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I cannot seem to get the correct angle. I've done this problem multiple times and still get -75 degrees.

All I do is 162cos(200t - 30) ->

convert to phasor domain (162cos(-30) + 162i*sin(-30)) = Vs

To get I_norton, I divide this result by the impedance of the resistor and inductor (10 + 50 * 10^(-3) * j * 200). Right? Why is this not yielding the correct answer?

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1 Answer 1

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Treat this like a voltage divider: Now,

\$ R = 10 \Omega \$

\$ X_L = j \omega L = j200 \cdot 50 \cdot 10^{-3} = j10 .\Omega \$

\$ X_C = {{-j} \over { \omega C} } = {{-j} \over {200 \cdot 500 \cdot 10^{-6}} } = {{-j} \over {10^{-1}} } = -j10 . \Omega \$

\$ V_{ab} = V_C = V_{Thevenin} = V_T , \$

\begin{eqnarray*} V_T &=& V_S ∙{{X_C} \over {R + X_L + X_C}}\\ &=& 162 ∠ 60^∘ \cdot {{-j10} \over {10 + j10 + -j10}}\\ &=& 162 ∠ 60^∘ \cdot {{-j10} \over {10}}\\ &=& -j \cdot 162 ∠ 60^∘ \\ &=& 162 ∠ 60^∘ \cdot 1 ∠ {-90}^∘\\ &=& 162 ∠ (60^∘ - 90^∘)\\ &=& 162 ∠ {-30^∘} . V\\ \end{eqnarray*}

\$ Z_N = Z_{Norton} = Z_{Thevenin} = Z_T \$ is the impedance looking into the terminals ab, and shorting any voltage sources, and opening any current sources.

So,

\begin{eqnarray*} Z_N = Z_T &=& {X_C \parallel ({R + X_L}) }\\ &=& {-j10} \parallel (10 + j10)\\ &=& \lbrace (-j10)^{-1} + (10+j10)^{-1} \rbrace ^{-1} \\ &=& \lbrace{ {10}\over {100 - j100} }\rbrace ^{-1} \\ &=& 10 - j10 . Ω\\ &=& 10√2 ∠ {-45^∘}.Ω \\ \end{eqnarray*}

So,

\$ I_N = {{V_T} \over {Z_N}} = 11.46 ∠ 15^∘.A \space \blacksquare \$

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  • \$\begingroup\$ V thevenin is not equal to Vs \$\endgroup\$ Commented Nov 28, 2016 at 17:47
  • \$\begingroup\$ This answer is right. The norton current is the short circuit current of the thevenin equivalent (actually the last step in the answer), which is Vthevenin series with the norton/thevenin impedance. \$\endgroup\$
    – SuperGeo
    Commented Nov 29, 2016 at 2:23
  • \$\begingroup\$ No, this is wrong. V thevenin is not equal to Vs. My answer, was correct. There was a problem in the textbook after talking to my professor \$\endgroup\$ Commented Nov 29, 2016 at 3:45
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    \$\begingroup\$ Oh my bad, I think I understand now. Your polar coordinate indicates sin. I had assumed it indicated cos. sorry about that \$\endgroup\$ Commented Nov 29, 2016 at 5:47

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