# EE115, Chapter 17

17.2 The bulb light just until the current through it charges the capacitor to the source voltage. Since the source is DC,  the capacitor will charge to the up to the DC voltage in the first moment then the current through the circuit will be zero.

17.4

a) $Xc = Vac/I =10/0.02 = 500$ ohms

b) $Xc = 24/0.008 =3000$ ohms

c) $Xc = 15/(300*10^{-6}) =50000 ohm = 50$ kohm

d) $Xc = 100/(50*10^{-6}) =2*10^6 ohm =2$ Mohm

17.6

$Xc = 1/(2*\pi*f*C)$

a) $Xc = 1/(2*\pi*10*0.1^{-6}) =1.59*10^5 ohm =159$ kohm

b) $Xc = 1/(2*\pi*50*0.1*10^{-6}) =3.18*10^4 ohm = 31.8$ kohm

c) $Xc = 1/(2*\pi*200*0.1*10^{-6}) =7.96*10^3 ohm =7.96$ kohm

d) $Xc =1/(2*\pi*10^4*0.1*10^{-6}) =159$ ohm

17.8

$Xc = 1/(2*\pi*f*C)$

$C =1/(2*\pi*f*Xc)$

a) $C = 1/(2*\pi*318.3*1000) =5*10^{-7} F =0.5 \mu F$

b) $C = 1/(2*pi*1591*1000) = 1*10^-7 F =0.1 \mu F$

c) $C = C(a)/10 =5*10^{-8} F = 50$ nF

d) $C = 1/(2*\pi*6366*1000) = 2.5*10^{-8} F = 25$ nF

17.10

$Xc = Vac/I = 10/0.002 = 5000$ ohm

$C =1/(2*\pi*f*Xc) =1/(2*pi*3183*5000) =1*10^{-8} F = 10$ nF

17.12

$I_0 = U/Xc = U*(2*\pi*f*C)$

a) $I= I_0/2 = 50/2 =25$ mA

b) $I = 2*I_0 =100$ mA

c) $I = I_0*2 = 100$ mA

d) $I = I_0/2 = 25$ mA

17.14

$Xc = 1/(2*\pi*f*C)$

a) $Xc =338.6$ ohm

b) $Xc = 13.26$ ohm

c) $Xc = 636.6$ ohm

d) $Xc = 1.447 kohm =1447$ ohm

17.16

$Xc = 1/(2*\pi*f*C)$

$f = 1/(2*\pi*C*Xc)$

a)  $f = 795.77$ Hz

b) $f =1000.3$ Hz

c) $f =3048.8$ Hz

d) $f =31.83$ Hz

17.18

$Xc = Xc1+Xc2$ when capacitors are in parallel.

a) $Xc =100+400 =500$ ohm

b) $Xc =1.8+1.2 =3$ kohm

c) $Xc =15+6+10 = 31$ ohm

d) $Xc =2.5+10 +2+1 =15.5$ kohm

17.20

$I = Vac/Xc = Vac*(2*\pi*f*C)$

a) f decreases it means I decreases

b) f increases it means I increases

17.22

$Xc = 1/(2*\pi*f*C)$,   $f =1591$ Hz
a) $Xc1 = 1000.3$ ohm

$Xc2 = 2000.69$ ohm

$Xc3 =5001.7$ ohm

b) $Xc(tot) = Xc1+Xc2+Xc3$

$Xc = 8002.69$ ohm

c) $I = Vac/Xc =120/8002.69 =15 mA =0.015$ A

d) $Vc1 = I*Xc1 =0.015*1000.3 = 15$ V

$Vc2 =I*Xc2 =0.015*2000.69 = 30$ V

$Vc3 =0.015*5001.7 = 75$ V

17.24

$1/Xc = 1/Xc1+1/Xc2+1/Xc3$

$Xc = 20$ ohm

$I = Vac/Xc = 120/20 =6$ A

$Ic1 =Vac/Xc1 =120/120 =1$ A

$Ic2 =Vac/Xc2 =120/60 = 2$ A

$Ic3 = Vac/Xc3 =120/40 = 3$ A

17.26

$C = 1/(2*\pi*f*Xc)$

$C1 =2.08*10^{-7} F = 0.2 \mu F$

$C2 =4*10^{-7} F = 0.4 \mu F$

$C3 =6.2*10^{-7} F = 0.62 \mu F$

$Ceq =C1+C2+C3 =1.22 \mu F =1.22*10^{-6} F$

17.28.

$C = Q/U = i*t/U$

$i = U/t *C = 10*0.33*10^{-6} = 3.3*10^{-6} A =3.3 \mu A$

17.30

$C = Q/U = i*t/U$

$i = C*U/t$

a) $i =100*10^{-6}*10 =10^{-3} A =1$ mA

b) $i = 100*10^{-6}*1 =10^{-4} A =0.1$ mA

c) $i =100*10^{-6}*50 =5*10^{-3} A =5$ mA

17.32

$C = Q/U = i*t/U = i/(U/t) = 0.015/150 =1*10^{-4} F = 100 \mu F$

17.34.

Take an alternating voltage source where you know Vac and frequency. Place the voltage source in series with the capacitor and an amper-meter (that measures current through capacitor).

You know $Xc = 1/(2*\pi*f*C)$

$I = Vac/Xc = Vac*(2*\pi*f*C)$

Then

$C = (2*\pi*f*Vac)/I$