Calculating Total Current in a Parallel RL Circuit

In summary: The Attempt at a SolutionIf you have a coil of N loops with the same area that's within a uniform magnetic field, each loop then has magnetic flux \Phi through it.
  • #1
Nikitin
735
27

Homework Statement


There's a DC supplying a current V over resistance R and inductor L in parallel. The circuit looks like this except it's DC http://sub.allaboutcircuits.com/images/02262.png

I need to find the total current.

The Attempt at a Solution



I know that the current through the first resistor is IR = V/R per ohms law, however I'm stuck on the current through the inductor. I first guessed it could be calculated from the impedance IL = V/z = V/iωL, but since ω is the angle frequency and equals zero for DC, I got stuck.

I think I need to apply faraday's law to find the current through the inductor, IL:

V = -ε = d(flux)/dt = L*dIL/dt, and calculate IL from the differential equation.

However, I can't find a physical reason as to why I would want to do that. Is it perhaps because the voltage drives a changing magnetic flux through the inductor?

thanks
 
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  • #2
basically, I need somebody to explain why this equation works.
10269.png
.

Why does the voltage drop over an inductor divided by itself inductance equal the change in current?
 
  • #3
Nikitin said:

Homework Statement


There's a DC supplying a current V over resistance R and inductor L in parallel. The circuit looks like this except it's DC http://sub.allaboutcircuits.com/images/02262.png

I need to find the total current.

For that circuit, with a constant voltage supply, the current through the supply will tend to infinity as t→∞, assuming ideal components. The voltage across the inductor will be constant and since L > 0, the instantaneous rate of change of current through it will be some nonzero number of constant sign, i.e. the magnitude of the current through the inductor will increase without bound.

Nikitin said:
Why does the voltage drop over an inductor divided by itself inductance equal the change in current?
It's 'instantaneous rate of change'.

If you know Faraday's law then you know how the voltage across the inductor is related to the flux linking its turns. Do you know an expression that relates the current through the inductor to the flux linkage produced by it?
http://en.wikipedia.org/wiki/Inductance#Inductance_of_a_solenoid
 
  • #4
Magnetic flux = L*current...

I know that a changing current produces a changing magnetic flux. But how do I link this to a voltage drop over the inductor producing a changing magnetic flux?

I don't understand how using the equation ε = -d(flux)/dt = -LdI/dt links the voltage drop to the changing current, since the epsilon is representing induced voltage, not the voltage drop?
 
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  • #5
Ohhh, but the voltage has to fall over the parallel circuit. So ε = -d(flux)/dt = -LdI/dt is the induced counter-voltage as the current changes, and since it's the only thing causing the voltage to drop, v = ε = -d(flux)/dt = -LdI/dt.

However, won't the current be in the wrong direction if this was true? I hate that minus-sign, it always confuses me.
 
  • #6
  • #7
With regards to emf vs. voltage, there are times where it's useful to make the distinction, but in terms of work done in moving charge through an emf or voltage, there's no difference.

Nikitin said:
However, won't the current be in the wrong direction if this was true? I hate that minus-sign, it always confuses me.
You need to be aware of the sign convention you've used in assigning reference polarities. ε is just a number (a scalar), does a positive value of ε mean that the top terminal of your inductor is at a higher potential than its lower terminal? Does it mean the opposite? On your circuit diagram there are reference directions assigned for current (there are no reference polarities shown for the element voltages), so it's clear what a positive value for IL means.

You use Faraday's law to derive element laws (equations that relate the voltage across an element to the current through it). For an ideal inductor:

V = L*di/dt is valid for the passive sign convention.
V = -L*di/dt is valid for the active sign convention.

http://en.wikipedia.org/wiki/Passive_sign_convention#Sign_conventions

I posted recently on this here:
https://www.physicsforums.com/showpost.php?p=4384249&postcount=28

In that post, if v was assigned a reference polarity according to the passive sign convention, then v = -ε = L*di/dt.
 
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  • #8
Nikitin said:
BTW, I never fully understood why you multiply the flux by N, like they did in wikipedia:
dfd651f5a6b43f962ea7d113d9cd81af.png


Like, why is the "total flux" equal to flux multiplied by the number of times the wires go around it? I don't fully understand it.
The changing flux links N turns and you get an induced emf for each turn, which are in series, so their emf's add.

Edit:
I reread your question and realized that I didn't really answer what you asked me.

Say you have a coil of N loops with the same area that's within a uniform magnetic field. Each loop then has magnetic flux [itex]\Phi[/itex] through it and since there are N loops, the total flux through the coil is [itex]N\Phi[/itex].
 
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  • #9
Sorry, I'm really mentally tired right now so excuse me if I making an obvious mistake.

By convention, NΦ = L*I instead of Φ = L* I.

Isn't it kind of stupid that L*I gives out the flux multiplied by N, instead of the actual flux (just Φ) going through the coil? Is there some physical reason for this?
 
  • #10
Nikitin said:
Sorry, I'm really mentally tired right now so excuse me if I making an obvious mistake.

By convention, NΦ = L*I instead of Φ = L* I.

Isn't it kind of stupid that L*I gives out the flux multiplied by N, instead of the actual flux (just Φ) going through the coil? Is there some physical reason for this?

In this expression NΦ = L*I, Φ is the magnetic flux through the surface bounded by one loop of the coil. If you were to define Φ = L*I, then the element law for the ideal inductor (using the passive sign convention) would be:

v = N*L*di/dt

Doesn't it seem silly to have two numbers in there when one would do?
 
  • #11
ok! thanks for the help! + rep
 

1. How do I calculate the total current in a parallel RL circuit?

To calculate the total current in a parallel RL circuit, you will need to use the formula I = V/R, where I represents current, V represents voltage, and R represents resistance. First, calculate the total resistance of the circuit by adding the individual resistances of each branch. Then, divide the voltage by the total resistance to find the total current.

2. What is the difference between a parallel RL circuit and a series RL circuit?

In a parallel RL circuit, the components (resistor and inductor) are connected in parallel to each other, meaning they have the same voltage but different currents. In a series RL circuit, the components are connected in series, meaning they have the same current but different voltages.

3. How does the inductor affect the total current in a parallel RL circuit?

The inductor in a parallel RL circuit affects the total current by creating a phase shift between the current and voltage. This means that the total current will not be equal to the sum of the individual branch currents, but it can be calculated using the total resistance and voltage of the circuit.

4. Can I use Ohm's Law to calculate the total current in a parallel RL circuit?

Yes, you can use Ohm's Law to calculate the total current in a parallel RL circuit. However, you will need to use the total resistance of the circuit, which is the sum of the individual resistances in each branch, instead of the individual resistance of a single component.

5. How can I verify my calculation of the total current in a parallel RL circuit?

You can verify your calculation of the total current in a parallel RL circuit by using Kirchhoff's Current Law (KCL). KCL states that the sum of the currents entering a node (or junction) in a circuit must equal the sum of the currents leaving the node. Therefore, you can check if your calculated total current matches the sum of the individual branch currents at a particular node.

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