by J. Blum, CDSC, Thailand

- Explain how a capacitor is built and what it can be used for.
- Explain why an initial current flows while charging. When does it stop?

Two circular parallel metal plates with the radius *r* = 24 cm, are arranged in a distance
*d* = 1.2 mm as a plate capacitor. The applied voltage is 240 V.

- Calculate the capacitance of this arrangement and the stored charge
*Q*._{C} - Calculate the electric field strength
*E*between the plates and the stored energy*E*in the electric field._{C} - The force of attraction between oppositely charged capacitor plates equals $F = 0,5 \cdot E \cdot Q$.
Determine the force
*F*, which one plate exerts to the other. - Compare the result with the force of the two point charges $Q_C = 316.8~nC$ and $-Q_C$ in the same distance.

- A capacitor is built out of two metal plates (area
*A*) and a certain distance*d*in between. The material in between is called dielectric. - If a voltage is applied electrons repell each other and move to the one side of the capacitor ...
capacitance: $C=\epsilon \frac{A}{d}=1.32~nF$

charge: $Q_C=VC=316.8~nC$

Electric field strength: $E=\frac{E}{d}=200~000~V/m$

Energy: $E_{el}=\frac{1}{2}CV^2=38.02\cdot 10^{-6}J$

- Force of attraction between plates: $F= 0,5 \cdot E \cdot Q=31.68~mN$
- Force of two point charges: $F=k\cdot \frac{Q_1 \cdot Q_2}{d^2}=626,69~N$

Knut has bought capacitors by Conrad electronics with value 10 µF/25 V.

- How does he has to connect six capacitors in order to get the maximum capacity? Calculate the capacitance and the maximum voltage.
- How does he has to connect six capacitors in order to get the maximum voltage? Calculate the the maximum voltage and capacitance.
- Design a circuit for 20 µF and 50 V
- Which advantage have electrolytic capacitors in comparision with ceramic capacitors?

In parallel: $C=6 \cdot 10~\mu F= 60~\mu F$, $V=25~V$

In series: $V=6 \cdot 25~V=150~V$, $C=1/(1/C_1+~...~+1/C_6)$$=1,67~\mu F$

Respectively 2 in series ($50~V$ and $5~\mu F$)and then 4 times parallel.

Beside a high capacity, electrolytic capacitors have a high maximum voltage.

A capacitor with 47 µF is connected to 24 V with a series resistance of 2,1 kΩ.

- How high is the time constant and how long does charging last?
- Calculate the capacitor's voltage after 1.5 τ.
- After which time the voltage reaches 90 % of its maximum?
- How high is the maximum inrush current?
- After which time the current is down to 30 % of its maximum value?
- Draw the
*t-V*- und_{C}*t-I*-diagram. Plot all calculated values._{C}

time conatant: $\tau =R\cdot C = 98,7~ms$

charging time: $5\tau =493,5~ms$

time: $2\tau=197,4~ms$

capacitors voltage: $V_C(1,5\tau)=V_0(1-e^{-t/\tau})=$$24~V(1-e^{-1,5\tau /\tau})=18,64~V$

questioned voltage: $V_C=90\% \cdot V_0=21,6~V$

$V_0(1-e^{-t/\tau})=V_C~~|:V_0$

$1-e^{-t/\tau}=V_C/V_0~~|-1 ~|\cdot(-1)$

$e^{-t/\tau}=1-V_C/V_0~~|ln(...)$

$-t/\tau=ln(1-V_C/V_0)~~|\cdot (-\tau)$

$t=-\tau\cdot ln(1-V_C/V_0)~~|\text{insert}$

$t=227~ms$

inrush current: $I_0=V_0/R=24~V/2,1~k\Omega=$$11,42~mA$

questioned current: $I_C=30\% \cdot I_0=3,426~mA$

$V_0/R\cdot e^{-t/\tau})=I_C~~|\cdot (-R/V_0)$

$e^{-t/\tau})=I_C\cdot R/V_0~~||ln(...)$

$-t/\tau=ln(I_C\cdot R/V_0)~~|\cdot (-\tau)$

$t=-\tau\cdot ln(I_C\cdot R/V_0)~~|\text{insert}$

$t=118,91~ms$

diagram:

idea from students

A 1 kΩ load is supplied with 325 V. After power failure a capacitor shall supply the circuit for 5 s with at least 63,3 % the input voltage.

Which capacity is necessary?

Calculate the voltage: $\frac{100~\%}{325~V}=\frac{63,3~\%}{V_C}$, $V_C=205,725~V$

Calculate the time constant: $\tau=\frac{t}{-ln(V_C/V_0)}=10,934~s$

Calculate the capacitance: $C=\tau /R=10,934~mF$

Alternatively estimation out of diagram: $0,633 ~\hat{=}~ 0,5~\tau$

A RC-circuit is supplied with the following control voltage. Plot the signal for voltage and current,

- when the charging time of the capacitor is 1 s.
- when the charging time of the capacitor is 2 s.
- when the charging time of the capacitor is 4 s.

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