Question

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

1. What is the equivalent resistance of the circuit?
2. What is the voltage drop across each lamp?
3. What is the power dissipated in each lamp

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 = (V)/(R_T)`

Plug in the values:

`I = (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}`