After our PM exchanges, I will go ahead and solve the problem as an example.
I don't have schematic capture software here and don't feel like spending a bunch of time with PowerPoint or Paint or some other drawing program, so I will describe the schematic in words.
There are three components in parallel each oriented to that current flows vertically through them. The bottom nodes of all three components is tied to ground, Similarly, the top nodes of all three components are tied together and this node is v(t). The three nodes are:
Current source: 12cos(wt) A; w = 2000 r/s
Resistor: 200Ω
Inductor: 200mH
By the passive sign convention, the symbolic current in the current source, i(t), is positive upward and the symbolic current in the resistor, i_R(t), and inductor, i_L(t), are both downward.
Working in the phasor domain (a.k.a., the frequency domain and/or Fourier domain), we will replace the current source and resistor with their Thevenin equivalent as seen by the inductor. We will then solve for the node voltage, V(jw), across the inductor. With this, we will then go back to the original circuits to find the phasor current in all three components. Having found the voltage and currents, we will then translate to the time domain and determine the values of each at t=1ms and compute the instantaneous power at that time.
First, translate our components to the frequency domain:
\(
i(t) = I_S \cos(\omega t)\ \Rightarrow \ I(s) = I_S
I_S = 12 \angle 0^oA
\omega = 2000 r/s
\
R = 200\Omega \ \Rightarrow \ Z_R = 200\Omega
\
L = 200mH \ \Rightarrow \ Z_L = j \omega L
Z_L = j (2000 \frac{r}{s})(200mH)(\frac{Vs}{A})
Z_L = j 400 \Omega
\)
Next, let's convert the current source and resistor into a Thevenin equivalent circuit:
\(
V_{TH}= I_S \ Z_R
Z_{TH} = Z_R
\)
We can now find the node voltage by voltage division:
\(
V = V_{TH} \ \frac{Z_L}{Z_{TH}+Z_L}
\)
With this, we can go back to the original circuit and find the currents in the two passive components:
\(
I_R = \frac{V}{Z_R} = V_{TH} \ \frac{Z_L}{Z_R(Z_R+Z_L)}
I_R = I_S \ \frac{Z_L}{Z_R+Z_L}
\
I_L = \frac{V}{Z_L} = V_{TH} \ \frac{Z_L}{Z_L(Z_R+Z_L)}
I_L = I_S \ \frac{Z_R}{Z_R+Z_L}
\)
Which, of course, are nothing but the current division formulas which, frankly, we could have started with. But I can never just rattle it off and so I usually perform a more fundamental analysis.
As a quick sanity check, we verify that KCL is satistified:
\(
I = I_R + I_L \ ?
I = I_S \ \frac{Z_L}{Z_R+Z_L} \ + \ I_S \ \frac{Z_R}{Z_R+Z_L}
I = I_S \ \frac{Z_L+Z_R}{Z_R+Z_L}
I = I_S \ (check)
\)
We now have the voltages and the currents in each component, but in order to find power we have to go back to the time domain because power is nonlinear and our frequency domain work is only valid for linear relationships.
\(
I_R = I_S \ \frac{Z_L}{Z_R+Z_L}
I_R = I_S \ \frac{j \omega L}{R + j \omega L}
I_R = 12 \angle 0^o A \ \frac{j400 \Omega}{200\Omega + j400\Omega}
I_R = 12A \ \frac{j2}{1 + j2}
I_R = 12A \ \frac{j2}{1 + j2}\frac{1-j2}{1-j2}
I_R = 4.8(j)(1-j2) A
I_R = 4.8(2+j)A
I_R = 4.8 \sqrt{5} \angle 26.57^o A
\
i_R(t) = 4.8 sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) A
\)
and
\(
I_L = I_S\frac{Z_R}{Z_R+Z_L}
I_L = I_S\frac{R}{R + j \omega L}
I_L = 12 \angle 0^o A \frac{200 \Omega}{200\Omega + j400\Omega}
I_L = 12A \frac{1}{1 + j2}\frac{1-j2}{1-j2}
I_L = 2.4(1-j2)A
I_L = 2.4 \sqrt{5} \angle -63.43^o A
\
i_L(t) = 2.4 sqrt{5} cos(2000 \frac{r}{s} t \ - \ 63.43^o) A
\)
and then the node voltage
\(
V = V_{TH} \ \frac{Z_L}{Z_{TH}+Z_L}
V = I_S \ \frac{R j\omega L}{R+j \omega L}
V = 12 \angle 0^o A \ \frac{(200 \Omega)(j400 \Omega)}{(200+j400)\Omega}
V = 12A \ \frac{(j400 \Omega)}{(1+j2)}
V = 4800A\Omega \ \frac{j}{(1+j2)}\frac{1-j2}{1-j2}
V = 960(j)(1-j2)V
V = 960(2+j)V
V = 960 \sqrt{5} \angle 26.57^o V
\
v(t) = 960 \sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) V
\)
As a sanity check, note that the voltage across the resistor is in phase with the current through the resistor while the voltage across the inductor is 90deg ahead of the current through it. Had we made any math errors (and I made three), this check alone probably would have caught them. It would have caught all of mine, but I was able to catch them earlier because I estimated what I expected the angle to be in terms of which octant it should fall and all of my mistakes put it in the wrong octant.
