Determining Steady-State Current and Voltages in Inductive-Resistive Circuit

  • Thread starter ghostbuster25
  • Start date
  • Tags
    Current
In summary: From that point, the voltage starts to decline, and it does so until the inductor is completely discharged. So, in short, the voltage across the inductor at any given time is equal to the peak voltage of the ramp-up.
  • #1
i have an inductive-resistive circuit with Vbat = 6.0V, R = 10 ohm and L = 0.10mH

I need to determine the steady-state current and the magnitudes of the steady-state voltages across the resister and across the i ductor.

the switch has been closed at t=0s

is the equation i need i(t) = (Vbat/R)(1-e^-Rt/L)

and if t=0 do i just use 0 in place of t in the equation or do i use a time constant i worked out previously to be 0.01t?
 
Physics news on Phys.org
  • #2
A few points here:
what are you measuring the time constant in? you say 0.01t...what units are those? Units are very important when you want to give a physical meaning to your calculations.

Secondly, the steady state current is just the theoretical maximum of the current as it builds up as far as it can go. It's a simple relationship between the maximum potential of the battery and the resistance of the circuit.
In order to answer the second part about voltage across inductor and resistor, try thinking about what these two elements will be doing when you reach the steady state current...you don't even really need to calculate that part.

Hope that helps
 
  • #3
I forgot to say: the equation you gave gives the current as a function of time. It's not really necessary if you're trying to find the steady state current, but you can use that equation if you like. Just take time (t) to the limit and see what happens to the right-hand side of the equation.
 
  • #4
tomwilliam said:
A few points here:
what are you measuring the time constant in? you say 0.01t...what units are those? Units are very important when you want to give a physical meaning to your calculations.

Secondly, the steady state current is just the theoretical maximum of the current as it builds up as far as it can go. It's a simple relationship between the maximum potential of the battery and the resistance of the circuit.
In order to answer the second part about voltage across inductor and resistor, try thinking about what these two elements will be doing when you reach the steady state current...you don't even really need to calculate that part.

Hope that helps

thanks for quick reply, this is due tommorow :(

I got the 0.001t because previously in the question it had asked me to work out the time constant in the circuit.

I have found out how to work outthe time when the current reaches 50% of steady state value. I believe that to use the equation 0.5=e^-Rt/L

I am really confused about this though
 
  • #5
ok...firstly, the time constant units should be in seconds...so if you calculated 0.01s you should check what your starting units are in. HINT: if you travel 3 kilometers in 1 second you can't call that 3 m/s...


imax = Vbat/R is the equation you need to find the steady state current. At this point, the voltage will be fully across the resistor. That should be enough information to help you find the voltage across the resistor and the inductor.

To find the time at which the current reaches 50% of its steady state value, first you need to find the maximum current (steady state current) using the formula I showed you above. Then, use the formula that you gave in your original post, which shows i as a function of time, and put them together. You know that half of the steady state value (on the left hand side) will be equal to the equation for i as a function of time at time t_half. Then just rearrange the equation to find t_half. Make sense?
 
  • #6
So the steady state will just be 6.0V/10ohm = 0.6V, the magnitudes across the resistor and the inductor i am not to sure about. Is it as simple as Vresistor = I/V
 
  • #7
is the current across the inductor i(t)=I(e(^-tR/L) ?
 
  • #8
Not quite. Your value is right, your units are wrong. You don't measure current in V.

On the magnitudes of the voltage across the resistor and inductor...use Ohm's law: V=iR
At steady state, the voltage will be all across the resistor.
 
  • #9
sorry its 0.6A
So the magnitude over the resistor should be V=0.6A/10ohm=0.06V

i think i have the wrong equation for the inductor
 
  • #10
"is the current across the inductor i(t)=I(e(^-tR/L) ?"

The VOLTAGE across the inductor changes as a function of time. What are you looking for, current or voltage? There is a graph here that shows how voltage across the inductor changes as a function of time:
http://www.allaboutcircuits.com/vol_1/chpt_16/3.html
 
  • #11
I am looking for the voltage across the inductor. I can't seem to find in my text how to work this out.
 
  • #12
The point is that the voltage across the inductor and the resistor change as a function of time. You can't answer the question you posed unless you say at what point you're interested in.
If you look at the graph of inductor voltage / time in that link I posted, you'll see what happens to the voltage across the inductor over time. Then ask yourself what steady state means. That should give you an answer without needing much calculation. You can also use that answer to deduce the voltage across the resistor.
 

1. What is a steady-state current and voltage?

In an inductive-resistive circuit, the steady-state current and voltage refer to the values of current and voltage that remain constant over time once the circuit has reached a stable state. This means that the current and voltage are no longer changing and have reached a consistent level.

2. How do you determine the steady-state current and voltage in an inductive-resistive circuit?

To determine the steady-state current and voltage, you will need to use the equations for inductors and resistors. First, calculate the total resistance of the circuit by adding up the resistances of all components. Then, use the equation V = IR to calculate the voltage across each component. Finally, use the equation V = L(di/dt) to calculate the inductive voltage, which is the voltage caused by the changing current in the inductor. The steady-state current can then be calculated by dividing the total voltage by the total resistance.

3. What factors affect the steady-state current and voltage in an inductive-resistive circuit?

The steady-state current and voltage in an inductive-resistive circuit are affected by the values of the inductance and resistance in the circuit, as well as the frequency of the alternating current. Higher inductance and resistance will result in higher steady-state current and voltage values, while a higher frequency will result in lower steady-state values.

4. How does the presence of a capacitor in an inductive-resistive circuit affect the steady-state current and voltage?

If a capacitor is present in the circuit, it will have a counteracting effect on the inductor. This means that the capacitor will store energy when the current is increasing and release energy when the current is decreasing. As a result, the presence of a capacitor will cause the steady-state current and voltage to be lower than what would be expected without the capacitor.

5. Can the steady-state current and voltage change over time in an inductive-resistive circuit?

No, the steady-state current and voltage in an inductive-resistive circuit should remain constant over time as long as the circuit does not experience any changes. However, if there are any changes in the circuit, such as a change in frequency or a component failure, the steady-state values may also change.

Suggested for: Determining Steady-State Current and Voltages in Inductive-Resistive Circuit

Replies
7
Views
869
Replies
8
Views
200
Replies
13
Views
688
Replies
8
Views
725
Replies
21
Views
265
Replies
1
Views
784
Replies
2
Views
770
Replies
52
Views
2K
Replies
6
Views
941
Back
Top