Question

The electric bulb is rated 220v and 100w. Its resistance is:

a. 444 ohm

b. None of the above

O c. 2.2 ohm

d. 22 ohm​

Answer 1

Answer:

$484 Ω$

Step-by-step explanation:

P = V^2/R

If V = 220 V we have

100 W = 220^2/R

R = 220^2/100 Ω = 484 Ω. This is the resistance of the bulb.

When V = 110 V, power consumed = V^2/R = 110^2/484 = 25 W.

So, 25 W power is consumed when it is operated on 110 V.

Answer 2

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When at least two quantities of Ohm's law can be determined, the third can be calculated.

100 Watts tells us the product of current and voltage, being the Joule’s law power equation.

220 Volts expresses the voltage multiplier.

Therefore, with the product being 100 W and one of the multipliers being 220 V, the value for current can be calculated under Joules law:

1. wattage = current • voltage
2. current = voltage ÷ wattage
3. current = 100 W ÷ 220 V
4. current = 0.4545 amperes

With current determined, Ohm’s law can be used to calculate resistance.

1. resistance = voltage ÷ current
2. resistance = 220 volts ÷ 0.4545 amperes
3. resistance = 484.0484 ohms

Logically, if two of three quantities of a specific ratio are known, then the third quantity can be calculated.

A permutation of Joule's law, its simple ratio I = E ÷ P does not show us all three quantities for the Ohm’s law ratio I = E ÷ R but does utilise two of the three.

Likewise, Ohm's law does not show all three quantities for the simple permutation of the Joule's law ratio, I = E ÷ P.

Specifications for the bulb tell its full story with a little help from relevant equations that form bases for electronic circuit function and load operation!