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Engineering Analysis 3

Lynch — Fri, 22 May 2026 18:37:41 GMT

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	\| Salgado \|\| \|\| Banerjee \|\| Banerjee \|\| ~~Prof. Lynch~~ \|\| Meem \|\| Meem \|\| Meem \|\| Salgado		\| Salgado \|\| \|\| Banerjee \|\| Banerjee \|\| \|\| Meem \|\| Meem \|\| Meem \|\| Salgado
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Simulating circuits

Lynch — Thu, 21 May 2026 16:24:53 GMT

Analysis of the Example Circuit

← Older revision		Revision as of 11:24, 21 May 2026
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	=== After a long time at <math>v_{in} = 0</math> V ===		=== After a long time at <math>v_{in} = 0</math> V ===

	The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math> (the capacitor acts as an open circuit). By KVL, <math>0 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 0</math>. In summary,		The circuit is in steady state (no state variable is changing), so <math>i_L^\prime = v_C^\prime = 0</math>. Since <math>Cv_C^\prime = i_L</math>, <math>i_L = 0</math> (the capacitor acts as an open circuit). By KVL, <math>0 = v_L + v_R + v_C = Li_L^\prime + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 0</math>. In summary,

	{\| class="wikitable"		{\| class="wikitable"
	\|+		\|+
	\|-		\|-
	\| <math>i_L</math>\|\| <math>i_L^'</math> \|\| <math>v_C</math> \|\| <math>v_C^'</math>		\| <math>i_L</math>\|\| <math>i_L^\prime</math> \|\| <math>v_C</math> \|\| <math>v_C^\prime</math>
	\|-		\|-
	\| 0 \|\| 0 \|\| 0 \|\| 0		\| 0 \|\| 0 \|\| 0 \|\| 0
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	=== Immediately after the switch to <math>v_{in} = 5</math> V at t=0 ===		=== Immediately after the switch to <math>v_{in} = 5</math> V at t=0 ===

	The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. Otherwise there would be an infinite current or an infinite voltage, respectively. So <math>v_C = i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = L i_L^' + R i + 0 = L i_L^'</math> or <math>i_L^' = 5/L = 5/0.001 = 5000</math>A/s. Also <math>v_C^' = i_L/C =0</math>. (Note the '''second derivative''' of <math>v_C</math> is '''not''' zero.)		The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. Otherwise there would be an infinite current or an infinite voltage, respectively. So <math>v_C = i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = L i_L^\prime + R i + 0 = L i_L^\prime</math> or <math>i_L^\prime = 5/L = 5/0.001 = 5000</math>A/s. Also <math>v_C^\prime = i_L/C =0</math>. (Note the '''second derivative''' of <math>v_C</math> is '''not''' zero.)

	{\| class="wikitable"		{\| class="wikitable"
	\|+		\|+
	\|-		\|-
	\| <math>i_L</math>\|\| <math>i_L^'</math> \|\| <math>v_C</math> \|\| <math>v_C^'</math>		\| <math>i_L</math>\|\| <math>i_L^\prime</math> \|\| <math>v_C</math> \|\| <math>v_C^\prime</math>
	\|-		\|-
	\| 0 \|\| 5000 A/s \|\| 0 \|\| 0		\| 0 \|\| 5000 A/s \|\| 0 \|\| 0
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	(In the circuit above, 1 ms is a "long time," as it is several times the time constant of decay.)		(In the circuit above, 1 ms is a "long time," as it is several times the time constant of decay.)

	The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,		The circuit is in steady state (no state variable is changing), so <math>i_L^\prime = v_C^\prime = 0</math>. Since <math>Cv_C^\prime = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^\prime + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,

	{\| class="wikitable"		{\| class="wikitable"
	\|+		\|+
	\|-		\|-
	\| <math>i_L</math>\|\| <math>i_L^'</math> \|\| <math>v_C</math> \|\| <math>v_C^'</math>		\| <math>i_L</math>\|\| <math>i_L^\prime</math> \|\| <math>v_C</math> \|\| <math>v_C^\prime</math>
	\|-		\|-
	\| 0 \|\| 0 \|\| 5 V \|\| 0		\| 0 \|\| 0 \|\| 5 V \|\| 0
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	=== Immediately after the switch to <math>v_{in} = 0</math> V ===		=== Immediately after the switch to <math>v_{in} = 0</math> V ===

	The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. So <math>i_L = 0</math> and <math>v_C = 5</math>V. By KVL, <math>0 = v_L + v_R + v_C = L i_L^' + R i_L + 5</math> or <math>i_L^' = -5/L = -5/0.001 = -5000</math>A/s. Also <math>v_C^' = i_L/C =0</math>.		The capacitor does not allow its voltage to change discontinuously and the inductor does not allow its current to change discontinuously. So <math>i_L = 0</math> and <math>v_C = 5</math>V. By KVL, <math>0 = v_L + v_R + v_C = L i_L^\prime + R i_L + 5</math> or <math>i_L^\prime = -5/L = -5/0.001 = -5000</math>A/s. Also <math>v_C^\prime = i_L/C =0</math>.

