Pho.rs: Hysteresis in a direct current circuit

A1 ^0.50 Write the equation for the $I$ with the solutions corresponding to the intersection points of curves (1) and (2).

1 \[\frac{A}{I} = 4 U_0 \frac{I}{I_0} \left(1 - \frac{I}{I_0} \right)\]

0.50

A2 ^0.50 Determine the slope $\mathrm{d}U/\mathrm{d}I$ of curve (2) at a given current $I$. Express your answer in terms of $A$ and $I$.

1 \[-\frac{A}{I^2}\]	0.50
2 \[\frac{A}{I^2}\]	0.30

A3 ^1.00 Determine the slope $\mathrm{d}U/\mathrm{d}I$ of curve (1) at a given current $I$. Express your answer in terms of $I_0$, $U_0$, and $I$.

1 \[4\frac{U_0}{I_0} \left( 1 - 2\frac{I}{I_0} \right)\]	1.00
2 correct answer without simplification, e.g. \[4U_0 \frac{I}{I_0} \left( 1 - \frac{I}{I_0} \right) - 4U_0 \frac{I}{I_0} \left( - \frac{I}{I_0} \right)\]	0.50
3 \[ 4 \frac{U_0}{I_0} \left(1 - \frac{I}{I_0} \right)\]	0.30
4 \[ 4 U_0 \frac{I}{I_0} \left(1 - \frac{I}{I_0} \right)\]	0.30

A4 ^1.50 Determine the value of $A$ for which curve (1) transitions smoothly into curve (2) as illustrated above (i.e., the resulting $I(U)$ curve is differentiable at the junction). Find the corresponding junction current $I_{\mathrm{c}}$. Express both answers in terms of $I_0$ and $U_0$.

1 Expression from A1 is used	0.20
2 Expressions from A2 and A3 are used, e.g. \[-\frac{A}{I^2} = 4 \frac{U_0}{I_0} \left(1 - \frac{2I}{I_0} \right)\]	0.40
3 Getting \[1-\frac{I}{I_0} = \frac{2I}{I_0}-1\]	0.20
4 \[I_c = \frac{2}{3}I_0\]	0.20
5 \[A=\frac{16}{27} U_0 I_0\]	0.50

B1 ^1.50 Write an expression for the current $I$ in the circuit as a function of $\mathcal{E}$. Express your answer in terms of $I_0$ and $U_0$.

1 Concept of load line is explicitely used for parabola	0.30
2 Quadratic equation is obtained, e.g.: \[4 \left( \frac{I}{I_0} \right)^2 - 5 \frac{I}{I_0} + \frac{\mathcal{E}}{U_0}=0\]	0.40
3 Solving quadratic equation \[I = I_0 \frac{5 \pm \sqrt{25 - 16 \frac{\mathcal{E}}{U_0}}}{8}\]	0.40
4 Only root with « - » is chosen \[I = I_0 \frac{5 - \sqrt{25 - 16 \frac{\mathcal{E}}{U_0}}}{8}\]	0.40

B2 ^2.00 Determine the value of $\mathcal{E}_\uparrow$. Express your answer in terms of $I_0$ and $U_0$.

1 Deciding that the point is the point at which the line touches the parabola for $\mathcal{D}=0$	1.00
2 \[ \mathcal{E}_\uparrow =\frac{25}{16} U_0\]	1.00

B3 ^2.00 Determine the value of $\mathcal{E}_\downarrow$. Express your answer in terms of $I_0$ and $U_0$.

1 Concept of load line is explicitely used for hyperbola	0.50
2 Quadratic equation is obtained: \[ \left( \frac{I}{I_0} \right)^2 - \frac{\mathcal{E}}{U_0} \frac{I}{I_0} + \frac{A}{U_0 I_0} = 0\]	0.50
3 \[ \mathcal{E}_\downarrow = U_0 \frac{8}{\sqrt{27}} \]	1.00

B4 ^1.00 Determine the internal resistance of the source $R_0$ for which, during a gradual increase and subsequent decrease of $\mathcal{E}$, no sudden jumps in the current occur?

1 Decide that the abrupt occurs when $-\mathrm{d}U/\mathrm{d}I > R_0$	0.60
2 \[R_0 \geq \frac{4}{3} \frac{U_0}{I_0}\]	0.40

Hysteresis in a direct current circuit

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