This problems includes a figure of a circuit with a switch, aresistor and a capa

Question

This problems includes a figure of a circuit with a switch, aresistor and a capacitor all in series (and in that order from thepositive side of the power source).
Switch S is closed at time t=0, to begin charging an initiallyuncharged capacitor of capacitance C=10 Farads through aresistor of resistance R=20 MOhms.
A) What is the time constant of the circuit? B) At what time is the electric potential across the capacitorequal to twice that across the resistor? C) A capacitor is added to decrease the time constant by half.What capacitance needs to be added, and is it in series or parallelto the original capacitor?
For part A, I used =RC, which gave a value of 200.
For part B, I initially used the equationVe-t/=Vo(1-e^-t/) toisolate t, but I feel that perhaps this is not the correct equationfor this problem. This is where I really need help.
For part C, I used this series of equations: Ceq=C1+ C2 (/2)=200/2=100 and=RCeqso Ceq=/R=100/2.0x10^7=5.0x10^-6 Ceq=1/(1/C1 +1/C2)=5.0x10^-6 so, C=1.0x10^-5 Capacitors in series. This seemed to work, but I was unsure if I was correct aboutthe Capacitors being in series.

Switch S is closed at time t=0, to begin charging an initiallyuncharged capacitor of capacitance C=10 Farads through aresistor of resistance R=20 MOhms.
A) What is the time constant of the circuit? B) At what time is the electric potential across the capacitorequal to twice that across the resistor? C) A capacitor is added to decrease the time constant by half.What capacitance needs to be added, and is it in series or parallelto the original capacitor?
For part A, I used =RC, which gave a value of 200.
For part B, I initially used the equationVe-t/=Vo(1-e^-t/) toisolate t, but I feel that perhaps this is not the correct equationfor this problem. This is where I really need help.
For part C, I used this series of equations: Ceq=C1+ C2 (/2)=200/2=100 and=RCeqso Ceq=/R=100/2.0x10^7=5.0x10^-6 Ceq=1/(1/C1 +1/C2)=5.0x10^-6 so, C=1.0x10^-5 Capacitors in series. This seemed to work, but I was unsure if I was correct aboutthe Capacitors being in series.
(/2)=200/2=100 and=RCeqso Ceq=/R=100/2.0x10^7=5.0x10^-6 Ceq=1/(1/C1 +1/C2)=5.0x10^-6 so, C=1.0x10^-5 Capacitors in series. This seemed to work, but I was unsure if I was correct aboutthe Capacitors being in series.

Explanation / Answer

You are on the right track! Please see below. A) What is the time constant of the circuit? Since
VC(t) = Vb(1 - e -t/RC) = RC= (10x10-6 x 20 x 106 )
= 200 s
B) At what time is the electric potential across the capacitorequal to twice that across the resistor? or since in general Vb= VC (t)   +VR(t) while VR(t) =0.5Vc(t) we have Vb= 1.5VC (t)  and sinceVC(t) = Vb(1 - e -t/RC) Vb= 1.5 Vb(1 - e -t/RC) wehave since 1.5 = 3/2 2/3 = (1 - e -t/RC)   and 1/3 = e -t/RC then t is t= -RCln(1/3) t= 200 ln(3) t= 220 s
t= -RCln(1/3) t= 200 ln(3) t= 220 s C) A capacitor is added to decrease the time constant by half.What capacitance needs to be added, and is it in series or parallelto the original capacitor? Definetly series and of the same value ( C=C) since
Cequivalent = C C/ ( C+ C) = 0.5C
Now '=0.5RC = 100 s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.