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.


Expert's answer

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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