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Exercise E1 Conversion: Energy, Power and Area

A Tesla Model 3 has an average power consumption of $ {{16~{\rm kWh}}\over{100~{\rm km}}}$ and an usable battery capacity of $60~{\rm kWh}$. Solar panels produces per $1~\rm m^2$ in average in December $0.2~{{{\rm kWh}}\over{{\rm m^{2}}}}$. The car is driven $50~{\rm km}$ per day. The size of a distinct solar module with $460~{\rm W_p}$ (Watt peak) is $1.9~{\rm m} \times 1.1~{\rm m}$.

1. What is the average power consumption of the car per day?

Solution

\begin{align*} {{W}\over{l}} &= {{16~{\rm kWh}}\over{100~{\rm km}}} = 0.16 {{~{\rm kWh}}\over{~{\rm km}}} \\ W &= 50~{\rm km} \cdot 0.16 {{~{\rm kWh}}\over{~{\rm km}}} = 8~{\rm kWh} \end{align*}

Result

\begin{align*} W = 8~{\rm kWh}\end{align*}

2. How many square meters (=$\rm m^2$) of solar panels are needed on average in December?

Solution

\begin{align*} A = {{8~{\rm kWh}}\over{0.2 {{~{\rm kWh}}\over{~{\rm m^{2}}}}}} = 40~{\rm m^{2}} \end{align*}

Result

\begin{align*} A=40~{\rm m^{2}} \end{align*}

3. How many panels are at least needed to cover this surface?

Solution

\begin{align*} A_{\rm panel} &= 1.9~{\rm m} \cdot 1.1~{\rm m} = 2.1~{\rm m^{2}} \\ n_{\rm panels} &= {{A }\over{ A_{\rm panel }}} = {{40~{\rm m^{2}}}\over{2.1~{\rm m^{2}/panel}}} \\ &= 19.04~{\rm panels} \rightarrow 20~{\rm panels} \end{align*}

Result

\begin{align*} n_{\rm panels} = 20~{\rm panels} \end{align*}

4. What is the combined $\rm kW_p$ of the panels you calculated in 3. ?

Solution

\begin{align*} P &= 20~{\rm panels} \cdot 460~{\rm kW_p / panel} \\ &= 9'200~{\rm kW_p} \end{align*}

Result

\begin{align*} P = 9'200~{\rm kW_p} \end{align*}

Exercise E2 Industrial Sensor Interface: Buffered Measurement Node

A 12 V industrial sensor module feeds a buffered measurement node through a resistor T-network. The capacitor at the output node is used to smooth the signal and to provide a stable voltage for a short measurement cycle. At first, the measurement electronics are disconnected. Once the capacitor is fully charged, a switch closes and the measurement load is connected.

Data: \begin{align*} U &= 12~{\rm V} \\ R_1 &= 2~{\rm k\Omega} \\ R_2 &= 10~{\rm k\Omega} \\ R_3 &= 3.33~{\rm k\Omega} \\ C &= 2~{\rm \mu F} \\ R_L &= 5~{\rm k\Omega} \end{align*}

Initially, the capacitor is uncharged and the switch $S$ is open.

1. What is the voltage across the capacitor after it is fully charged?

Solution

With the switch open, no DC current flows through $R_3$ after the capacitor is fully charged. Therefore, there is no voltage drop across $R_3$, and the capacitor voltage is equal to the divider voltage of $R_1$ and $R_2$:

\begin{align*} U_C(\infty) &= U \cdot \frac{R_2}{R_1 + R_2} \\ &= 12~{\rm V} \cdot \frac{10~{\rm k\Omega}}{2~{\rm k\Omega} + 10~{\rm k\Omega}} \\ &= 12~{\rm V} \cdot \frac{10}{12} \\ &= 10~{\rm V} \end{align*}

Result

\begin{align*} U_C(\infty) = 10~{\rm V} \end{align*}

2. How long does the charging process take?

Solution

For the charging process, the capacitor sees the Thevenin resistance of the left-hand network.

First, calculate the parallel combination of $R_1$ and $R_2$:

\begin{align*} R_1 \parallel R_2 &= \frac{R_1 R_2}{R_1 + R_2} = \frac{2~{\rm k\Omega} \cdot 10~{\rm k\Omega}}{2~{\rm k\Omega} + 10~{\rm k\Omega}} \\ &= \frac{20}{12}~{\rm k\Omega} \approx 1.67~{\rm k\Omega} \end{align*}

Thus, the Thevenin resistance is

\begin{align*} R_{\rm th} &= R_3 + (R_1 \parallel R_2) \\ &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\ &\approx 5.00~{\rm k\Omega} \end{align*}

The charging time constant is

\begin{align*} \tau_1 &= R_{\rm th} C \\ &= 5.00~{\rm k\Omega} \cdot 2~{\rm \mu F} \\ &= 10~{\rm ms} \end{align*}

In practice, the capacitor is considered fully charged after about $5\tau_1$:

\begin{align*} t_{\rm charge} &\approx 5\tau_1 = 5 \cdot 10~{\rm ms} = 50~{\rm ms} \end{align*}

