Question

A 20 ohm lamp and a 5 ohm lamp are connected in series and placed across a potential difference of 50 V.

Answer 1

Hi there!

1.

Since the two resistors are in series, we can simply add:
$R_T = R_1 + R_2 + ... R_n$

$R_T = 20 + 5 = \boxed{25 \Omega}$

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:
$V = IR\\\\I = \frac{V}{R_T}$

Plug in the values:
$I = \frac{50}{25} = 2 A$

Now,

we can use Ohm's law to find the individual voltage for each lamp.

20 Ohm lamp:
$V = 2 * 20 = \boxed{40 V}$

5 Ohm lamp:
$V = 2 * 5 = \boxed{10 V}$

3.

To solve, we can use the power equation.

$P (\text{Watts})= IV$

Plug in the values for each.

20 Ohm lamp:
$P = 2 * 40 = \boxed{80 W}$

5 Ohm lamp:
$P = 2 * 10 = \boxed{100 W}$