How Do You Calculate Voltage Drops in a Simple Circuit?

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Homework Help Overview

The discussion revolves around calculating voltage drops in a simple circuit involving a light bulb and two connecting wires. The circuit includes a 6V ideal voltage source, a light bulb with a resistance of 30 Ohms, and wires with a resistance of 1.0 Ohm each.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the total resistance in the circuit and how it relates to the current. There are attempts to apply Ohm's law (V=IR) to find the current and voltage drops. Some participants question the resistance values assigned to the wires and the light bulb.

Discussion Status

There is a mix of attempts to calculate current and voltage drops, with some participants providing calculations while others seek clarification on resistance values. The conversation reflects a productive exploration of the problem, with guidance offered on considering the circuit's total resistance.

Contextual Notes

Participants are navigating assumptions about the resistances of the wires and the overall circuit configuration. There is a noted confusion regarding the division of total wire resistance and its implications for voltage drop calculations.

dasblack
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Homework Statement


Assume the wire resistance is 1.0 Ohms, the power supply is an ideal 6V voltage source, and the light bulb resistance is 30 Ohms.
In the circuit below:
a). A lightbulb is connected to the power supply using two such wires. Calculate the current (in A) in the circuit.
b) Calculate the voltage drop (in V) across each wire.

2.jpg


Homework Equations


V=IR
R=dL/A


The Attempt at a Solution


a.) I = 6/1 = 6ohms
b.) V = 6 x 1 = 6V

I have no clue how to do this problem. Professor hasn't taught this material yet.

Thanks
 
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If the light bulb is connected with the two wires. Then the resistances are in series.
Find the total resistance. The total pd=6V.
So the total current is simply I=V/R.

The sum of the voltage drops across the wires and bulb will add up to be 6. So just use V=IR and you'll get it.
 
Consider this circuit as an ideal voltage source connected in series with 3 resistances (wire segment, light bulb, wire segment). Maybe redrawing the circuit will help. Ask yourself what the total resistance of the circuit is, and how this relates to the current. Then ask yourself how this current relates to the voltage drops across each individual element.
 
So for a.) I = 6/32 => 0.19 Amps

b.)V = 0.19 x 32 => 6 volts ?
 
a) is correct. For b) think of the individual wire elements. From the picture, it looks like each wire would have a resistance of 0.5 ohms. 6 volts is the drop in potential across the total resistance.
 
mplayer said:
a) is correct. For b) think of the individual wire elements. From the picture, it looks like each wire would have a resistance of 0.5 ohms. 6 volts is the drop in potential across the total resistance.

I see what you meant by the total resistance. Shouldn't each wire have a resistance of 1 ohm and not 0.5? From part a 2 ohms for the wires and 30 for the bulb
 
dasblack said:
I see what you meant by the total resistance. Shouldn't each wire have a resistance of 1 ohm and not 0.5? From part a 2 ohms for the wires and 30 for the bulb

Yes, you're right, each piece of wire is 1 ohm. I thought it was referring to the wire as a whole, and wanted you to divide the total wire resistance by 2 in order to split it up. Sorry for the confusion :/

Now for part b) you need to find the voltage drop across each piece of wire. You have the current through each piece, and you know the resistance of each piece.
 

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