	{\| class="wikitable"		{\| class="wikitable"
	\|+		\|+
	\|-		\|-
	\| <math>i_L</math>\|\| <math>i_L^'</math> \|\| <math>v_C</math> \|\| <math>v_C^'</math>		\| <math>i_L</math>\|\| <math>i_L^\prime</math> \|\| <math>v_C</math> \|\| <math>v_C^\prime</math>
	\|-		\|-
	\| 0 \|\| -5000 A/s \|\| 5 V \|\| 0		\| 0 \|\| -5000 A/s \|\| 5 V \|\| 0
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	<math>		<math>
	v_C^' = (1/C) i_L = 625,000 i_L		v_C^\prime = (1/C) i_L = 625,000 i_L
	</math>		</math>

	By KVL, <math>v_{in} = v_L +v_R + v_C = L i_L^' + R i_L + v_C</math>, or		By KVL, <math>v_{in} = v_L +v_R + v_C = L i_L^\prime + R i_L + v_C</math>, or

	<math>		<math>
	i_L^' = -(R/L) i_L - (1/L) v_C + (1/L) v_{in}		i_L^\prime = -(R/L) i_L - (1/L) v_C + (1/L) v_{in}
	</math>		</math>

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	<math>		<math>
	i_L^{''} = -(R/L) i_L^' - (1/L) v_C^'		i_L^{\prime\prime} = -(R/L) i_L^\prime - (1/L) v_C^\prime
	</math>		</math>

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	<math>		<math>
	i_L^{''} + (R/L) i_L^' + (1/LC) i_L = 0.		i_L^{\prime\prime} + (R/L) i_L^\prime + (1/LC) i_L = 0.
	</math>		</math>

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	<math>		<math>
	v_C^{''} + (R/L) v_C^' + (1/LC) v_C = 0.		v_C^{\prime\prime} + (R/L) v_C^\prime + (1/LC) v_C = 0.
	</math>		</math>

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	</math>		</math>

	To solve for the constants <math>A,B</math> for <math>i_L(t)</math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^'(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response		To solve for the constants <math>A,B</math> for <math>i_L(t)</math> for the case where we switch the input voltage from 5V to 0, we use the initial conditions <math>i_L(0) = 0, i_L^\prime(0) = -5000</math>A/s. This gives us <math>A=0</math> and <math>B = -5000/23,848 = -0.21</math>A, for a response

	<math>		<math>
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	</math>		</math>

	The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for <math>v_C(t)</math> using the initial conditions <math>v_C(0) = 5, v_C^'(0) = 0</math>.		The time <math>t</math> for one-quarter of a cycle of oscillation satisfies <math>23,848 t = \pi/2</math>, or <math>t = 66 \mu</math>s. Plugging this time <math>t</math> into the response, we get <math>i_L = -128</math>mA, which agrees with the plot. We could do a similar calculation to get the particular solution for <math>v_C(t)</math> using the initial conditions <math>v_C(0) = 5, v_C^\prime(0) = 0</math>.

File:EA3-LTspice2.zip

Lynch — Thu, 21 May 2026 10:48:53 GMT

Lynch uploaded File:EA3-LTspice2.zip

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Simulating circuits

Lynch — Thu, 21 May 2026 10:48:29 GMT

← Older revision		Revision as of 05:48, 21 May 2026
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	The Windows interface is fairly intuitive to use, but the MacOS interface is not. Below is a brief tutorial to get you started on the Mac; it should be easy to replicate these steps on Windows. The Mac instructions below refer to "ctl-click" and "two-finger click." "ctl-click" means pressing the control key then clicking, and "two-finger click" means clicking with two fingers on the trackpad.		The Windows interface is fairly intuitive to use, but the MacOS interface is not. Below is a brief tutorial to get you started on the Mac; it should be easy to replicate these steps on Windows. The Mac instructions below refer to "ctl-click" and "two-finger click." "ctl-click" means pressing the control key then clicking, and "two-finger click" means clicking with two fingers on the trackpad.

