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The output of a automobile alternator is typically about 14V D.C. The headlight circuit draws about 4 Amps of current. Using Ohm's Law (V = IR), the resistance of the headlight bulb can be calculated. Suppose your alternator quits working as you are driving home. The battery voltage now supplying energy to the headlight bulb drops to 12V D.C. Assume the bulb resistance R remains constant (it won’t, but that is another story).

Required:
How much current (in Amps) now flows through the bulb?

Respuesta :

hamzaahmeds

Answer:

I = 3.42 A

Explanation:

Ohm's law gives a relationship between the potential difference and the current passing through a conductor. The formula of Ohm's Law is given as follows:

V = IR

where,

V = Voltage

I = Current

R = Resistance

Initially:

V = 14 volts

I = 4 A

R = ?

Therefore,

14 volts = (4 A)R

R = (14 volts)/(4 A)

R = 3.5 Ω

Now, when V becomes 12 volts and R remains constant:

V = IR

12 volts = I(3.5 Ω)

I = (12 volts)/(3.5 Ω)

I = 3.42 A

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