Evaluating these at t = 1ms yields:
\(
2000 \frac{r}{s} 1ms \frac{180^o}{\pi} = 114.59^o
\
v(1ms) = 960 \sqrt{5} cos(114.59^o \ + \ 26.57^o) V
v(1ms) = 960 \sqrt{5} cos(141.16^o) V
v(1ms) = 960 \sqrt{5} cos(141.16^o) V
v(1ms) = 960 \sqrt{5}(-0.7789)V
\
i_R(1ms) = 4.8 sqrt{5} cos(114.59^o \ + \ 26.57^o) A
i_R(1ms) = 4.8 sqrt{5} cos(141.16^o) A
i_R(1ms) = 4.8 sqrt{5}(-0.7789)A
\
i_L(1ms) = 2.4 sqrt{5} cos(2000 \frac{r}{s} t \ - \ 63.43^o) A
i_L(1ms) = 2.4 sqrt{5} cos(114.59^o - \ 63.43^o) A
i_L(1ms) = 2.4 sqrt{5} cos(51.16^o) A
i_L(1ms) = 2.4 sqrt{5}(0.6271)A
\
i_S(1ms) = 12 cos(141.59^o)A
i_S(1ms) = 12 (-0.4161)A
\)
We can now calculate the powers at t=1ms
\(
P_R(t) = v_R(t)i_R(t)
P_R(1ms) = [960 \sqrt{5}(-0.7789)V][4.8 sqrt{5}(-0.7789)A]
P_R(1ms) = (960)(4.8)(5)(-0.7789)^2W
P_R(1ms) = \ +13.98kW
\
P_L(t) = v_L(t)i_L(t)
P_L(1ms) = [960 \sqrt{5}(-0.7789)V][2.4 sqrt{5}(0.6271)A]
P_L(1ms) = (960)(2.4)(5)(-0.7789)(0.6271)W
P_L(1ms) = (960)(2.4)(5)(-0.7789)(0.6271)W
P_L(1ms) = \ -5.63kW
\
P_S(t) = v_S(t)i_S(t)
P_S(1ms) = [960 \sqrt{5}(-0.7789)V][12(-0.4161)A]
P_S(1ms) = (12)(960)(\sqrt{5})(-0.7789)(-0.4161)W]
P_S(1ms) = \ +8.35kW
\)
As a final sanity check, we know that the power delivered by the current source has to equal the sum of the powers consumed by the resistor and inductor.
\(
P_S = P_R + P_L \ ?
P_S = +13.98kW \ + \ -5.63kW
P_S = +8.35kW \ (check)
\)
Now, it is always possible I have made a mistake that survived all of my units and sanity checks. The most likely such mistake is that I simply solved the wrong problem or made a mistake in setting up the problem.
Notice that I worked symbolically for as far as I could and I kept things like sqrt(5) around as long as I could. This not only reduces roundoff error in the final result, but it makes going back and checking for where you made a mistake a lot easier and correcting it once it is found much easier.
It also makes it easier to leverage the results to answer other questions. For instance, if we were then asked to find the power delivered by the current source as a function of time, we can do so in just a couple of lines.
\(
P_S(t) = v_S(t)i_S(t)
P_S(t) = [960 \sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) V][12 cos(2000 \frac{r}{s} t)A]
P_S(t) = 25.76 cos(2000 \frac{r}{s} t \ + \ 26.57^o) cos(2000 \frac{r}{s} t)kW
\)
We could then apply a trig identity to simplify this further.
