Answer
(a) $71mA$
(b) $V_{42}=3.0V$
$V_{17}=1.2V$
$V_{110}=7.8V$
Work Step by Step
(a) As the resistors are connected in series so we can find the equivalent resistance as
$R=R_1+R_2+R_3$
$\implies R=42+17+110=169\Omega$
Now we can find the current
$I=\frac{V}{R}$
We plug in the known values to obtain:
$I=\frac{12}{169}=0.710A=71mA$
(b) We can find the potential difference across each resistor as
$V_{42}=IR=(0.710)(42)=3.0V$
$V_{17}=IR=(0.710)(17)=1.2V$
$V_{110}=IR=(0.710)(110)=7.8V$