Result

\begin{align*} \tau_1 = 10~{\rm ms} \\ t_{\rm charge} \approx 50~{\rm ms} \end{align*}

3. Draw the time-dependent capacitor voltage.

Solution

The charging equation of the capacitor is

\begin{align*} u_C(t) &= U_C(\infty)\left(1 - e^{-t/\tau_1}\right) \end{align*}

With

\begin{align*} U_C(\infty) &= 10~{\rm V} \\ \tau_1 &= 10~{\rm ms} \end{align*}

the time-dependent capacitor voltage is

\begin{align*} u_C(t) = 10\left(1 - e^{-t/(10~{\rm ms})}\right)~{\rm V} \end{align*}

It starts at $0~{\rm V}$ and rises exponentially toward $10~{\rm V}$.

Result

\begin{align*} u_C(t) = 10\left(1 - e^{-t/(10~{\rm ms})}\right)~{\rm V} \end{align*}

4. Once the capacitor is fully charged, the switch is closed and the load resistor is connected. What is the load voltage in the stationary state?

Solution

Use the Thevenin equivalent of the left-hand network as seen from the capacitor/load node:

\begin{align*} U_{\rm th} &= 10~{\rm V} \\ R_{\rm th} &= 5.00~{\rm k\Omega} \end{align*}

After closing the switch, the load voltage is the divider voltage of $R_{\rm th}$ and $R_L$:

\begin{align*} U_L(\infty) &= U_{\rm th} \cdot \frac{R_L}{R_{\rm th} + R_L} \\ &= 10~{\rm V} \cdot \frac{5~{\rm k\Omega}}{5~{\rm k\Omega} + 5~{\rm k\Omega}} \\ &= 10~{\rm V} \cdot \frac{1}{2} \\ &= 5~{\rm V} \end{align*}

Result

\begin{align*} U_L(\infty) = 5~{\rm V} \end{align*}

5. How long does it take until this stationary state is reached?

Solution

After the switch is closed, the capacitor sees the equivalent resistance

\begin{align*} R_{\rm eq} &= R_{\rm th} \parallel R_L \\ &= 5.00~{\rm k\Omega} \parallel 5.00~{\rm k\Omega} \\ &= 2.50~{\rm k\Omega} \end{align*}

The new time constant is

\begin{align*} \tau_2 &= R_{\rm eq} C \\ &= 2.50~{\rm k\Omega} \cdot 2~{\rm \mu F} \\ &= 5~{\rm ms} \end{align*}

In practice, the new stationary state is reached after about $5\tau_2$:

\begin{align*} t_{\rm settle} &\approx 5\tau_2 = 5 \cdot 5~{\rm ms} = 25~{\rm ms} \end{align*}

Result

\begin{align*} \tau_2 = 5~{\rm ms} \\ t_{\rm settle} \approx 25~{\rm ms} \end{align*}

6. Draw the time-dependent load voltage.

Solution

At the instant the switch closes, the capacitor voltage cannot jump. Therefore, the initial load voltage is

\begin{align*} u_L(0^+) = 10~{\rm V} \end{align*}

The final stationary value is

\begin{align*} u_L(\infty) = 5~{\rm V} \end{align*}

Thus, the transient load voltage is

\begin{align*} u_L(t) &= u_L(\infty) + \left(u_L(0^+) - u_L(\infty)\right)e^{-t/\tau_2} \\ &= 5~{\rm V} + \left(10~{\rm V} - 5~{\rm V}\right)e^{-t/(5~{\rm ms})} \\ &= 5 + 5e^{-t/(5~{\rm ms})}~{\rm V} \end{align*}

So the load voltage starts at $10~{\rm V}$ and decays exponentially to $5~{\rm V}$.

Result

\begin{align*} u_L(t) = 5 + 5e^{-t/(5~{\rm ms})}~{\rm V} \end{align*}

Exercise E3 Hall-Sensor Test Bench: Air-Core Calibration Coil

In a production test bench for Hall sensors, a small air-core coil is used to generate a reproducible magnetic field. An air-core design is chosen because it avoids hysteresis and remanence effects of iron cores. The coil is wound as a short, single-layer cylindrical coil.

Data: \begin{align*} l &= 22~{\rm mm} \\ d &= 20~{\rm mm} \\ d_{\rm Cu} &= 0.8~{\rm mm} \\ N &= 25 \\ \rho_{\rm Cu,20^\circ C} &= 0.0178~{\rm \Omega\,mm^2/m} \end{align*}

For the inductance of the short air-core coil, use Wheeler's approximation: \begin{align*} L[{\rm \mu H}] \approx \frac{r^2 N^2}{9r + 10l} \end{align*} with $r$ and $l$ entered in inches.

The coil is then connected to a DC supply.

