Comments on: The FYS1120 Weekly Assignment Post #3 http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/ The home of FYS1120 Mon, 23 Dec 2013 11:26:13 +0000 hourly 1 https://wordpress.org/?v=4.9.3 By: Mikael B Steen http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comment-60 Thu, 25 Nov 2010 19:46:20 +0000 http://mindseye.no/?p=284#comment-60 Thank you for that Filip! I’m sure that is the correct answer then 🙂 I’ll correct it.

]]> By: Filip Sund http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comment-59 Thu, 25 Nov 2010 19:21:28 +0000 http://mindseye.no/?p=284#comment-59 The answer both I and the “official” solution got to 4b) were ~1.812 ohm:

[latex]R = \frac{\varepsilon}{I} = \frac{\varepsilon}{I_1 + I_2} = \frac{\varepsilon}{I_4 + I_5}[/latex]

where I_1 and I_2 are the currents through the upper resistances, I_4 and I_5 through the lower.

]]> By: Mikael B Steen http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comment-30 Tue, 12 Oct 2010 08:55:19 +0000 http://mindseye.no/?p=284#comment-30 Hey Andreas! 🙂 Ah, thank you. I must have overseen that we actually had an internal ressistance. I’ll fix that right up.

]]> By: Andreas V. Solbrå http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comment-27 Mon, 11 Oct 2010 14:25:18 +0000 http://mindseye.no/?p=284#comment-27 Hey!

Good lookin’ answers, dude. However, in 2), you need to account for an inner resistance r in the battery (this only gives you an added resistance r, as in a series circuit). The answers i got were then
2a) I_0 = eps/(R + r)
I_inf = 0
2b) I_0 = eps/r
I_inf = eps/(R + r)

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