Hi there!
1.
Since the two resistors are in series, we can simply add:
[tex]R_T = R_1 + R_2 + ... R_n[/tex]
[tex]R_T = 20 + 5 = \boxed{25 \Omega}[/tex]
2.
In series, the potential difference of each resistor (lamp) ADDS UP. We can begin by finding the current through the circuit using Ohm's law:
[tex]V = IR\\\\I = \frac{V}{R_T}[/tex]
Plug in the values:
[tex]I = \frac{50}{25} = 2 A[/tex]
Now,
we can use Ohm's law to find the individual voltage for each lamp.
20 Ohm lamp:
[tex]V = 2 * 20 = \boxed{40 V}[/tex]
5 Ohm lamp:
[tex]V = 2 * 5 = \boxed{10 V}[/tex]
3.
To solve, we can use the power equation.
[tex]P (\text{Watts})= IV[/tex]
Plug in the values for each.
20 Ohm lamp:
[tex]P = 2 * 40 = \boxed{80 W}[/tex]
5 Ohm lamp:
[tex]P = 2 * 10 = \boxed{100 W}[/tex]
