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