1. Calculate the coil resistance $R$ at room temperature.

Solution

First, determine the copper cross-sectional area:

\begin{align*} A_{\rm Cu} &= \frac{\pi}{4} d_{\rm Cu}^2 = \frac{\pi}{4}(0.8~{\rm mm})^2 \\ &= 0.503~{\rm mm^2} \end{align*}

The mean length of one turn is approximately the circumference:

\begin{align*} l_{\rm turn} &\approx \pi d = \pi \cdot 20~{\rm mm} = 62.83~{\rm mm} \end{align*}

Thus, the total wire length is

\begin{align*} l_{\rm Cu} &= N \cdot l_{\rm turn} = 25 \cdot 62.83~{\rm mm} \\ &= 1570.8~{\rm mm} = 1.571~{\rm m} \end{align*}

Now calculate the resistance:

\begin{align*} R &= \rho_{\rm Cu}\frac{l_{\rm Cu}}{A_{\rm Cu}} \\ &= 0.0178~{\rm \Omega\,mm^2/m}\cdot\frac{1.571~{\rm m}}{0.503~{\rm mm^2}} \\ &\approx 0.0556~{\rm \Omega} \end{align*}

Result

\begin{align*} R \approx 0.0556~{\rm \Omega} = 55.6~{\rm m\Omega} \end{align*}

2. Calculate the coil inductance $L$.

Solution

First convert radius and coil length to inches:

\begin{align*} r &= 10~{\rm mm} = \frac{10}{25.4}~{\rm in} \approx 0.394~{\rm in} \\ l &= 22~{\rm mm} = \frac{22}{25.4}~{\rm in} \approx 0.866~{\rm in} \end{align*}

Using Wheeler's approximation:

\begin{align*} L[{\rm \mu H}] &\approx \frac{r^2 N^2}{9r + 10l} \\ &\approx \frac{(0.394)^2 \cdot 25^2}{9\cdot 0.394 + 10\cdot 0.866} \\ &\approx 7.94~{\rm \mu H} \end{align*}

Result

\begin{align*} L \approx 7.94~{\rm \mu H} \end{align*}

3. Which DC voltage $U$ must be applied so that the stationary coil current becomes $I = 1~\rm A$? How large is the current density $j$ in the copper wire?

Solution

In the stationary DC state, the inductance no longer affects the current. Then only the ohmic resistance remains:

\begin{align*} U &= RI \\ &= 0.0556~{\rm \Omega} \cdot 1~{\rm A} \\ &= 0.0556~{\rm V} \end{align*}

The current density is

\begin{align*} j &= \frac{I}{A_{\rm Cu}} \\ &= \frac{1~{\rm A}}{0.503~{\rm mm^2}} \\ &\approx 1.99~{\rm A/mm^2} \end{align*}

Result

\begin{align*} U \approx 55.6~{\rm mV} \\ j \approx 1.99~{\rm A/mm^2} \end{align*}

4. How much energy is stored in the coil in the stationary state?

Solution

The magnetic energy stored in an inductor is

\begin{align*} W_{\rm mag} &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.94\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &\approx 3.97\cdot 10^{-6}~{\rm J} \end{align*}

Result

\begin{align*} W_{\rm mag} \approx 3.97~{\rm \mu J} \end{align*}

5. Sketch the coil current $i(t)$ when the coil is switched on.

Solution

A coil does not allow its current to jump instantly. Therefore:

\begin{align*} i(0) &= 0 \\ i(\infty) &= I = 1~{\rm A} \end{align*}

The current rises exponentially:

\begin{align*} i(t) &= I\left(1-e^{-t/\tau}\right) \end{align*}

with the time constant

\begin{align*} \tau = \frac{L}{R} \end{align*}

So the sketch starts at $0~\rm A$, rises steeply at first, and then approaches $1~\rm A$ asymptotically.

Result

\begin{align*} i(t) = 1\left(1-e^{-t/\tau}\right)~{\rm A} \end{align*} with an exponential rise from $0~\rm A$ to $1~\rm A$.

6. How long does it take until the current has practically reached its stationary final value?

Solution

First compute the time constant:

\begin{align*} \tau &= \frac{L}{R} = \frac{7.94~{\rm \mu H}}{0.0556~{\rm \Omega}} \\ &\approx 1.43\cdot 10^{-4}~{\rm s} = 0.143~{\rm ms} \end{align*}

A useful engineering rule is: after about $5\tau$, the current has reached more than $99\%$ of its final value.

\begin{align*} t_{\rm practical} &\approx 5\tau \\ &\approx 5 \cdot 0.143~{\rm ms} \\ &\approx 0.714~{\rm ms} \end{align*}

Mathematically, the exact final value is reached only after infinite time.

Result

\begin{align*} \tau \approx 0.143~{\rm ms} \\ t_{\rm practical} \approx 0.714~{\rm ms} \end{align*}

7. How large is the energy dissipated as heat in the coil resistance during the current build-up?

Solution

During the switching-on process, part of the supplied energy is stored in the magnetic field and part is dissipated as heat in the winding resistance.

For an RL switch-on process, the heat loss during current build-up is equal to the finally stored magnetic energy:

\begin{align*} W_{\rm loss} &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.94\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &\approx 3.97\cdot 10^{-6}~{\rm J} \end{align*}

Result

\begin{align*} W_{\rm loss} \approx 3.97~{\rm \mu J} \end{align*}