Answer
(a) $0.875A, 0.25A$
(b) $0.50A$
Work Step by Step
(a) We know that
$I=I_1+I_2$.......eq(1)
$-I_1(4.0\Omega)+9.0V-I(5.0\Omega)=0$
This simplifies to:
$I_1=2.25A-1.25I$......eq(2)
We apply the loop rule at DCABD to obtain:
$9.0V-6.0V+I_2(2.0\Omega)-I_1(4.0\Omega)=0$
This simplifies to:
$I_2=2.0I_1-1.5A$......eq(3)
Substituting eq(2) in eq(3), we obtain:
$I=2.25A-1.25I+3.0A-2.5I$
$\implies I=\frac{5.25A}{4.75}$
$I=1.1A$
We plug in this value in eq(2) to obtain:
$I_1=2.25A-1.25(1.1A)=0.875A$
Similarly $I_2=3.0A-2.5(1.1A)$
$I_2=0.25A$
(b) We know that
$9.0V-6.0V-I(2.0\Omega)-I(4.0\Omega)=0$
This simplifies to:
$I=0.50A$
