Fysik opg 420-422

Opgaver i fysik som skal udskrives, til Onsdag d. 21. Februar 2007.

Opgave 420

A
LaTex: \begin{eqnarray} U_0 &=& 9 V\\ I_2 &=& 0,2 A\\ R_2 &=& 22 \Omega\\ U_2 &=& A\cdot R_2 = 4,4 V\\ \end{eqnarray}
B
LaTex: U_1 = U_0 - U_2 = 4,6 V
C
LaTex: \begin{eqnarray} R_1 &=& 10 \Omega\\ I_1 &=& \frac{U_1}{R_1} = 0,46 A\\ \end{eqnarray}
D
LaTex: \begin{eqnarray} I_3 &=& I_1 -I_2 = 0,26 A\\ R_3 &=& \frac{U_2}{I_3} = 16,92 \Omega\\ \end{eqnarray}
E
LaTex: \begin{eqnarray} A &=& 0 V\\ B &=& 0 V\\ C &=& 4,4 V\\ D &=& 4,4 V\\ E &=& 9 V\\ \end{eqnarray}
F
LaTex: \begin{eqnarray} A &=& -4,4 V\\ B &=& -4,4 V\\ C &=& 0 V\\ D &=& 0 V\\ E &=& 4,6 V\\ \end{eqnarray}

Opgave 421

Billede:Fysik_opg_421.png

LaTex: \begin{eqnarray} U_0 &=& 9 V\\ U &=& 0,72 V\\ I &=& 130 mA = 0,13 A\\ P &=& I \cdot U = 93,6 mW\\ U_r &=& U_0 - U = 8,28 V\\ R &=& \frac{U}{I} = 63,69 \Omega\\ \end{eqnarray}

Opgave 422

a
LaTex: \begin{array}{l} U_P = U_0 - R_i \times I \\ U_P = 12V - 0,05\Omega \times 40A \\ U_P = 10V \\ \end{array}
b
LaTex: \begin{array}{l} R = \frac{U}{I} - R_i \\ R = \frac{{12V}}{{40A}} - 0,05\Omega \\ R = 0,25\Omega \\ \end{array}
c
LaTex: p_y = R\cdot I^2 = 0,25\cdot 40^2 = 400 W
d
LaTex: p_i = R_i\cdot I^2 = 0,05\cdot 40^2 = 80 W
e
LaTex: \eta = \frac{p_y}{p_i+p_y} = 0,83 = 83\percent
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