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I have this for Q4. a):

enter image description here

I have been struggling to calculate this as my numbers do not seem to be correct. When do I include the 90° lead that voltage has over current due to the inductor? Here's my attempt:

enter image description here

I understand that part b is pretty much the same but Xc = 1/wC and I want just the branch, not total current.

Any help or pointers with this would be great as I am a bit stuck on this one.

Thanks

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  • \$\begingroup\$ w value used in formula for XL is wrong. w=1000 \$\endgroup\$ – Harshad D Aug 25 '16 at 17:39
  • \$\begingroup\$ Um. Just curious. Isn't your step (1) in 4a wrong? \$\endgroup\$ – jonk Aug 25 '16 at 17:40
  • \$\begingroup\$ Maybe this will help you understand all the places you have gone wrong. electrical4u.com/rl-parallel-circuit \$\endgroup\$ – Harshad D Aug 25 '16 at 17:47
  • \$\begingroup\$ I was under the impression that t = time? If so wouldn't I need 1/t = frequency for 2*pi*frequency? \$\endgroup\$ – Christopher Dyer Aug 25 '16 at 18:00
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You attempt has many issues. First, let's clear up what \$\omega\$ is. With an expression in the form of \$ \sin(at) \$, one period or one cycle is when: $$ at_{period} = 2\pi $$ $$\Rightarrow a = 2\pi\frac{1}{t_{period}} = 2\pi\times{freq} = \omega $$ Therefore, \$ \omega \$ is simply the coefficient in front of \$ t \$, which is \$10^3\$ in 4a).

For the circuit 4a): $$ I_{total} = \frac{V}{Z_{total}} $$ $$ Z_{total} = \frac{1}{\frac{1}{Z_R}+\frac{1}{Z_L}} $$ (\$ Z_{total} \$ is not \$ Z_R + Z_L \$ as in your step 3.)

The equation is equivalent to: $$ I_{total} = \frac{V}{Z_R} + \frac{V}{Z_L} = I_R + I_L $$ I will use this alternative representation because the intermediate quantities are slightly more interesting. $$ I_R = \frac{100\angle 50^\circ}{5} = 20\angle 50^\circ $$ You are looking for an answer with time dependence, use the impedance of the inductor \$Z_L = j\omega L\$ which has the time related phase information (don't use reactance). $$ I_L = \frac{100\angle 50^\circ}{(j\omega L = \omega L\angle 90^\circ)} = \frac{100}{1000\times0.02} \angle(50-90)^\circ = 5\angle{-40}^\circ $$ Finally, $$ I_{total} = 20\angle 50 + 5\angle {-40} = 20.6\angle 35.96 $$ (You need to look up how to add two numbers with phase angles. You cannot just add the amplitudes and the phase angles.) $$ I_{total} = 20.6\sin(10^3t+35.96^\circ)\space A $$

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  • \$\begingroup\$ Excellent answer. Helped confirm some issues that I had. \$\endgroup\$ – Christopher Dyer Aug 30 '16 at 12:14

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