# Series and parallel circuit with multiple resistors?

Series and parallel circuit with multiple resistors??

## 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?

I=V/R
ΔV=-IR

## 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

• phys1.pdf
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gneill
Mentor

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.

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

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:

$I = \frac{V}{Ra + Rb}$

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

$Va = I Ra= V \frac{Ra}{Ra + Rb}$

Similarly, the voltage across Rb is

$Va = I Rb= V \frac{Rb}{Ra + Rb}$

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.

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

gneill
Mentor

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

Yes.

Thank you so much for the clarification!