I'm studying Electronics test questions for the third of three job interviews . I've run
into questions that stumped me before but always eventually see what I overlooked.
I have one now that should be simple but I can't see what I'm missing:
Question: What is the true power consumed in a 30V series RLC circuit if Z=20 and R=10 ohms.
I know
Z = SQUAREROOT( R^2 + [(X(L) -X(C)]^2)
and
P(true) = E x I x COS(theta)
and
(theta)= arctan[I(L)/I(R))]
and
PF=COS(theta)
but since the values for L and C are not given I'm drawing a blank on how to solve it.
I know it has nothing to do with arduinos but it can't hurt to discuss it can it ?
frequency is not given. (the answer is given though so I'll know when the solution is correct. I used the answer to reverse engineer the answer so I know theta and of course COS(theta). Question is from IndiaBix.com Test 4. (I did 1,2,3 & 5 already)
I worked through the pandemic. I've never worked remote, ever in my life, and would not no matter how much the job payed.
The company I worked at for 2 years just ran out of funding so they furloughed all the workers. The company I'm interviewing for is just down the street. (72) Because
we were furloughed we still have healthcare, in case they call us back.
I think there's a mistake in the question .
COS(theta) should be R/Z but that doesn't work because theta is 7505 degrees and
cos(theta) = 0.25.
The true power (answer) is 22.5W.
P(true)= E x I x COS(theta)
= 30 x 3 x 0.2
= 22.5W
Using COS(theta) = R/Z and using the given R=10, and solving for Z using the
theta = 75.5 degrees from reverse engineering the question from the answer,
I get Z=40, which works for COS(theta) = R/Z if theta = 75.5 degrees.
cos(75.5) = 0.25
Given R=10, 10/Z = 0.25 => Z=40 (NOT 20).
current = 30/20 = 1.5A, therefore voltage across resistor is 10 * 1.5 = 15V, so real power in the resistor is 15 x 1.5 = 22.5W
apparent power is 30 x 1.5A = 45W, and hey presto apparent/real = 0.5 = cos(theta).
I don't get all this 75.5degrees or the 0.25 at all.
The question just gives you abs(Z) and R. That and the fact its a series circuit (current same everywhere) is all you need.
Given only: Z=20, R=10, V=30, series RLC cos(theta) = R/|Z| = 10/20 = 0.5 = > ACOS(cos(theta))= theta = 60 degrees. Current: I = Z/R=30/20 = 1.5A
.'. voltage across resistor: V= I x R=1.5 x 10 = 15V
.'. Real Power: (in the resistor) : P= I x V = 1.5 x 15 = 22.5W
Also expressed as True Power (TP) = I^2 x R = 1.5A x 10 ohms = 22.5W Solution: 22.5W Answer: 22.5W
That's the only thing that makes sense based on what's given.
Thanks Mark.
I can sleep now...
L isn't given, neither is C, so can you clarify what you mean by "cos(theta) = R/L" ?
Given only Z=20, R=10, V=30, series RLC, how do I get to cos(theta) = R/L ?