A multi loop curcuit- Resistors in parallel and series

In summary: You'll need to calculate the voltage drops across each of the resistors and then calculate the current using Ohm's Law for each individual resistor.In summary, the battery in the given diagram has negligible internal resistance and is connected to four resistors. Two resistors with values of 2Ω and one resistor with a value of 4Ω are connected in parallel, while the remaining resistor with a value of 3Ω is connected in series with the others. To find the equivalent resistance of the network, the parallel resistors can be combined using the equation 1/Req = 1/R1 + 1/R2 + 1/R3. This results in a Req of 0.8Ω. Adding
  • #1
clickcaptain
32
0

Homework Statement



The battery in the figure below has negligible internal resistance. Find
l_51ef2b6852a62056a601d62c5837c0c9.jpg


(a) the current in each resistor:
(b) the power delivered by the battery:

Homework Equations



V = IR

Resistor Connected in series

Req = R1 + R2 + R3

Resistor Connected in Parallel

1/Req = 1/R1 + 1/R1 + 1/R3

The Attempt at a Solution



I think that the resistors 3omega and vertical-2omega are connected in series, and the other 2 are connected in parallel.

So the ones in Parallel..

1/Req1 = 1/2 + 1/4
1/Req1 = 3/4
Req1 = 1.333

The ones in Series

Req2 = 3 + 2 = 5

so...

Req = Req1 +Req2 =1.333 + 5 = 6.333But to find the Current, I you just use Ohm's law

I = V/r

so the 3.0-omega resistor would be

I = 6/3 = 2

...but that's wrong and I am at a loss to know why?? thanks for the help!
 
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  • #2
Unfortunately the 2Ω, the 2Ω and the 4Ω are all in parallel. They share a common node on each side. That's what || means.

Find the equivalent of those 3 together and then you can add in series.
 
  • #3
The problem that you are having comes from the diagram you are using. When you see something like this you should redraw it in an easier way to see it. Start with the battery, and connect it to the first resister. Then notice that the connection splits in three, so this means you can redraw those three resisters in parallel. Finally they reconnect at the end, and the single wire returns to the battery.

When you draw it this way, you will see that all the 2 ohm resistors and the 4 ohm are in parallel, and the 3 ohm resistor is in series with them.
 
  • #4
So would the current for the first resistor be I = V/R = 6 V/ 2 = 3 ?
 
  • #5
You have to determine the Req for the network, in order to determine I.
 
  • #6
I got Req= 3.8

parallel = 1/Req = 1/2 + 1/2 + 1/4 = 5/4
Req-p = 0.8

so Req = 0.8 + 3 = 3.8 A

Then do I use this in V = IR?
 
  • #7
That would be the way to do it.

Except it's 3.8Ω
 
  • #8
clickcaptain said:
Then do I use this in V = IR?

But be careful, since the current isn't nesseceraly the same in every resistor.
 

Related to A multi loop curcuit- Resistors in parallel and series

1. What is a multi loop circuit?

A multi loop circuit is a type of electrical circuit that contains multiple loops or branches for the flow of current. This means that there are multiple paths for the current to travel through, allowing for more complex connections between components.

2. What is the difference between resistors in parallel and series?

Resistors in parallel are connected side by side, allowing for multiple paths for the current to flow through. This decreases the overall resistance in the circuit. In contrast, resistors in series are connected end to end, creating only one path for the current to flow through. This increases the overall resistance in the circuit.

3. How do I calculate the total resistance in a multi loop circuit?

The total resistance in a multi loop circuit depends on the configuration of the resistors. If the resistors are connected in parallel, the total resistance can be calculated using the formula 1/Rt = 1/R1 + 1/R2 + 1/R3 + ..., where Rt is the total resistance and R1, R2, R3, etc. are the individual resistances. If the resistors are connected in series, the total resistance can be calculated by adding the individual resistances together.

4. What happens to the current in a multi loop circuit with resistors in parallel?

In a multi loop circuit with resistors in parallel, the current will divide and flow through each resistor based on its individual resistance. This means that the current will be higher in branches with lower resistance and lower in branches with higher resistance.

5. How are voltage and current affected in a multi loop circuit with resistors in series?

In a multi loop circuit with resistors in series, the voltage will be divided between each resistor, but the current will remain the same throughout the circuit. This means that the total voltage across all resistors will equal the sum of the individual voltage drops, but the current will be the same at any point in the circuit.

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