# A hair dryer draws 1 200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are oper

Question

A hair dryer draws 1 200 W, a curling iron draws 800 W, and an electric light fixture draws 500 W. If all three of these appliances are operating in parallel on a 120-V circuit, what is the total current drawn?

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1 year 2021-08-31T21:38:09+00:00 1 Answers 5 views 0

Therefore,

The Total current drawn is 20.83 Ampere.

Explanation:

Given:

Voltage, V = 120 V

Power of Appliances,

P1 = 1200-W Hair dryer,

P2 = 800-W Curling iron,

P3 = 500-W Light

To Find:

Current  flowing in the line, I = ?

Solution:

We have Power formula,

$$Power = Voltage\times Current\\P=V\times I$$

Current for hair dryer,

$$P_{1}=V\times I_{1}$$

Substituting the values we get

$$1200=120\times I_{1}\\I_{1}=\dfrac{1200}{120}=10\ Ampere$$

Similarly for curling iron,

$$P_{2}=V\times I_{2}$$

$$800=120\times I_{2}\\I_{2}=\dfrac{800}{120}=6.6667\ Ampere$$

Similarly for electric light

$$P_{3}=V\times I_{3}$$

$$500=120\times I_{3}\\I_{3}=\dfrac{500}{120}=4.1667\ Ampere$$

As all three of these appliances are operating in parallel on a 120-V circuit,  the total current will be given as

$$I=I_{1}+I_{2}+I_{3}$$

Substituting the values we get

$$I=10+6.6667+4.1667=20.8334\approx 20.83\ Ampere$$

Therefore,

The Total current drawn is 20.83 Ampere.