2
\$\begingroup\$

I need to solve these two exercises but I really have no idea where to start. How should I proceed to solve this kind of problems?

I tried to find something on my teacher's files but haven't understood how to solve the exercises because there are no examples.

What is the difference between I(RMS) and I(with the dot)?

Text

\$\endgroup\$
2
  • \$\begingroup\$ For figure 11 are you able to write the expression for U in terms of (1) i1 and R and XL and (2) i2 and XC?. I=i1+i2 is in phase with U. Start from the above. \$\endgroup\$
    – AJN
    Nov 11, 2020 at 18:17
  • \$\begingroup\$ @AJN no I can't do that. I dont understand also the difference between I(RMS) and I (with the dot) \$\endgroup\$
    – user266743
    Nov 11, 2020 at 18:26

1 Answer 1

1
\$\begingroup\$

Based on the problem saying,

$$\dot{U} \text{ and } \dot{I} \text{ are in phase.}$$

It seems clear to me that the dot over the variables indicates phasor quantities. Their magnitudes are given in rms.

So, your solution will be something like this,

$$\dot{U} = U_{rms}\angle \theta° $$ $$\dot{I} = I_{rms}\angle \theta° $$

or, if you set the θ to be your reference then,

$$\dot{U} = U_{rms}\angle 0° $$ $$\dot{I} = I_{rms}\angle 0° $$

\$\endgroup\$
3
  • \$\begingroup\$ Thank you, it seems so strange that the answer is so simple. For the problem 1 I just need to add I1, I2 and that's it. So my question is: why even bother to write a circuit? Should I do something with Xc, R, XL? \$\endgroup\$
    – user266743
    Nov 12, 2020 at 6:43
  • \$\begingroup\$ Your welcome. Yes, you have little more to do. $$\dot{i_1} = \frac{U_{rms}\angle 0°}{R+jX_L} \text{ and } \dot{i_2} = \frac{U_{rms}\angle 0°}{-jX_C} $$ You need to take the given information and find the effective (rms) value of i. \$\endgroup\$ Nov 12, 2020 at 15:10
  • \$\begingroup\$ Here is an additional hint. If U and I are in phase, then the reactive component (at -90°) of the i1 current must exactly match the reactive component of i2 (all reactive at +90°) so that only a resistive component of current I remains. \$\endgroup\$ Nov 12, 2020 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.