Answer to Question #322877 in Electric Circuits for chellibyn

Question

21.For the network shown in Figure below, determine the current flowing in each branch.

Answer 1

$R_1=R_2=R_3=1\Omega$

$R_2 ,R_3$

Parallel combination

$R_3=\frac{R_2R_3}{R_2+R_3}=\frac{1\times1}{1+1}=0.5\Omega$

$R_{eq}=0.5+1=1.5\Omega$

Now current calculate

$I=\frac{2}{1.5}=1.33A$

Now $R_2 ,R_3$ Same current flow same

$I_2=I_3=\frac{1.33}{2}=0.66A$