I don't have schematic capture software here and don't feel like spending a bunch of time with PowerPoint or Paint or some other drawing program, so I will describe the schematic in words.
There are three components in parallel each oriented to that current flows vertically through them. The bottom nodes of all three components is tied to ground, Similarly, the top nodes of all three components are tied together and this node is v(t). The three nodes are:
Current source: 12cos(wt) A; w = 2000 r/s
Resistor: 200Ω
Inductor: 200mH
By the passive sign convention, the symbolic current in the current source, i(t), is positive upward and the symbolic current in the resistor, i_R(t), and inductor, i_L(t), are both downward.
Working in the phasor domain (a.k.a., the frequency domain and/or Fourier domain), we will replace the current source and resistor with their Thevenin equivalent as seen by the inductor. We will then solve for the node voltage, V(jw), across the inductor. With this, we will then go back to the original circuits to find the phasor current in all three components. Having found the voltage and currents, we will then translate to the time domain and determine the values of each at t=1ms and compute the instantaneous power at that time.
First, translate our components to the frequency domain:
\(
i(t) = I_S \cos(\omega t)\ \Rightarrow \ I(s) = I_S
I_S = 12 \angle 0^oA
\omega = 2000 r/s
\
R = 200\Omega \ \Rightarrow \ Z_R = 200\Omega
\
L = 200mH \ \Rightarrow \ Z_L = j \omega L
Z_L = j (2000 \frac{r}{s})(200mH)(\frac{Vs}{A})
Z_L = j 400 \Omega
\)
Next, let's convert the current source and resistor into a Thevenin equivalent circuit:
\(
V_{TH}= I_S \ Z_R
Z_{TH} = Z_R
\)
We can now find the node voltage by voltage division:
\(
V = V_{TH} \ \frac{Z_L}{Z_{TH}+Z_L}
\)
With this, we can go back to the original circuit and find the currents in the two passive components:
\(
I_R = \frac{V}{Z_R} = V_{TH} \ \frac{Z_L}{Z_R(Z_R+Z_L)}
I_R = I_S \ \frac{Z_L}{Z_R+Z_L}
\
I_L = \frac{V}{Z_L} = V_{TH} \ \frac{Z_L}{Z_L(Z_R+Z_L)}
I_L = I_S \ \frac{Z_R}{Z_R+Z_L}
\)
Which, of course, are nothing but the current division formulas which, frankly, we could have started with. But I can never just rattle it off and so I usually perform a more fundamental analysis.
As a quick sanity check, we verify that KCL is satistified:
\(
I = I_R + I_L \ ?
I = I_S \ \frac{Z_L}{Z_R+Z_L} \ + \ I_S \ \frac{Z_R}{Z_R+Z_L}
I = I_S \ \frac{Z_L+Z_R}{Z_R+Z_L}
I = I_S \ (check)
\)
We now have the voltages and the currents in each component, but in order to find power we have to go back to the time domain because power is nonlinear and our frequency domain work is only valid for linear relationships.
\(
I_R = I_S \ \frac{Z_L}{Z_R+Z_L}
I_R = I_S \ \frac{j \omega L}{R + j \omega L}
I_R = 12 \angle 0^o A \ \frac{j400 \Omega}{200\Omega + j400\Omega}
I_R = 12A \ \frac{j2}{1 + j2}
I_R = 12A \ \frac{j2}{1 + j2}\frac{1-j2}{1-j2}
I_R = 4.8(j)(1-j2) A
I_R = 4.8(2+j)A
I_R = 4.8 \sqrt{5} \angle 26.57^o A
\
i_R(t) = 4.8 sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) A
\)
and
\(
I_L = I_S\frac{Z_R}{Z_R+Z_L}
I_L = I_S\frac{R}{R + j \omega L}
I_L = 12 \angle 0^o A \frac{200 \Omega}{200\Omega + j400\Omega}
I_L = 12A \frac{1}{1 + j2}\frac{1-j2}{1-j2}
I_L = 2.4(1-j2)A
I_L = 2.4 \sqrt{5} \angle -63.43^o A
\
i_L(t) = 2.4 sqrt{5} cos(2000 \frac{r}{s} t \ - \ 63.43^o) A
\)
and then the node voltage
\(
V = V_{TH} \ \frac{Z_L}{Z_{TH}+Z_L}
V = I_S \ \frac{R j\omega L}{R+j \omega L}
V = 12 \angle 0^o A \ \frac{(200 \Omega)(j400 \Omega)}{(200+j400)\Omega}
V = 12A \ \frac{(j400 \Omega)}{(1+j2)}
V = 4800A\Omega \ \frac{j}{(1+j2)}\frac{1-j2}{1-j2}
V = 960(j)(1-j2)V
V = 960(2+j)V
V = 960 \sqrt{5} \angle 26.57^o V
\
v(t) = 960 \sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) V
\)
As a sanity check, note that the voltage across the resistor is in phase with the current through the resistor while the voltage across the inductor is 90deg ahead of the current through it. Had we made any math errors (and I made three), this check alone probably would have caught them. It would have caught all of mine, but I was able to catch them earlier because I estimated what I expected the angle to be in terms of which octant it should fall and all of my mistakes put it in the wrong octant.
