Checking Voltage Drop while Supplying 6A Current

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SUMMARY

This discussion focuses on calculating voltage drop while supplying a 6A current using a voltmeter with a maximum voltage of 20V and a resistance of 50,000 Ohms. The calculations for series and parallel circuits are outlined, with the series circuit yielding a current of 2.39E-4 A and the parallel circuit resulting in a current of 23.99 A. A correction is provided for the equivalent resistance formula, clarifying that the correct formula is 1/Req = (1/R1 + 1/R2), leading to an equivalent resistance of 1.99992 Ohms.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with series and parallel circuit configurations
  • Knowledge of equivalent resistance calculations
  • Experience with using voltmeters
NEXT STEPS
  • Review Ohm's Law applications in circuit analysis
  • Study the principles of series and parallel circuits
  • Learn about voltmeter specifications and their impact on measurements
  • Explore advanced circuit analysis techniques, such as Thevenin's and Norton's theorems
USEFUL FOR

Electrical engineering students, hobbyists working on circuit projects, and professionals involved in circuit design and analysis will benefit from this discussion.

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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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