Practice Problems: RC Circuits Solutions
1. (easy) A 200Ω resistor, a 5000μF capacitor, a switch, and a 10 v battery are in series in a single circuit loop. Determine the initial and steady state currents. How long wiil the circuit take to reach steady state (approximately).
Find the time constant:
τ = RC = 200(5000x10-6) = 1.0 seconds
Steady state will essentially be established at 5 seconds.
The initial current:
I = ε/R = 10/200 = 0.05 A
At steady state the current will go to zero.
2. (easy) A pair of 2 F capacitors (in parallel) is in series with another pair of 1 F capacitors (in parallel). What is the equivalent capacitance of this configuration?
The two 2 F capacitors are equal to one 4F capacitor. The two 1 F capacitors are equal to one 2 F capacitor. The combination of the those, in series, through reciprocal addition is 1.33 F.
3. (moderate) Analyze the circuit below to find the following:
a. The current in the loop after three time constants
b. The energy stored in each capacitor at steady state
4. A capacitor (C = 3.2x10-7 F) is charged as electrons move from one plate to the other to achieve a potential difference of 500 v. How many electrons were moved in order for the capacitor to achieve this charge?
First find the potential energy stored in the capacitor:
E = ½CV2
E = 0.5(3.2x10-7)(500)2
E = 0.04 J
Now find the total charge moved:
E = QV/2
0.04 = Q(500)/2
Q = 1.6x10-4 C
Finally, find the number of electrons moved:
#electrons = Total charge moved/charge per electron
#elecrons = 1.6x10-4/1.6x10-19
#electrons = 1x1015 electrons
5. (moderate) Analyze the circuit below to find the charge stored on each capacitor at steady state. Additionally, find the ammeter reading at 3.0 milliseconds.
The voltage across each capacitor at steady state is 120 volts.
For the 2μF capacitor: Q = CV = 2x10-6(120) = 2.4x10-4coulombs
For the 3μF capacitor: Q = CV = 3x10-6(120) = 3.6x10-4coulombs
To find the current at 3.0 ms we need to know the max current and the time constant:
The equivalent resistance for the two resistors is 1200Ω and the equivalent capacitance is 5μF.
Io = ε/Req = 120/1200 = 0.1 A
τ = RC = 1200(5μF) = 0.006 s
I = Ioe-t/RC = 0.1(e-0.003/0.006)) = 0.061 A
6. (moderate) A 3 μF (air-filled) capacitor is connected to a 12 volt battery. Calculate the amount of energy is stored in the E-field. Additionally, If a piece of polystyrene (k = 2.56) is inserted into the capacitor, how much energy will remain in the E-field? Why does it decrease? During the insertion of the dielectric material, will it feel an attraction for the plates or a repulsion?
The initial energy stored in the capacitor (prior to insertion):
Eo = ½CoVo2
Eo = ½(3x10-6)(12)2
Eo = 0.00022 J
Upon insertion of the dielectric, the capacitance will increase, but the voltage will decrease.
C = kCo
C = 2.56(3 μF) = 7.68 μF
k = Vo/V
2.6 = 12/V
V = 4.6 volts
E = ½CV2
E = ½(7.68x10-6)(4.6)2
E = 0.000081 J
During the insertion, work is done by the E-field on the charges in the dielectric material. This causes the field to lose energy. The dielectric feels an attractive force as it enters the field. The positive ends of the molecules in the dielectric are attracted to the negative plate and the negative ends are attracted to the positive plate.