Answer to Question #277969 in Other for Yming

Question

A 0.2µF capacitor has a charge of 20 µC. Find the voltage and energy.

2. If the energy stored in a 1/8 F capacitor is 25 J, find the voltage and charge.

3. Find the maximum and minimum values of capacitance that can be obtained

from ten 1µF capacitors.

4. Suppose you have a 9V battery, a 2μF capacitor, and a 7.40μF capacitor.

(a) Find the charge and energy stored if the capacitors are connected to the

battery in series. (b) Do the same for a parallel connection.

5. In an open-heart surgery, a much smaller amount of energy will defibrillate

the heart. (a) What voltage is applied to the 8µF capacitor of a heart

defibrillator that stores 40 J of energy? (b) Find the amount of stored charge.

6. If the current through a 1mH inductor is 𝑖(௧) = 20 cos 100𝑡 𝑚𝐴, find the

terminal voltage and the energy stored.

7. Find the maximum and minimum values of inductance that can be obtained

using ten 10mH inductors.

Answer 1

A.

$$Q=CU=>U=\dfrac{Q}{C}=\dfrac{20\times10^{-6}C}{0.2\times10^{-6}F}=100V$$

$$E=\dfrac{Q^2}{2C}=\dfrac{(20\times10^{-6}C)^2}{2(0.2\times10^{-6}F)}=0.001J$$

2.

$$E=\dfrac{Q^2}{2C}=>Q=\pm\sqrt{2CE}$$

$$Q=\pm\sqrt{2(\dfrac{1}{8}F)(25J)}=\pm2.5C$$

$$U=\dfrac{Q}{C}=\dfrac{\pm 2.5C}{\dfrac{1}{8}F}=\pm20V$$

3.

Series capacitors

$$\dfrac{1}{C_{TS}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...+\dfrac{1}{C_{10}}$$

$$=\dfrac{1}{10C_1}=\dfrac{1}{10\times 10^{-6}F}$$

$$C_{TS}=10^5F$$

Parallel  capacitors

$$C_{TP}=C_1+C_2+...+C_{10}=10C_1$$

$$=10\times 10^{-6}F=10^{-5}F$$

The maximum value of capacitance that can be obtained is $10^5F.$

The minimum value of capacitance that can be obtained is $10^{-6}F$ (when we use only one capacitor).

If we have to use all ten 1µF capacitors together then the maximum value of capacitance that can be obtained is $10^5F,$ and the minimum value of capacitance that can be obtained is $10^{-5}F.$

4.

(a) Series capacitors

$$\dfrac{1}{C_{TS}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{1}{2\times 10^{-6}F}+\dfrac{1}{7.4\times 10^{-6}F}$$

$$=\dfrac{4.7}{7.4\times 10^{-6}F}$$

$$Q_S=C_{TS}U=\dfrac{7.4\times 10^{-6}F}{4.7}\cdot9V$$

$$\approx14.17\times 10^{-6} C=14.17\mu C$$

$$E_S=\dfrac{C_{TS}U^2}{2}=\dfrac{7.4\times 10^{-6}\cdot(9V)^2}{2(4.7)}$$

$$\approx63.766\times 10^{-6} J=63.766\mu J$$

(b) Parallel  capacitors

$$C_{TP}=C_1+C_2=2\times 10^{-6}F+7.4\times 10^{-6}F$$

$$=9.4\times 10^{-6}F$$

$$Q_P=C_{TP}U=9.4\times 10^{-6}F\cdot9V$$

$$=84.6\times 10^{-6} C=84.6\mu C$$

$$E_P=\dfrac{C_{TP}U^2}{2}=\dfrac{9.4\times 10^{-6}\cdot(9V)^2}{2}$$

$$=0.3807\times 10^{-3} J=0.3807m J$$

5.

(a)

$$E=\dfrac{CU^2}{2}=>U=\pm\sqrt{\dfrac{2E}{C}}$$

$$=\pm\sqrt{\dfrac{2(40J)}{8\times 10^{-6}F}}=\pm(\sqrt{10}\times10^3)V$$

(b)

$$Q=CU=\pm(\sqrt{10}\times10^3)V\cdot(8\times 10^{-6}F)$$

$$=\pm(8\sqrt{10}\times10^{-3} )C$$

6.

i.

$$V(t)=L\dfrac{di}{dt}=10^{-3}(-20(100)\times10^{-3}\sin(100t))V$$

$$=-2\sin(100t)mV$$

ii.

$$E=\dfrac{Li^2}{2}=\dfrac{10^{-3}(20\cos(100t)\times10^{-3})^2}{2}J$$

$$=0.2\cos^2(100t)\mu J$$

7.

Series inductances

$$L_{TC}=L_1+L_2+...+L_{10}$$

$$=10(0.01H)=0.1H$$

Parallel  inductances

$$\dfrac{1}{L_{TP}}=\dfrac{1}{L_1}+\dfrac{1}{L_2}+...+\dfrac{1}{L_{10}}$$

$$=10(\dfrac{1}{0.01H})$$

$$L_{TP}=\dfrac{0.01H}{10}=0.001H=1mH$$

The maximum possible inductance is $0.1H=100mH.$

The mimimum possible inductance is $0.001H=1mH.$