Calculating Resistance in a Circuit using Kirchhoff's Laws

In summary, the conversation discussed a circuit problem involving finding the total resistance between two points and the ammeter reading when a battery was placed between those points. There were some attempts made to solve the problem, but the key step of using delta-star conversion was not known. The conversation then delved into discussing the use of delta-star conversion and how it simplifies the circuit. Finally, an alternative approach using KCL at the nodes was suggested.
  • #1
physicsStudent00

Homework Statement


1) what is the total resistance of this circuit between points A and B?

2) if the horizontal resistor was replaced with an ammeter what would it read if a 6V battery was placed between points A and B? let R = 20 ohms

upload_2017-10-5_16-31-12.png

upload_2017-10-5_16-31-12.png

Homework Equations




The Attempt at a Solution


had multiple attempts, the solution for 1) is 7R/5 but I've never got that, I've been getting 7R/3 and then I'm yet to move to part 2 because I've been getting the resistance for part 1 wrong, which is most likley to be with how I am imagining the current moving through the circuit
 
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  • #2
Try doing part 2. Working on it may provide insight for part 1.
 
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  • #3
Asymptotic said:
Try doing part 2. Working on it may provide insight for part 1.
so for part 2 the circuit is split into 2 parallel paths each of equal resistance, therefore the current will be the same in each of the parallel paths, meaning that:
if the total resistance is 30ohms,
I=V/R=6/30=0.2A
so the current in each path is 0.2/2=0.1A
the ammeter will read 0.1A
 
  • #4
physicsStudent00 said:
had multiple attempts, the solution for 1) is 7R/5 but I've never got that, I've been getting 7R/3
Have you studied delta- star conversion? That is the key step in solving part 1.

physicsStudent00 said:
so for part 2 the circuit is split into 2 parallel paths each of equal resistance, therefore the current will be the same in each of the parallel paths,
This is not true. There is an ammeter in place of the horizontal resistance R.
 
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  • #5
cnh1995 said:
Have you studied delta- star conversion? That is the key step in solving part 1.This is not true. There is an ammeter in place of the horizontal resistance R.
i don't have any idea about a delta-star conversion, so the ammeter would provide a pathway for the current to flow across, also it would have some resistance, so therefore it creates an extra path. the current I am guessing would flow through the ammeter, from the left to the right? as it goes from the path of lowest resistance?
 
  • #6
physicsStudent00 said:
as it goes from the path of lowest resistance?
That is one approach one should give up in circuit analysis. Current flows according to Ohm's law and follows KCL , not caring whether the path is of least resistance or highest resistance.

physicsStudent00 said:
i don't have any idea about a delta-star conversion
You should look it up. It is a simple and very useful method for simplification of networks.

For part 2, draw the circuit diagram and simplify the network. Find the total current and use current division rule to get the ammeter current.
 
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  • #7
physicsStudent00 said:
also it would have some resistance, so therefore it creates an extra path
I missed this part in your post. You can assume the ammeter resistance to be zero in case it is not specified. (Even if it is specified, its value is very small).
 
  • #8
No need to delta-Y conversion. Imagine you connect a battery of known emf to points A, B, and write the current law for the nodes P, Q, S in terms the potentials U1, U2, and ε assuming zero potential at the negative terminal of the battery.

upload_2017-10-5_16-31-15.png

In case b) the ammeter can be considered ideal, that is, of zero internal resistance. That would mean U1=U2, the upper pair of resistors are connected in parallel, and so are the lower pair of resistors. What is the resultant then? What is the input current?
 
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  • #9
ehild said:
No need to delta-Y conversion. Imagine you connect a battery of known emf to points A, B, and write the current law for the nodes P, Q, S in terms the potentials U1, U2, and ε assuming zero potential at the negative terminal of the battery.

View attachment 212349
In case b) the ammeter can be considered ideal, that is, of zero internal resistance. That would mean U1=U2, the upper pair of resistors are connected in parallel, and so are the lower pair of resistors. What is the resultant then? What is the input current?
with the delta Y conversion it makes a lot more sense now, thank you very much for the help.
 
  • #10
physicsStudent00 said:
with the delta Y conversion it makes a lot more sense now, thank you very much for the help.
Have you solved the problem with the delta-Y conversion? Here is an other approach which uses KCL at the nodes P, Q, S

Node P: I - (ε-U1)/R - (ε-U2)/(2R) = 0
Node Q: (ε-U1)/R - (U1-U2)/R - U1/(2R) = 0
Node S: U1/(2R) + U2/R - I = 0
Solve for I. The total resistance is RT = ε/I.
upload_2017-10-6_10-8-52.png
 

1. What is resistance in a circuit?

Resistance in a circuit is the measure of how difficult it is for electric current to flow through a material. It is measured in Ohms (Ω) and is determined by the material, length, and cross-sectional area of the wire or component in the circuit.

2. How does resistance affect current in a circuit?

According to Ohm's Law, the current in a circuit is directly proportional to the voltage and inversely proportional to the resistance. This means that as resistance increases, the current decreases, and vice versa.

3. What factors can affect resistance in a circuit?

The main factors that affect resistance in a circuit are the material of the wire or component, the length of the wire, and the cross-sectional area of the wire. Temperature and the presence of other components in the circuit can also affect resistance.

4. How do you calculate the total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of the individual resistances. This can be calculated using the formula: Rtotal = R1 + R2 + R3 + ... + Rn, where R represents resistance in Ohms.

5. How do you calculate the total resistance in a parallel circuit?

In a parallel circuit, the total resistance is not simply the sum of the individual resistances. Instead, it is calculated using the formula: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn. To find the total resistance, take the reciprocal of the sum of the reciprocals of each resistance.

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