Problem about steady-state circuit analysis

[e0ne199]

Homework Statement

hello everyone, i have a problem about calculating power absorbed by elements in a steady-state circuit

Relevant Equations

the question is about power so the equation i am about to use is P=V*I and P = I^2*R

Here is the question, and my solution is for question (a) :

knowns :
I = 12 cos 2000t = 12 ∠0°
L=0.2 H
R=200Ω
ω=2000, X_L = j*0.2*2000 = j400Ω
Z_t = (j400*200)/(200+j400) = 178.9 ∠26.57°Ω

i want to find the power absorbed by R, so :
I_R = (Z_t /R)*I = (178.9 ∠26.57°Ω/200)*12 ∠0° = 10.73∠26.57°A
P = I²*R = (10.73∠26.57°)²*200 = 23040∠53.14°watt
convert to time domain result :
P=23040 cos (2000t + 53.14°) watt

t = 1 ms, P=23040 cos (2000 (0.001) + 53.14°)

P = 13,170 kW

my answer is obviously wrong since the right answer for question (a) is 13.98 kW, and i know if i want to continue to the next question it will also be a wrong answer too, so would you like to help me point out which is wrong in my solution? any response is appreciated...

 

[anorlunda]

The inductor absorbs zero energy over an entire cycle. What about at specific times within a cycle?

 

[e0ne199]

The inductor absorbs zero energy over an entire cycle. What about at specific times within a cycle?

it is when t=1ms and the answer i made above is for the resistor, could you please point out what it is wrong with my answer?

 

[cnh1995]

could you please point out what it is wrong with my answer?

As you are asked to consider a specific instant of time, you can't use phasor algebra here. You need to consider the instantaneous values for voltage and current at t= 1ms.

 

[e0ne199]

As you are asked to consider a specific instant of time, you can't use phasor algebra here. You need to consider the instantaneous values for voltage and current at t= 1ms.

could you give me an example?

 

[cnh1995]

If you knew the current through R at t= 1ms, how would you calculate the power being absorbed in R at that instant?

 

[e0ne199]

If you knew the current through R at t= 1ms, how would you calculate the power being absorbed in R at that instant?

i need to get the current or the voltage across the R value...do you mean i have to set everything to t=1ms?
for example, I(t=1ms) = 12 cos (2000*0.001) = 11.99V?

 

[e0ne199]

hello anyone?

 

[cnh1995]

I(t=1ms) = 12 cos (2000*0.001) = 11.99V?

Yes, you should begin from there.
(Be careful with the units.)
So,

If you knew the current through R at t= 1ms, how would you calculate the power being absorbed in R at that instant?

 

[e0ne199]

Yes, you should begin from there.
(Be careful with the units.)
So,

btw how about the inductor?
should i make it time-based first : 400 cos (2000t + 90°) and put t=1ms into that equation too? or do i have to find the Z equivalent first and convert it to time based equation?

 

[cnh1995]

You can solve this problem in a number of ways.

Have you studied differential equations and Laplace Transform?

 

[e0ne199]

You can solve this problem in a number of ways.

Have you studied differential equations and Laplace Transform?

yes i have learned that but how do you use that to this problem?
i have done what you suggested but i am still unable to reach the conclusion, probably because i don't know how to treat the inductor in that circuit, i hope you would like to help me a little bit more about it

 

[cnh1995]

Ok, it is a bit difficult for me to suggest anyone approach as it can be solved in a number of ways.

I can think of only one method which does not involve calculus. It may not be the best approach, but might be easier to follow if you don't want to use much math.
Find the steady state rms current (and its phase) through R using current division rule. You can use the reactance of the inductor in this calculation.
Once you get that, you can find the instantaneous current and power absorbed in it using simple trigonometry.

So, start with the current division rule and find Irms through R.

 

[e0ne199]

Ok, it is a bit difficult for me to suggest anyone approach as it can be solved in a number of ways.

I can think of only one method which does not involve calculus. It may not be the best approach, but might be easier to follow if you don't want to use much math.
Find the steady state rms current (and its phase) through R using current division rule. You can use the reactance of the inductor in this calculation.
Once you get that, you can find the instantaneous current and power absorbed in it using simple trigonometry.

