Checking Voltage Drop while Supplying 6A Current

AI Thread Summary
The discussion focuses on measuring voltage drop while supplying a 6A current from a battery using a voltmeter with a resistance of 50,000 Ohms. The calculations involve determining current flow in both series and parallel circuits, with the series equation yielding a current of approximately 2.39E-4 A and the parallel circuit calculation resulting in about 23.99 A. A correction is noted regarding the formula for equivalent resistance in parallel circuits, emphasizing that the correct expression is 1/Req = (1/R1 + 1/R2). The user seeks confirmation on the accuracy of their calculations and understanding of the concepts. Overall, the thread highlights the importance of correct formula application in circuit analysis.
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Homework Statement


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.
See attachment****

This connection is in series so V= I (R1+R2)
V / (R1+R2) = I
12 / (2+ 50000) = I
2.39E-4 = I

How much current (in A) would flow through the headlight for Circuit b?

The connection is in parallel so Req = (1/ R1 + 1/R2)
Req = (1/2 + 1/50000) = .50002

I = V / R
I = 12 / .50002
I= 23.99

Homework Equations


I want to check if what I have done is right.


The Attempt at a Solution


Is right bellow the questions.

Thanks
 

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You have your formula for Req wrong.

It should be 1/Req= (1/R1 + 1/R2)

So, Req = 1.99992.
 

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