Voltage Drop across multiple resistors

In summary: I am assuming he doesn't want me to subract anything.In summary, the task is to solve for all unknown currents and voltages in a circuit with a voltage source of 12V and seven resistors. Using the equation V=IR, the unknown currents are found by setting the voltage across the two loops to equal 0 and using three equations with three unknowns. The resulting values for the currents are It=48mA, I1=24mA, and I2=24mA. To find the voltage across each resistor, simply multiply the corresponding current with the resistance value.
  • #1
talaroue
303
0

Homework Statement


Solve for all unkown i's and V's
EDIT: there is a seventh resistor that is right under the negative side of the source.

Vs=12V
Interpetation2.jpg



Homework Equations


V=IR

The Attempt at a Solution



I got the I's by finding the voltage across the entire 2 loops to equal 0, then just simply had 3 equations, and 3 uknowns and went from there and got them to be

It=48mA I1=24mA I2=24mA

I don't understand to find the voltages now...
Do I simply just do (I1*R1)-Vs=V1?
 
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  • #2
talaroue said:
I don't understand to find the voltages now...
Do I simply just do (I1*R1)-Vs=V1?
It's not clear to me what you are asked to find. The voltage drop across each resistor? Or the voltage at certain points with respect to some reference?
 
  • #3
All it asks for is Solve for all unkown i's and V's
 
  • #4
talaroue said:
All it asks for is Solve for all unkown i's and V's
Then just find the voltage drops across each resistor.
 
  • #5
which is just R*I=V right? It just doesn't make sense that some times I have to have a different I for each resistor and then sometimes like this one I only need 2. for the whole circuit. Do you understand what I mean?

I will post how I found I my next post.
 
  • #6
talaroue said:
which is just R*I=V right?
Yes.
It just doesn't make sense that some times I have to have a different I for each resistor and then sometimes like this one I only need 2. for the whole circuit. Do you understand what I mean?
Realize that current must be the same throughout a single branch of the circuit. There are only three independent branches, thus only three currents needed to describe this circuit. Also, since the resistors are all the same, there'll only be three different voltage drops.
 
  • #7
Solving for I's...

it=i1+i2

with i1 going to the loop to the left, and i2 going to the loop to the right.

Where I came up with the equations...
12-R1i1-R6i1-R5i1-R7i=0

12-R2i2-R3i2-R4i2-R7i=0

then plugging in it=i1+i2 for in in both and then just combinding like terms I got these two equations

12-400i1-100i2=0
12-100i1-400i2=0
multiplying the top equation by (-4) I2 cancels out leaving just I1 so i just solved and got

I1=36/1500=24mA then plugged that back into the equations about to get I2 I got 24mA for I2 as well.

I=I1+I2=24mA+24mA=48mA

so then for the voltage across each resistor is simply the corresponding I to the resistor for example for R1...

V=IR
V=?
I=I1=24mA
R=100 Ohms

thats all?
I don't have to subtract anything?
 
  • #8
So basically the only time a new current is created, i guess you could say is if it splits from one to 2 or more?

Why don't you have to subtract anything?
 
  • #9
talaroue said:
Solving for I's...

it=i1+i2

with i1 going to the loop to the left, and i2 going to the loop to the right.

Where I came up with the equations...
12-R1i1-R6i1-R5i1-R7i=0

12-R2i2-R3i2-R4i2-R7i=0

then plugging in it=i1+i2 for in in both and then just combinding like terms I got these two equations

12-400i1-100i2=0
12-100i1-400i2=0
multiplying the top equation by (-4) I2 cancels out leaving just I1 so i just solved and got

I1=36/1500=24mA then plugged that back into the equations about to get I2 I got 24mA for I2 as well.

I=I1+I2=24mA+24mA=48mA

so then for the voltage across each resistor is simply the corresponding I to the resistor for example for R1...

V=IR
V=?
I=I1=24mA
R=100 Ohms
Looks good.
thats all?
I don't have to subtract anything?
It depends on what is meant by "find all the unknown Vs". Does your instructor usually take a particular point as being 0 V? (For example, on one side of the voltage supply. But that's arbitrary.)

talaroue said:
So basically the only time a new current is created, i guess you could say is if it splits from one to 2 or more?
Right.
 
  • #10
I am starting to understand this! He doesn't really state but on other problems he has a voltage for the resistor listed. just like on one of my other problems.
 

FAQ: Voltage Drop across multiple resistors

1. What is voltage drop?

Voltage drop is the decrease in electrical potential energy that occurs as current flows through a component or series of components in a circuit. It is measured in volts (V).

2. How does voltage drop occur across multiple resistors?

When multiple resistors are connected in a series, the voltage drop across each resistor is proportional to its resistance. This means that the more resistance a resistor has, the greater the voltage drop will be across it.

3. How do you calculate voltage drop across multiple resistors?

The voltage drop across each resistor in a series circuit can be calculated using Ohm's law: V = I x R, where V is the voltage drop, I is the current flowing through the resistor, and R is the resistance of the resistor. The total voltage drop across the series of resistors is equal to the sum of the individual voltage drops.

4. What factors can affect voltage drop across multiple resistors?

The main factors that can affect voltage drop across multiple resistors are the total resistance of the circuit, the current flowing through the circuit, and the individual resistances of each resistor. Changes in any of these factors can result in a change in the voltage drop across the resistors.

5. How can voltage drop be minimized in a circuit?

To minimize voltage drop in a circuit, you can either decrease the resistance of the components or increase the current flowing through the circuit. This can be achieved by using thicker wires with lower resistance, or by using a power source with a higher output voltage. Additionally, connecting resistors in parallel instead of in series can also reduce the overall voltage drop in a circuit.

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