# series circuit regarding rms current

Discussion in 'Homework Help' started by squirby, Nov 29, 2009.

1. ### squirby Thread Starter Member

Aug 21, 2009
15
0
hey guys. just wondering if anyone could help me solve this question. i got an answer but i'm not sure.

"The series combination of a 100 ohm resistor and 200 mH inductor must not absorb more than 5W of power at any instant. Assuming a sinusoidal current with ω = 400 rad/s, what is the largest rms current that can be tolerated?"

my answer for the rms current was 20.2484A.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
You may have lost track of the decimal point. Power may be calculated using I^2*R. For a current of an even 20 amps, I^2 is 400. Times 100 (the resistor alone) is 40,000 watts.

3. ### mattc82 Member

Mar 13, 2009
22
0
I worked it and got roughly 27.6 mA < -39° which would make since due to the inductor.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Keep in mind also that if the power is "true" power (5Watts) then the frequency and inductance are irrelevant in this situation.

The pure inductance can't dissipate power so the series resistor is the sole "sink" of power in the circuit.

Hence you should be able to find Irms directly, from the relationship P=(Irms^2)R

Are you sure the question didn't ask for the largest total applied rms voltage?

5. ### lmartinez Active Member

Mar 8, 2009
224
6
The instantaneous power delivered to an RL seris circuit with a sinusoidal voltage source can be calculated by:

p = Vrms*Irms*(cos θ + cos(2*ω*t + θ)

However, if you are referring to the average power, then the last term on the above given equation goes away. The average power can then be calculated by:

p = Vrms*Irms*(cos θ )

based on the information you provided, it can not be confirmed that the value of 20.2484 is correct. Please advice

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Let's put some numbers in

P=5W & R=100Ω

Irms=√(5/100)=.2236A - the largest allowable rms current value to stay within 5W dissipation.

Isn't that the end of the matter?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
It might also be argued that the question does actually ask for the Irms value which gives a peak instantaneous power of 5W - a remote possibility.

In this case the instantaneous power would take the form

$p(t)=100Ipk^{2}sin^{2}(\omega t)$

So if p(t) max = 5W

Hence

$p(t) max = 5 = 100Ipk^{2}$
$Ipk^{2}=.05$
$Ipk=0.2236$
$Irms=0.1581$

8. ### squirby Thread Starter Member

Aug 21, 2009
15
0
hey guys. thanks for the replies. i redid my question and got 62.48mA as the rms current. i got this by getting the phasor form for the series combination of the resistor and inductor and using power formula.

or is t_n_k correct in that the resistor is the sole sink of the current and this was a trick question in that you could totally ignore the inductor and the frequency?

thanks guys.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Well if the rms current is 62.48mA then the power in the resistor is 0.39Watts - which is a long way short of 5W.