Ohm's Law and current

  • #1
I am currently stuck on these two problems dealing with ohm's law. I understand that a current is equal to 1/r(va-vb) but am lost in the question somehow. I do not know how to post the two figures so I will try my best to explain it. The figure for problem 1(a) has a 12 v car battery with the ammeter hooked up to a 12 v. car battery and headlight with the headlight being furthest from the right. The figure for problem 1(b) has the ammeter farthest from the right of the car battery and the headlight is not hooked directly to the battery but rather the wires from the ammeter. The figure for question #2 are identical but replace an ampmeter with a voltmeter. I apologize if these questions are too vague and appreciate any help/guidance with the problems...

1.The manufacturer of a 12V car headlight specifies it will draw a current of 6A. You would like to check this claim with an ammeter designed to measure currents up to 10A and having a resistance of 0.1 Ohms. Which of the two figures below represents a circuit where the ammeter correctly measures the current in the headlight?
a) How much current (in A) would flow in the ammeter for Circuit a?
b) How much current (in A) flows through the ammeter for Circuit b?

2.You would like to check if the battery voltage drops while it is supplying a current of 6A. You use a voltmeter designed to measure voltages up to 20V and having a resistance of 50,000 Ohms. Which of the two circuits below should be used?
a)How much current (in A) would flow through the headlight for Circuit a?
b)How much current (in A) would flow through the headlight for Circuit b?
 

Answers and Replies

  • #2
Find the resistance

It would be nice if you included the figures that are being referenced. But I don't know how to do it either. Advice from others?

Lacking that, I would find the resistance of the headlight. The voltage across it is 12V and the current it draws is 6A. Then add that resistance to the figures instead of the headlight. Then I would replace the meters with the resistance also. Now look at the result and see what makes sense. It should be obvious at that point.
 
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