So, start with the current division rule and find Irms through R.

so do i have to find total impedance of the resistor and inductor? about the inductor, do i have to start calculating its reactance by assume its reactance at t=1ms?

 

[anorlunda]

ω=2000, XL = j*0.2*2000 = j400Ω

Inductive impedance is valid only for complete cycles. For less than one complete cycle you must use LdI/dt. and solve the differential equations.

 

[Ibitoye]

Homework Statement:: hello everyone, i have a problem about calculating power absorbed by elements in a steady-state circuit
Relevant Equations:: the question is about power so the equation i am about to use is P=V*I and P = I^2*R

Here is the question, and my solution is for question (a) :

View attachment 246566

knowns :
I = 12 cos 2000t = 12 ∠0°
L=0.2 H
R=200Ω
ω=2000, X_L = j*0.2*2000 = j400Ω
Z_t = (j400*200)/(200+j400) = 178.9 ∠26.57°Ω

i want to find the power absorbed by R, so :
I_R = (Z_t /R)*I = (178.9 ∠26.57°Ω/200)*12 ∠0° = 10.73∠26.57°A
P = I²*R = (10.73∠26.57°)²*200 = 23040∠53.14°watt
convert to time domain result :
P=23040 cos (2000t + 53.14°) watt

t = 1 ms, P=23040 cos (2000 (0.001) + 53.14°)

P = 13,170 kW

my answer is obviously wrong since the right answer for question (a) is 13.98 kW, and i know if i want to continue to the next question it will also be a wrong answer too, so would you like to help me point out which is wrong in my solution? any response is appreciated...

the issue in your solution is that you computed the Ir(resistor current) wrongly.
Compute for I_R = (X_L/X_L+R)*I, you should be just fine
and also avoid getting a single phasor expresion for your power, because power is not a linear response.

 

[Steve4Physics]

Here’s one possible strategy that (I believe) is fairly straightforward...

Find the total complex impedance, ##Z_T##.
Find ##V(t)##, the source voltage (the voltage across R and L) using ##V(t) = I(t) Z_T##.
Evaluate ##V(0.001)## (source voltage at t=1ms).
Evaluate ##I(0.001)## (source current at t=1ms).
_________________

a) The instantaneous power absorbed by R at t=1ms is ##P = \frac {V(0.001)^2}{R}##.

b) Find the current through R at t=1ms using ##I_R = \frac {V(0.001)}{R}##
Using Kirchhoff’s 1st law, the current through L at t=1ms is ##I_L = I(0.001) - I_R##
The instantaneous power power absorbed by L at t=1ms is ##P = V(0.001) \times I_L##.

c) For instantaneous power absorbed by the source at t=1ms, use the answers from parts a) and b), applying conservation of energy.

Take care with signs (+, -).

(Edit: Typo' corrected.)

 

[boo]

Easiest way to solve this is to use phasors to calculate both current and voltage through resistor and inductor. (Of course the resistor and inductor voltages will be the same). Once you have their magnitude and phase use the following formulas for instantaneous power:

P(resistor) = 0.5*Vm*Im*(1 + cos(2wt + 2*theta_v))
P(inductor) = 0.5*Vm*Im*cos(2wt + 2*theta_v - 90)

where Vm and I am are resistor voltage and current (peak not RMS) in the first formula and inductor voltage and current in the second. and here w = 2000 rad/sec and t = 1 ms.

since you will likely have the angle theta_v in degrees remember to convert the 2wt term into degrees as well by multiplying by 180/pi.

If you don't want to use these formulas just calculate the magnitude and phase of V and I for the circuit elements, convert to the time domain and use P = 0.5* v(t)*i(t). that's what these formulas reduce to anyways after some trigonometric manipulation.

 

[boo]

In case it's unclear above Vm is the voltage phasor magnitude and theta_v is the voltage phasor angle (phase).

 

[phongsakb]

i need to get the current or the voltage across the R value...do you mean i have to set everything to t=1ms?
for example, I(t=1ms) = 12 cos (2000*0.001) = 11.99V?

12cos(2 rad) = -4.99376

 

[phongsakb]