Inductors and capacitors in AC circuits are kind of like resistors in DC circuits, except we use a more general concept of Impedance (Z) rather than resistnace. Circuits with impedance can be solved using ohms law, just like with resistors: I*Z=V. The only catch is that Z is a complex number; hence consists of 2 components, R and j*X, j is the imaginary unit, and X is the 'rectance'. To solve an AC circuit, convert every thing to impedances, then solve with 'ohms law'. The impedance of an inductor is 2*j*pi*L*f where f is the frequency. But in this question you dont even need to explicitly calculate this!
so for part 1) you can simply use ohms law across the resistor, so we have 64v across an 8 ohm resistor, so the current through the resistor (hence the inductor) must be 8A.
2) Here we use Z*I=V, so we get 100=Z*8, so Z is 12.5 ohms
3) This ones a bit more tricky. An impedance that is purely reactive, doesnt absorb any power averaged over time. During the first half cycle of the sinusiod, It absorbs power, but during the second half cycle, it delivers power back to the supply. So the average power is 0! So it follows from this, that the only power absorbed by the circuit, is that which is absorbed by the resistor, which can can calcultae easy, its just V*I, across the resistor, which is 512W. BUT, the question may be asking for the 'apparent power'. This is more complicated, as it consists of two components, the 'real' power into the resistor, and the so-called 'reactive power' into the inductor. If this is the case we can find the overall power into the circuit as a complex number, and it will be the current^2 multiplied by the complex impedance of the circuit. This gives you a complex answer for the power, whos real component is the power absorbed by the resistor, and the imaginary component is the reactive power into the inductor.
4)as above
5) Not sure how this should be different to q3....seems like a silly question.
6) Power factor of the load is the ratio of the real part of the power, to the magnitude of the apparent power. So in this case, the magnitude of the apparent power is simply aplied voltage * current into the circuit, i.e. 800 watts, and the real power we have already calculated, as 512 watts, so the power factor is 0.64. This can also be calculated from the angle (argument) of the complex impednce of the circuit. (Complex numbers do have a use!)
7) once again, as no details are given about the source, this would probably be the same as q6, it seems like a rather silly question
Hope this helps!
so for part 1) you can simply use ohms law across the resistor, so we have 64v across an 8 ohm resistor, so the current through the resistor (hence the inductor) must be 8A.
2) Here we use Z*I=V, so we get 100=Z*8, so Z is 12.5 ohms
3) This ones a bit more tricky. An impedance that is purely reactive, doesnt absorb any power averaged over time. During the first half cycle of the sinusiod, It absorbs power, but during the second half cycle, it delivers power back to the supply. So the average power is 0! So it follows from this, that the only power absorbed by the circuit, is that which is absorbed by the resistor, which can can calcultae easy, its just V*I, across the resistor, which is 512W. BUT, the question may be asking for the 'apparent power'. This is more complicated, as it consists of two components, the 'real' power into the resistor, and the so-called 'reactive power' into the inductor. If this is the case we can find the overall power into the circuit as a complex number, and it will be the current^2 multiplied by the complex impedance of the circuit. This gives you a complex answer for the power, whos real component is the power absorbed by the resistor, and the imaginary component is the reactive power into the inductor.
4)as above
5) Not sure how this should be different to q3....seems like a silly question.
6) Power factor of the load is the ratio of the real part of the power, to the magnitude of the apparent power. So in this case, the magnitude of the apparent power is simply aplied voltage * current into the circuit, i.e. 800 watts, and the real power we have already calculated, as 512 watts, so the power factor is 0.64. This can also be calculated from the angle (argument) of the complex impednce of the circuit. (Complex numbers do have a use!)
7) once again, as no details are given about the source, this would probably be the same as q6, it seems like a rather silly question
Hope this helps!