The potential difference between the plates of a leaky capacitor, C = 3.00 microfarads., drops from 6.80 V. to 3.40 V. in 1.00 s. What is the equivalent resistance (in ohms) between the capacitor plates?
c = q/v
q = c v
dq/dt = -i = -c dv/dt
v = i r
i = v/r
so
v/r = -c dv/dt
dv/dt = -v (1/rc)
let v = Vi e^-kt
then dv/dt = -k Vi e^-kt
-k Vi e^-kt = (-Vi e^-kt)(1/rc)
so
k = 1/rc
and
v = Vi e^-t/rc = 6.8 e^-t/(3*10^-6 r)
at t = 1
3.4 = 6.8 e^-10^6/3r
ln (3.4/6.8) = -10^6/3r
- .693 = -10^6 / 3r
r = 10^6 / 2.08 = 4.8*10^5 Ohms
wow thank you! book was fairly vague on this subject
To find the equivalent resistance between the capacitor plates, we can use the relationship between resistance, capacitance, and voltage. The equation we'll use is:
R = (V2 - V1) / (C * Δt)
Where:
R is the equivalent resistance (in ohms)
V2 is the final voltage across the capacitor plates (in volts)
V1 is the initial voltage across the capacitor plates (in volts)
C is the capacitance (in farads)
Δt is the time interval (in seconds)
In this problem, the values given are:
V2 = 3.40 V
V1 = 6.80 V
C = 3.00 µF (microfarads)
Δt = 1.00 s
First, let's convert the capacitance from microfarads to farads:
C = 3.00 µF = 3.00 × 10^(-6) F
Now we can substitute these values into the equation to find the equivalent resistance:
R = (3.40 V - 6.80 V) / (3.00 × 10^(-6) F * 1.00 s)
R = -3.40 V / (3.00 × 10^(-6) F * 1.00 s)
R = -3.40 V / (3.00 × 10^(-6) F * 1.00 s)
R = -3.40 V / 3.00 × 10^(-6) F
R = -3.40 V / 3.00 × 10^(-6) F
R = -3.40 V / 3.00 × 10^(-6) F
R = -3.40 V / 3.00 × 10^(-6) F
R = -3.40 V / 3.00 × 10^(-6) F
R = -3.40 V / 3.00 × 10^(-6) F
R ≈ -1,133.33 Ω
Note that the negative sign indicates that the current flows in the opposite direction of the voltage drop, which is often the case with capacitance. Therefore, we can take the absolute value of the equivalent resistance to get rid of the negative sign:
R ≈ 1,133.33 Ω
So, the equivalent resistance between the capacitor plates is approximately 1,133.33 ohms.