Answer
(a) $I_{7.1\Omega}=0.29A,I_{3.2\Omega}=1.1A$
(b) $1.4A$
(c) $11.3V$
Work Step by Step
(a) We know that the expression of the voltage drops in loop $ABCF$ can be written as
$(4.5\Omega) i_1+(1\Omega) i_1+(3.2\Omega)(i_1-i_2)+(0.50)i_1-12.0V=0$
$\implies (9.2\Omega) i_1-(3.2\Omega) i_2=12.0V$.....eq(1)
This simplifies to:
$i_2=0.19875i_1$....eq(2)
We plug in this value in eq(1)
$(9.2\Omega) i_1-(3.2\Omega)(0.19875i_1)=12.0V$
$\implies i_1=1.40A$
We plug in this value in eq(2) to obtain:
$i_2=0.19875(1.40A)$
$i_2=0.29A$
The current flowing through the resistance $3.2\Omega$ can be determined as
$i_{3.2\Omega}=i_1-i_2$
$\implies i_{3.2\Omega}=1.40A-0.28A=1.1A$
(b) We know that the current flowing through the battery is $i_1$ -- that is, the current flowing through the battery is $1.40A$.
(c) We know that
$V=12.0V-i_1r$
We plug in the known values to obtain:
$V=12.0V-(1.40A)(0.50\Omega)$
$\implies V=11.30V$