Circuit Impedance and real energy used.

[learningbeforeearning]

I have been asked a question and am unsure of where to start as I am brand new to this topic.

The question is:
A 230V a.c inductive a.c load has a phase angle of 41.4Deg and a resistance of 15Ohms. Calculate the circuit impedance and the real energy used after a quarter of an hour.

Any assistance is appreciated.

 

[poni]

Do you know what they want and all the terms used( ie resistance and impedance ,just to mention two)?,

 

[runningpiglet]

I have been asked a question and am unsure of where to start as I am brand new to this topic.

The question is:
A 230V a.c inductive a.c load has a phase angle of 41.4Deg and a resistance of 15Ohms. Calculate the circuit impedance and the real energy used after a quarter of an hour.

Any assistance is appreciated.

It sounds the kind of question posed to students so you might find textbooks have guidance and examples. As a clue I'd suggest you think what might happen if the phase shift were 45 degrees - then the reactive impedance would be j15 ohms i.e. the same as the resistance and the magnitude of the impedance 21.21ohms. Since the phase angle is lower than 45deg then you already know that the reactive impedance will be somewhat less than 15ohm. You probably won't be able to work it out in your head but it is not a very hard calculation. Knowing the impedance you can next work out the current drawn and from that the heat burned in the resistance. The energy answer could be expressed in mega joules or kilowatthours - do they say which units they want?

 

[BorisJ]

It sounds the kind of question posed to students so you might find textbooks have guidance and examples. As a clue I'd suggest you think what might happen if the phase shift were 45 degrees - then the reactive impedance would be j15 ohms i.e. the same as the resistance and the magnitude of the impedance 21.21ohms. Since the phase angle is lower than 45deg then you already know that the reactive impedance will be somewhat less than 15ohm. You probably won't be able to work it out in your head but it is not a very hard calculation. Knowing the impedance you can next work out the current drawn and from that the heat burned in the resistance. The energy answer could be expressed in mega joules or kilowatthours - do they say which units they want?

Look at it a a phaser diagram with 15 ohms on the horizontal axis. Vertical axis is inductive impedance and is tan41.4 x 15= 13.22 ohms. From pythagorus we can work out the circuit impedance (sq root of x2 + y2) and it comes to 20 ohms. 230/20 gives us a current of 11.5 amps. Power = i2 r 11.5x11.5 x15 = 1983.75 watts. Energy used in 1/4 of an hour 0.25 of that = 496watts.