Question #275637

1. 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

1.



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)}=\pm2.5CU=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}\cdot9V14.17×106C=14.17μC\approx14.17\times 10^{-6} C=14.17\mu CES=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}FQP=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 CEP=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 minimum possible inductance is 0.001H=1mH.0.001H=1mH.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS