Answer
a) $R_{eq}=1350\Omega$
b) $V_{990\Omega}=8.82V$
$V_{680\Omega}=3.18V$
$V_{750\Omega}=3.18V$
c) $I_{750\Omega}=0.00424A$
$I_{680\Omega}=0.00467A$
$I_{990\Omega}=0.00891A$
Work Step by Step
a) The $750\Omega$ and $680\Omega$ resistors are in parallel and in series with the $990\Omega$ resistor.
$R=\frac{1}{\frac{1}{750\Omega}+\frac{1}{680\Omega}}+990\Omega=1350\Omega$
b) $V=IR$
The current through the $990\Omega$ resistor can be found using the Ohm's Law with the equivalent resistance. The voltage across the $990\Omega$ can be found using the Ohm's Law again.
$I=\frac{V}{R}=\frac{12.0V}{1350\Omega}=0.00891A$
$V_{990\Omega}=IR=(0.00891A)(990\Omega)=8.82V$
The voltage across $750\Omega$ and $680\Omega$ resistor is equal, but the sum of the currents is equal to the current through resistor $990\Omega$
$V_{680\Omega}=12.0V-8.82V=3.18V$
$V_{750\Omega}=3.18V$
c) $I_{750\Omega}=\frac{V_{750\Omega}}{R}=\frac{3.18V}{750\Omega}=0.00424A$
$I_{680\Omega}=\frac{V_{680\Omega}}{R}=\frac{3.18V}{680\Omega}=0.00467A$
$I_{990\Omega}=\frac{V_{990\Omega}}{R}=\frac{8.82V}{990\Omega}=0.00891A$
