• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Series and parallel circuit with multiple resistors?

  • Thread starter conov3
  • Start date
34
0
Series and parallel circuit with multiple resistors??

1. Homework Statement

If R1 = 27Ω and R2 = 7Ω :
a. What is the equivalent resistance of the circuit?
b. What is the current from the battery?
c. What is the voltage drop across the R1 resistor?

2. Homework Equations

I=V/R
ΔV=-IR

3. The Attempt at a Solution

Attached.. I am unsure if I did it correct, I got the first two I believe
Any help would be appreciated!
 

Attachments

gneill

Mentor
20,489
2,615
Re: Series and parallel circuit with multiple resistors??

Your answers for (a) and (b) look okay (although you didn't show any details for (a)).

For (c) I don't see an answer shown. I would suggest a voltage divider approach, since you have two resistances in series with given total voltage drop.
 
34
0
Re: Series and parallel circuit with multiple resistors??

Would I be able to break it up to 4.5V going both ways.. say V=IR
I=4.5V/6ohms to get .75A
That is where I get confused with how to find the voltage drop with more than one resistor
 

gneill

Mentor
20,489
2,615
Re: Series and parallel circuit with multiple resistors??

Would I be able to break it up to 4.5V going both ways.. say V=IR
I=4.5V/6ohms to get .75A
That is where I get confused with how to find the voltage drop with more than one resistor
No! Voltage does not split between parallel branches, it is common (equal) for both. Current divides between branches as you've seen.

The voltage from the top to the bottom of BOTH branches is 9V. That means if you're analyzing the left branch (containing R1), you can completely ignore the other branch provided you know the branch voltage. In this case you do know the branch voltage, it's 9V.

If you have two resistors in series, say Ra and Rb, and a voltage across them, V, then the current flowing is:

[itex] I = \frac{V}{Ra + Rb} [/itex]

Thus by Ohm's law the voltage that appears across Ra is:

[itex] Va = I Ra= V \frac{Ra}{Ra + Rb} [/itex]

Similarly, the voltage across Rb is

[itex] Va = I Rb= V \frac{Rb}{Ra + Rb} [/itex]

So as you can see, a voltage divider takes the "input" voltage and places across each resistor a fraction of the total voltage in proportion to that resistor's fraction of the total resistance. This is a very handy relationship, and is well worth memorizing.
 
34
0
Re: Series and parallel circuit with multiple resistors??

So does that mean voltage drop =9*(27/33) => 7.36V?
 

gneill

Mentor
20,489
2,615
Re: Series and parallel circuit with multiple resistors??

So does that mean voltage drop =9*(27/33) => 7.36V?
Yes.
 
34
0
Re: Series and parallel circuit with multiple resistors??

Thank you so much for the clarification!
 

Related Threads for: Series and parallel circuit with multiple resistors?

Replies
5
Views
1K
Replies
12
Views
2K
Replies
2
Views
689
Replies
7
Views
708
  • Posted
Replies
3
Views
2K
  • Posted
Replies
2
Views
6K
  • Posted
Replies
5
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top