×1=0.5Ω
Req=0.5+1=1.5ΩR_{eq}=0.5+1=1.5\OmegaReq=0.5+1=1.5Ω
Now current calculate
I=21.5=1.33AI=\frac{2}{1.5}=1.33AI=1.52=1.33A
Now R2,R3R_2 ,R_3R2,R3 Same current flow same
I2=I3=1.332=0.66AI_2=I_3=\frac{1.33}{2}=0.66AI2=I3=21.33=0.66A
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