problems – Elektromagnetisme http://elektromagnetisme.no The home of FYS1120 Mon, 20 Oct 2014 11:23:54 +0000 en-US hourly 1 https://wordpress.org/?v=4.9.3 28429679 The FYS1120 Weekly Assignment Post #3 http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/ http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comments Wed, 06 Oct 2010 05:38:57 +0000 http://mindseye.no/?p=284 Continue reading →]]> This week the problems focuses mainly on circuits and Kirchhoff’s laws.

Assignment #7

1)

a) I_1 = \frac{11}{3} A, \ \ I_2 = -\frac{4}{3} A, \ \ I_3 = -\frac{7}{3}A.

Indicating that current nr.2 and 3 is directed in the opposite way of what was assumed.

b) P_{\varepsilon_1} = \varepsilon _1 I_1 = 44 W

P_{dissapated} = R (I_1 ^2 + I_2 ^2 + I_3 ^2) = 41.33 W.

c) Of course the second voltage supply is also doing work on the charges in the current, so this is the reason that P_{dissapated} \neq P_{\varepsilon_1} In this case it is actually doing negative work.

2)

a) At t = 0, the moment when the switch is closed, there is no charge on the capacitor and therefore no potential difference over it. This means that the current at this time will be as if there were no capacitor in the circuit, so I = \frac{\varepsilon}{R}

As time goes to infinity, the capacitor gets charged up. At some point it will reach a potential difference equal to the potential difference of the voltage supply and the current will be zero.

b) At t = 0 it is a ressistance R in one subcircuit, while there is no ressistance in the other, so the current will go trough the circuit with no ressistance. Since there is no resistance I = \frac{\varepsilon}{0} = \infty,  which means that the capacitor gets charged up infinitely fast.

When the capacitor is fully charged, no current will go towards it and therefore all current will go trough the resistance. This means that we now have I = \frac{\varepsilon}{R}

3)

a) \tau = RC = 1 \times 10^{-6} s

b) q(t=30s) = \varepsilon C \left[ 1 - e^{-\frac{t}{\tau}} \right] = 1.2 \times 10^{-11} C

c) I(t=30s) = \frac{dq}{dt}(30s) = 0

4)

a) Here I used Kirchhoff’s laws to get 5 equations which I put in a matrix, did rowreduction and got I_1 = 6A, \ \ I_2 = 5A, \ \ I_3 = 1 A, \ \ I_4 = 7A, \ \ I_5 = 4 A.

b) R = \frac{\varepsilon}{I} = \frac{\varepsilon}{I_1 + I_2} = \frac{\varepsilon}{I_4 + I_5} = 1.812.. (correction: Filip Sund)

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