Question

Two appliances are connected in parallel to a 120-v battery and draw currents i1 = 3.0 a and i2 = 3.1

Answer 1

Initially they are connected in parallel, so they have the same voltage V=120 V at their ends. Therefore we can use Ohm's law to calculate the resistance of each appliance:
$R_1 = \frac{V}{I_1}= \frac{120 V}{3.0 A}=40 \Omega$
$R_2 = \frac{V}{I_2}= \frac{120 V}{3.1 A}=38.7 \Omega$

When they are connected in series, they are crossed by the same current I. The equivalent resistance of the circuit in this case is $R_{eq}=R_1+R_2 = 78.7 \Omega$, so we can use Ohm's law for the entire circuit to find the current in the circuit:
$I= \frac{V}{R_{eq}}= \frac{120 V}{78.7 \Omega}=1.52 A$