Question #277969

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.


1
Expert's answer
2021-12-13T16:30:24-0500

A.


Q=CU=>U=QC=20×106C0.2×106F=100VQ=CU=>U=\dfrac{Q}{C}=\dfrac{20\times10^{-6}C}{0.2\times10^{-6}F}=100V




E=Q22C=(20×106C)22(0.2×106F)=0.001JE=\dfrac{Q^2}{2C}=\dfrac{(20\times10^{-6}C)^2}{2(0.2\times10^{-6}F)}=0.001J

2.


E=Q22C=>Q=±2CEE=\dfrac{Q^2}{2C}=>Q=\pm\sqrt{2CE}

Q=±2(18F)(25J)=±2.5CQ=\pm\sqrt{2(\dfrac{1}{8}F)(25J)}=\pm2.5C

U=QC=±2.5C18F=±20VU=\dfrac{Q}{C}=\dfrac{\pm 2.5C}{\dfrac{1}{8}F}=\pm20V

3.

Series capacitors


1CTS=1C1+1C2+...+1C10\dfrac{1}{C_{TS}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...+\dfrac{1}{C_{10}}


=110C1=110×106F=\dfrac{1}{10C_1}=\dfrac{1}{10\times 10^{-6}F}


CTS=105FC_{TS}=10^5F

Parallel  capacitors


CTP=C1+C2+...+C10=10C1C_{TP}=C_1+C_2+...+C_{10}=10C_1

=10×106F=105F=10\times 10^{-6}F=10^{-5}F


The maximum value of capacitance that can be obtained is 105F.10^5F.

The minimum value of capacitance that can be obtained is 106F10^{-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 105F,10^5F, and the minimum value of capacitance that can be obtained is 105F.10^{-5}F.


4.

(a) Series capacitors


1CTS=1C1+1C2=12×106F+17.4×106F\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}


=4.77.4×106F=\dfrac{4.7}{7.4\times 10^{-6}F}


QS=CTSU=7.4×106F4.79VQ_S=C_{TS}U=\dfrac{7.4\times 10^{-6}F}{4.7}\cdot9V

14.17×106C=14.17μC\approx14.17\times 10^{-6} C=14.17\mu C

ES=CTSU22=7.4×106(9V)22(4.7)E_S=\dfrac{C_{TS}U^2}{2}=\dfrac{7.4\times 10^{-6}\cdot(9V)^2}{2(4.7)}

63.766×106J=63.766μJ\approx63.766\times 10^{-6} J=63.766\mu J

(b) Parallel  capacitors


CTP=C1+C2=2×106F+7.4×106FC_{TP}=C_1+C_2=2\times 10^{-6}F+7.4\times 10^{-6}F

=9.4×106F=9.4\times 10^{-6}F

QP=CTPU=9.4×106F9VQ_P=C_{TP}U=9.4\times 10^{-6}F\cdot9V


=84.6×106C=84.6μC=84.6\times 10^{-6} C=84.6\mu C

EP=CTPU22=9.4×106(9V)22E_P=\dfrac{C_{TP}U^2}{2}=\dfrac{9.4\times 10^{-6}\cdot(9V)^2}{2}

=0.3807×103J=0.3807mJ=0.3807\times 10^{-3} J=0.3807m J



5.

(a)


E=CU22=>U=±2ECE=\dfrac{CU^2}{2}=>U=\pm\sqrt{\dfrac{2E}{C}}

=±2(40J)8×106F=±(10×103)V=\pm\sqrt{\dfrac{2(40J)}{8\times 10^{-6}F}}=\pm(\sqrt{10}\times10^3)V

(b)

Q=CU=±(10×103)V(8×106F)Q=CU=\pm(\sqrt{10}\times10^3)V\cdot(8\times 10^{-6}F)

=±(810×103)C=\pm(8\sqrt{10}\times10^{-3} )C

6.

i.


V(t)=Ldidt=103(20(100)×103sin(100t))VV(t)=L\dfrac{di}{dt}=10^{-3}(-20(100)\times10^{-3}\sin(100t))V

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

ii.


E=Li22=103(20cos(100t)×103)22JE=\dfrac{Li^2}{2}=\dfrac{10^{-3}(20\cos(100t)\times10^{-3})^2}{2}J

=0.2cos2(100t)μJ=0.2\cos^2(100t)\mu J


7.

Series inductances


LTC=L1+L2+...+L10L_{TC}=L_1+L_2+...+L_{10}

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

Parallel  inductances


1LTP=1L1+1L2+...+1L10\dfrac{1}{L_{TP}}=\dfrac{1}{L_1}+\dfrac{1}{L_2}+...+\dfrac{1}{L_{10}}

=10(10.01H)=10(\dfrac{1}{0.01H})

LTP=0.01H10=0.001H=1mHL_{TP}=\dfrac{0.01H}{10}=0.001H=1mH

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

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



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