Bob asked in Science & MathematicsEngineering · 8 months ago

# Ohm's Law Current Problem?

2Suppose 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). How much current now flows through the bulb. (15 pts)

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• 8 months ago

This is a dumb question from your teacher ! Since car has more than one head light bulb. Driving car at night at least lights up two head lights, four sigh lights and two tail lights.

TOTAL CURRENT (exclude the car engine consume) = (12.6V-12V) / R (total resistance from bulbs)

As two unknowns from one equation cannot be solved.

• Anon8 months agoReport

Not Dumb, the kid neglected to post either resistance or wattage of the bulbs from previous question.Or, incomplete question.

• 8 months ago

The only intelligent answer is I=E/R. Or, 12 volts over Whatever resistance the bulbs and wires, switches, fuses present.

Without knowing this value, no accurate number possible. I will Assume the low beams are on,no wire/switch resistance; 2 at 55 Watts each (typical), P=E/I; the current is 9.1666 Amps.

• Anonymous
8 months ago

I = 12 ÷ R

It's this simple,

using the instructions.

• 8 months ago

Without knowing the Resistance, current or wattage of the bulbs the question cannot be answered except I= 12v/Rt.

Maybe this info was in Q#1.

• 8 months ago

So, assume the alternator provides 13.2 V to make sure it charges the 6-cell lead-acid storage battery in the car. Ohm's Law is V=IR. Assuming R remains constant, current is found by I=V/R. That says the current flowing will be directly proportional to the applied Voltage. The headlight lamp will see a Voltage reduction of 9% from 13.2V to 12V. That means the lamp will see 91% of the Voltage applied so the current will now be 91% of what it was, or I2 = .91I1. Sorry, can't use subscripts here for I2 and I1.

• Anon8 months agoReport

BS. Its obviously 12.6 volt battery.. Alt volts not a concern,, it's Battery voltage supplied AT this instant in time (It will steadily drop).

• 8 months ago

Depends on the resistance of the bulb. I = E/R = 12/R

• 8 months ago

Not enough information.

You didn't say what the voltage was with the alternator working.

Or more importantly, what the resistance of the bulb is.