Answer
a)
$R=77.2\Omega$
b)
$I_1=0.55A$
$I_2=0.18A$
$I_3=0.16A$
c) Smaller
Work Step by Step
(a) We know that
$R=(\frac{1}{\epsilon}-\frac{1}{R_2}-\frac{1}{R_3})^{-1}$
We plug in the known values to obtain:
$R=(\frac{0.88A}{12.0V}-\frac{1}{22\Omega}-\frac{1}{67\Omega})^{-1}$
$R=77.2\Omega$
(b) According to Ohm's law
$I=\frac{\epsilon}{R}$
For resistor $R_1$
$I_1=\frac{12.0V}{22\Omega}=0.55A$
For resistor $R_2$
$I_2=\frac{12.0V}{67\Omega}=0.18A$
For resistor $R_3$
$I_3=\frac{12.0V}{77.2\Omega}=0.16A$
(c) We know that the resistance and the current are inversely proportional. Thus, if the total current is grater than $0.88A$ at constant emf then the value of equivalent resistance will become less and therefore the new value of the unknown resistor will be less.
