Physics Circuit HELP!?
In the attached circuit drawing, if the capacitor has zero charge at the time zero, and the switch is closed, how many excess electrons will it have after 1.5 s?
--- I am stuck on this, thanks in advance!
- sLv 77 years agoFavorite Answer
The time constant for the switch closed is
τ = R5 * C
τ = 5E3 ohms * 6.0E-5 Farads
τ = 0.300 seconds
τ = 300E-3 seconds
τ = 300 milliseconds
So since it will be fully charged after 300 milliseconds, the charge is found once the voltage across the capacitor is determined as follows:
R1, R2, & R3 form the upper part of the voltage divider as 5K + 2.5K = 7.5K,
R4 forms the middle part of the voltage divider as 5K ohms and is the same voltage across the RC circuit,
and R6 forms the lower leg of the voltage divider as 5K ohms.
The voltage across R5 and C is the same as the voltage across R4 which is found as:
V123 = 12V * [ (5K + 5K) / (7.5K + 5K + 5K) ] = 12V * (10K / 17.5K) = 6.8571 V
V6 = 12V * [ 5K / (7.5K + 5K + 5K) ] = 3.4286 V
V4 = V123 - V6 =
V4 = 6.8571 V - 3.4286 V
V4 = 3.4286 V
and the voltage on the capacitor is 3.486V
q = C * V
q = 6.0E-5 F * 3.4286V
q = 17.1430E-6 Coulombs
q = 17.1430 microCoulombs
Therefore the number of excess electrons that are deposited (saturated after 300 milliseconds) is:
# of electrons = 17.1430E-6 Coulombs * (6.241E18 electrons / Coulomb)
# of electrons = 106.9895E12 electrons
# of electrons = 106.9895 Tera-electrons
This was actually a tricky 5-part problem ! ;-bSource(s): vast engineering experience.