Evaluating these at t = 1ms yields:
\(
2000 \frac{r}{s} 1ms \frac{180^o}{\pi} = 114.59^o
\
v(1ms) = 960 \sqrt{5} cos(114.59^o \ + \ 26.57^o) V
v(1ms) = 960 \sqrt{5} cos(141.16^o) V
v(1ms) = 960 \sqrt{5} cos(141.16^o) V
v(1ms) = 960 \sqrt{5}(-0.7789)V
\
i_R(1ms) = 4.8 sqrt{5} cos(114.59^o \ + \ 26.57^o) A
i_R(1ms) = 4.8 sqrt{5} cos(141.16^o) A
i_R(1ms) = 4.8 sqrt{5}(-0.7789)A
\
i_L(1ms) = 2.4 sqrt{5} cos(2000 \frac{r}{s} t \ - \ 63.43^o) A
i_L(1ms) = 2.4 sqrt{5} cos(114.59^o - \ 63.43^o) A
i_L(1ms) = 2.4 sqrt{5} cos(51.16^o) A
i_L(1ms) = 2.4 sqrt{5}(0.6271)A
\
i_S(1ms) = 12 cos(141.59^o)A
i_S(1ms) = 12 (-0.4161)A
\)
We can now calculate the powers at t=1ms
\(
P_R(t) = v_R(t)i_R(t)
P_R(1ms) = [960 \sqrt{5}(-0.7789)V][4.8 sqrt{5}(-0.7789)A]
P_R(1ms) = (960)(4.8)(5)(-0.7789)^2W
P_R(1ms) = \ +13.98kW
\
P_L(t) = v_L(t)i_L(t)
P_L(1ms) = [960 \sqrt{5}(-0.7789)V][2.4 sqrt{5}(0.6271)A]
P_L(1ms) = (960)(2.4)(5)(-0.7789)(0.6271)W
P_L(1ms) = (960)(2.4)(5)(-0.7789)(0.6271)W
P_L(1ms) = \ -5.63kW
\
P_S(t) = v_S(t)i_S(t)
P_S(1ms) = [960 \sqrt{5}(-0.7789)V][12(-0.4161)A]
P_S(1ms) = (12)(960)(\sqrt{5})(-0.7789)(-0.4161)W]
P_S(1ms) = \ +8.35kW
\)
As a final sanity check, we know that the power delivered by the current source has to equal the sum of the powers consumed by the resistor and inductor.
\(
P_S = P_R + P_L \ ?
P_S = +13.98kW \ + \ -5.63kW
P_S = +8.35kW \ (check)
\)
Now, it is always possible I have made a mistake that survived all of my units and sanity checks. The most likely such mistake is that I simply solved the wrong problem or made a mistake in setting up the problem.
Notice that I worked symbolically for as far as I could and I kept things like sqrt(5) around as long as I could. This not only reduces roundoff error in the final result, but it makes going back and checking for where you made a mistake a lot easier and correcting it once it is found much easier.
It also makes it easier to leverage the results to answer other questions. For instance, if we were then asked to find the power delivered by the current source as a function of time, we can do so in just a couple of lines.
\(
P_S(t) = v_S(t)i_S(t)
P_S(t) = [960 \sqrt{5} cos(2000 \frac{r}{s} t \ + \ 26.57^o) V][12 cos(2000 \frac{r}{s} t)A]
P_S(t) = 25.76 cos(2000 \frac{r}{s} t \ + \ 26.57^o) cos(2000 \frac{r}{s} t)kW
\)
We could then apply a trig identity to simplify this further.
