Parallel circuit multiple choice question

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Homework Help Overview

The discussion revolves around a multiple-choice question regarding the behavior of a simple series circuit when a second resistor is connected in parallel with an existing resistor and an ideal battery. The subject area is electrical circuits, specifically focusing on concepts of current, voltage, and resistance in parallel configurations.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason through the effects of adding a parallel resistor on current and voltage, expressing confusion about their professor's explanation. Some participants question the implications of using an ideal battery versus a current source, while others reflect on personal experiences with electrical circuits.

Discussion Status

Participants are exploring different interpretations of the question and the underlying principles of circuit behavior. Some guidance has been offered regarding the nature of ideal batteries and how they affect current and voltage in the circuit, but there is no explicit consensus on the correct answer among all participants.

Contextual Notes

The original poster's understanding is challenged by the professor's explanation, leading to a discussion about the assumptions made regarding the type of battery and the definitions of current and voltage in the context of the problem.

DrKitten_Paws
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This past Monday I took an exam and got a 99, woot woot! But I can't seem to understand why I got the following question wrong:

Complete the following statement: A simple series circuit contains a resistance R and an ideal battery. If a second resistor is connected in parallel with R,
A) the voltage across R will decrease.
B) the current through R will decrease.
C) the total current through the battery will increase.
D) the rate of energy dissipation in R will increase.
E) the equivalent resistance of the circuit will increase.

I picked B since I used the following reseasoning: Itotal = I1 + I2, therefore the current in R is decreased. My professor said that C is the correct answer. Although she tried to explain it, I couldn't understand it. Do you mind explaining it to me, please?

Thank you so much!
 
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This is not in my area at all, and I probably would have scored only "1" on that test (if "A" is the correct answer.) My knowledge of electrical stuff is almost non-existent. Other than adding a few outlets to my house and rebuilding a bunch of things like VCR's, my experience is nil. The only reason that I think of "A" is that I once had to run a 6VDC motor from a 12VDC power supply. A friend who knows about this kind of stuff told me to parallel wire a resistor with the same rating as the load of the motor to cut the voltage in half. He called it a "voltage divider". While I can't say for sure why, it worked.
 
DrKitten_Paws said:
This past Monday I took an exam and got a 99, woot woot! But I can't seem to understand why I got the following question wrong:

Complete the following statement: A simple series circuit contains a resistance R and an ideal battery. If a second resistor is connected in parallel with R,
A) the voltage across R will decrease.
B) the current through R will decrease.
C) the total current through the battery will increase.
D) the rate of energy dissipation in R will increase.
E) the equivalent resistance of the circuit will increase.

I picked B since I used the following reseasoning: Itotal = I1 + I2, therefore the current in R is decreased. My professor said that C is the correct answer. Although she tried to explain it, I couldn't understand it. Do you mind explaining it to me, please?

Thank you so much!

Welcome to the PF.

Your professor is correct. As long as the battery is an ideal battery (no internal source resistance), the output voltage is indepent of output current. So the voltage across the parallel combination of R//R is the same as it was across the initial single R. And since I=V/R, that means that the current through the first resistor will not change when the second R is put in parallel with it. All that happens is that twice as much current is drawn from the battery when you put the 2nd R in parallel with the first.

Make sense now?
 
I picked B since I used the following reseasoning: Itotal = I1 + I2, therefore the current in R is decreased.

I can see where you went wrong...

Your answer would be correct if an ideal battery was an current source. However an ideal battery is a voltage source.

For a constant voltage source the current Itotal isn't constant it depends on the load...

Itotal = Vbat/R1 + Vbat/R2
 

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