	Below is an example circuit with a square wave voltage source (changes from 0V to 5V at t = 0 and from 5V to 0V at t = 1 ms) and the result of the simulation. You can also download [[Media:EA3-~~LTspice~~.zip\|'''EA3-relevant sample circuits''']] to test.		Below is an example circuit with a square wave voltage source (changes from 0V to 5V at t = 0 and from 5V to 0V at t = 1 ms) and the result of the simulation. You can also download [[Media:EA3-LTspice2.zip\|'''EA3-relevant sample circuits''']] to test.

	[[File:EA3-LRC1-sim-result.jpg\|x300px\|A completed circuit and its simulation.]]		[[File:EA3-LRC1-sim-result.jpg\|x300px\|A completed circuit and its simulation.]]

Engineering Analysis 3

Lynch — Thu, 21 May 2026 10:23:18 GMT

← Older revision		Revision as of 05:23, 21 May 2026
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	* [https://othello.mech.northwestern.edu/ea3/book/summary/scientists.html '''Famous scientists (in EA3)''']		* [https://othello.mech.northwestern.edu/ea3/book/summary/scientists.html '''Famous scientists (in EA3)''']

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	===Homeworks===		===Homeworks===

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	* Lynch lecture notes: [[Media:EA3-Lynch-lecture-06-03-Spring2026.pdf\|Friday May 8]]		* Lynch lecture notes: [[Media:EA3-Lynch-lecture-06-03-Spring2026.pdf\|Friday May 8]]


⚫	<!--
	'''CLICK HERE FOR [[Engineering Analysis 3 (Archive 2025)]]'''		'''CLICK HERE FOR [[Engineering Analysis 3 (Archive 2025)]]'''
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Simulating circuits

Lynch — Tue, 19 May 2026 20:34:02 GMT

After a long time at v_{in} = 5 V

← Older revision		Revision as of 15:34, 19 May 2026
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	=== After a long time at <math>v_{in} = 5</math> V ===		=== After a long time at <math>v_{in} = 5</math> V ===

	(In the circuit above, 1 ms ~~classifies as~~ a "long time," as it is several times the time constant of decay.)		(In the circuit above, 1 ms is a "long time," as it is several times the time constant of decay.)

	The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,		The circuit is in steady state (no state variable is changing), so <math>i_L^' = v_C^' = 0</math>. Since <math>Cv_C^' = i_L</math>, <math>i_L = 0</math>. By KVL, <math>5 = v_L + v_R + v_C = Li_L^' + Ri_L + v_C = 0+0+v_C</math>, so <math>v_C = 5</math>. In summary,

Engineering Analysis 3

YunsooYoon — Fri, 08 May 2026 23:39:53 GMT

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User:YunsooYoon

Lynch — Fri, 08 May 2026 16:54:38 GMT

User account YunsooYoon was created by Lynch and password was sent by email

File:EA3-Lynch-lecture-06-03-Spring2026.pdf

Lynch — Fri, 08 May 2026 16:13:57 GMT

Lynch uploaded File:EA3-Lynch-lecture-06-03-Spring2026.pdf

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Engineering Analysis 3

Lynch — Fri, 08 May 2026 16:13:43 GMT

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	* [[Media:EA3-HW6-Spring2026.pdf\|HW6]], due Thursday May 14		* [[Media:EA3-HW6-Spring2026.pdf\|HW6]], due Thursday May 14

			===Other Materials===

			* Lynch lecture notes: [[Media:EA3-Lynch-lecture-06-03-Spring2026.pdf\|Friday May 8]]

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File:EA3-HW6-Spring2026.pdf

Lynch — Fri, 08 May 2026 02:00:36 GMT

Lynch uploaded File:EA3-HW6-Spring2026.pdf

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Engineering Analysis 3

Lynch — Fri, 08 May 2026 02:00:15 GMT

← Older revision		Revision as of 21:00, 7 May 2026
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	* [https://othello.mech.northwestern.edu/ea3/book/summary/scientists.html '''Famous scientists (in EA3)''']		* [https://othello.mech.northwestern.edu/ea3/book/summary/scientists.html '''Famous scientists (in EA3)''']

			===Homeworks===

			* [[Media:EA3-HW6-Spring2026.pdf\|HW6]], due Thursday May 14

	<!--		<!--