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I have the below question. I am not trying to ask the answer for the problem.

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But my question is that, we can see that 60deg is the phase shift.

But both the voltage and current across the passive element is 60deg. With respect to what is the phase shift 60deg.

Is the phase shift 60deg is with respect to AC Voltage source? Or is it with respect to what?

And for the power calculation, I should just multiply the values of V and I.

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If I said that the voltage was \$10\cos(100t)\$ and the current was \$5\sin(100t)\$, would you be asking about the phase angle? I mean; the phase angle (when not mentioned) is implicitly assumed to be 0° so there wouldn't be any worry there if it wasn't mentioned. Are you OK with that?

So, if both voltage and current are shifted by 60° (or any phase angle), does it really affect how you would answer the question? No, because both are time shifted by the same amount and, the load that converted the cosine voltage waveform into a sinusoidal current is still exactly the same.

With respect to what is the phase shift 60deg.

To answer your question; it matters not one bit what the 60° is relative to. And, it's exactly the same when phase angle is not mentioned (assumed to be 0°). Other (different) problems involving phase angle might require us to validly ask what it is respect to but not in this example.

And for the power calculation, I should just multiply the values of V and I

Unfortunately, the answers given are probably all wrong.

When we talk about voltages or currents and mention the "sin" or "cos" parts we are implying that the voltage or current is a peak value (not an RMS value). To calculate VI or watts we have to use RMS voltage and current values and, the apparent power for a 10 volt peak amplitude voltage and a 5 amp peak amplitude current is 25 VA. So, if the answers to the question are all using phasor notation (\$X\angle\$ for instance), \$X\$ should be an RMS value.

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  • \$\begingroup\$ Ok thank you. For example, if the phase shift is different for Voltage and Current, the phase shift would be caused by the load, am I correct? And in that case, what is the phase shift relative to ? \$\endgroup\$ – Newbie Mar 30 at 5:01
  • \$\begingroup\$ The phase difference is caused by the load. \$\endgroup\$ – Andy aka Mar 30 at 5:04
  • \$\begingroup\$ Thank you. I didn't know the answers were wrong. Could you please tell me how you got 25VA and should we just ignore the Sin and Cos while calculation of the Power? \$\endgroup\$ – Newbie Mar 30 at 5:17
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    \$\begingroup\$ 25VA = \$\dfrac{10}{\sqrt2} \times \dfrac{5}{\sqrt2}\$. And there is an implied 90 deg phase shift in the original question due to the sin cos thing so that's where the 90 deg in the answer should come from i.e. answer (b) is correct except it should be 25 and not 50. \$\endgroup\$ – Andy aka Mar 30 at 5:20
  • \$\begingroup\$ Thank you for the answer. \$\endgroup\$ – Newbie Mar 30 at 5:21
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The complex power of a system is given by:

$$\text{Q}=\text{V}_\text{RMS}\cdot\text{I}_\text{RMS}\cdot\sin\left(\varphi\right)\tag1$$

Where \$\varphi\$ is the absolute phasedifference between the voltage and the current.

Using your given functions, we have:

  • $$\text{V}\left(t\right)=10\cos\left(100t+60^\circ\right)=10\cos\left(100t+\frac{\pi}{3}\right)\tag2$$
  • $$\text{I}\left(t\right)=5\sin\left(100t+60^\circ\right)=5\cos\left(100t-\frac{\pi}{6}\right)\tag3$$

So:

$$\text{Q}=\frac{10}{\sqrt{2}}\cdot\frac{5}{\sqrt{2}}\cdot\sin\left(\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right)=25\cdot1=25\space\text{VAR}\tag4$$

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  • \$\begingroup\$ Better check the equivalent displacements for (2) and (3). Since they're the same, \$\sin{\phi}=0\$. \$\endgroup\$ – a concerned citizen Mar 30 at 18:23
  • \$\begingroup\$ This answer does not address the OP's question at all. \$\endgroup\$ – Elliot Alderson Mar 30 at 19:00

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