Calculate voltage drop by using kirchoff's law

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Discussion Overview

The discussion revolves around calculating the voltage drop across a resistor in a series circuit using Kirchhoff's law and Ohm's law. Participants explore the relationships between voltage, current, and resistance in the context of a specific circuit with a 30V battery and three resistors.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a circuit with a 30V battery and three resistors (8 ohms, 2 ohms, and 4 ohms) in series, seeking to calculate the voltage drop over the 8 ohm resistor.
  • Another participant suggests that the current can be calculated using Ohm's Law and that the total resistance can be summed mentally.
  • A participant questions whether the calculated voltage (17.136V) represents the voltage after passing through the resistor or the voltage drop itself.
  • It is noted that the voltage across a resistor is the product of the current through it and its resistance, and that larger resistors drop more voltage than smaller ones, in proportion to their resistance.

Areas of Agreement / Disagreement

Participants generally agree on the application of Ohm's Law to find the voltage across the resistor, but there is some uncertainty regarding the interpretation of the calculated voltage as either the voltage drop or the voltage after the resistor.

Contextual Notes

There are unresolved aspects regarding the initial calculations and the assumptions made about the circuit configuration and the application of Kirchhoff's law.

Who May Find This Useful

This discussion may be useful for students learning about circuit analysis, particularly those interested in applying Kirchhoff's laws and Ohm's law in series circuits.

Arno@praxispc
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Homework Statement



The circuit has a battery Voltage of 30V, a resistor of R1(8ohms), R2 (2Ohms) and R3 (4ohms) in series. Calculate the voltage drop over R1

Homework Equations



Please give me the proper equation of this? The one I've got says you take 30-8I-2I-4I=0
which gives you current of 2.142 and not the voltage.

The Attempt at a Solution



Now what i tried doing is this.
30V-8-2-4=0
30V=14
V=14/30
V=0.466
but I'm sure its wrong.
 
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You have the correct current, although you could just add the resistor values up using mental arithmetic and work out the current using Ohm's Law.

This current flows through an 8 ohm resistor, so what is the voltage across the resistor?
 
ok so what you are saying is that if i use ohms law V=I*R
I will have to use the 8 ohm resistor times by the current of 2.142= 17.136V
Is this the voltage after it went through the resistor or is this the amount of voltage drop?
 
The voltage across a resistor is equal to the current through it times the resistance of that resistor.

So, the 8 ohm resistor times the current of 2.142 Amps = 17.136 Volts.

If you do this for all the resistors, you will see that they add up to the supply voltage of 30 volts.

You should also see that the bigger resistors drop more of the the available voltage than the smaller ones, exactly in proportion to their resistance.
 

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