1. The manufacturer of a 12V car headlight specifies it will drawa current of 6A
ID: 1736308 • Letter: 1
Question
1.
The manufacturer of a 12V car headlight specifies it will drawa current of 6A. You would like to check this claim with an ammeterdesigned to measure currents up to 10A and having a resistance of0.1 Ohms. Which of the two figures below represents a circuit wherethe ammeter correctly measures the current in the headlight?B
Howmuch current (in A) would flow in the ammeter for Circuit a?C
Howmuch current (in A) flows through the ammeter for Circuit b?D
You would like to check if the battery voltage drops while itis supplying a current of 6A. You use a voltmeter designed tomeasure voltages up to 20V and having a resistance of 50,000 Ohms.Which of the two circuits below should be used?E
Howmuch current (in A) would flow through the headlight for Circuita?F
Howmuch current (in A) would flow through the headlight for Circuitb?Explanation / Answer
You always connect ammeters in series, if you connect themparallel they just act as resistors.
So when in series, the ammeter simply measures the currentgiven, so that would be 6 A.
So when an ammeter is set up in as a resistor you solvefor current with I = V / R, or in this case I = 12 V /.1 = 120 A.
When setting up a voltmeter you want to set it up in parallel,so it does act as a resistor.
You solve for current by using the same formula I = V / Ragain. In this case we have two R, so you add them together and sincethey are in parallel you just add them regularly: R1 + R2 = 2 +50,000 V is just 12 so: 12 / 50,002 = 2.39 E -4 A
For when the voltmeter is connected in series you have to addthe resistance inversely, but its the same idea. For R it equals- 1/R = (1/2) + (1/50,000) = 1.99 So when using the formula I = V / R = 12 / 1.99 = 6 A
I hope that's what you were looking for, because your picsdidn't show up when I viewed the question like I said however ifits the same as when I did it, it should be right.
So when in series, the ammeter simply measures the currentgiven, so that would be 6 A.
So when an ammeter is set up in as a resistor you solvefor current with I = V / R, or in this case I = 12 V /.1 = 120 A.
When setting up a voltmeter you want to set it up in parallel,so it does act as a resistor.
You solve for current by using the same formula I = V / Ragain. In this case we have two R, so you add them together and sincethey are in parallel you just add them regularly: R1 + R2 = 2 +50,000 V is just 12 so: 12 / 50,002 = 2.39 E -4 A
For when the voltmeter is connected in series you have to addthe resistance inversely, but its the same idea. For R it equals- 1/R = (1/2) + (1/50,000) = 1.99 So when using the formula I = V / R = 12 / 1.99 = 6 A
I hope that's what you were looking for, because your picsdidn't show up when I viewed the question like I said however ifits the same as when I did it, it should be right.
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