Transcript RC Circuits

RC Circuits
Voltage on the Capacitor
Initially there is no voltage across the capacitor,
since there is no charge on the plates. The
current is large and all the voltage drop is across
the resistor.
As the voltage builds on the capacitor current
falls to zero when the capacitor is completely
charged.
Approaching RC Circuits
Create a Kirchoff’s loop equation for the circuit.
Before we can solve this equation, all variables
must be expressed in terms of time and charge.
This can be done by relating the voltage across
the capacitor to its charge using the equation Q
= CV and relating the current through the circuit
with the charge on the capacitor using the
relationship I = dQ/dt
Cont’d
Now we should have a differential equation involving
various constants, Q and t. After separating variables,
indefinite integration yields an equation relating charge
to time with a constant of integration.
Use the initial conditions to eliminate the constant of
integration.
Now we should have charge as a function of time Q(t).
If we want to calculate current as a function of time, we
can pulg Q(t) into our previous formula relating current
and time I = dQ/dt.
We should be able to check that our new equation gives
the right current values at t = 0 and t = infinity.
RC Circuits
The current through the circuits discussed so far have been timeindependent, as long as the emf of the source is timeindependent. The currents in these circuits can be determined by
applying Kirchhoff's first and/or second rule.
A simple circuit in which the current is time dependent is the RC
circuit which consists of a resistor R connected in series with a
capacitor C. Applying Kirchhoff's second rule to the current loop I
gives
ξ – IR – Q/C = 0
Current
The current in the resistor and the capacitor is
the same.
I = dq/dt
ξ – (dq/dt)R – Q/C = 0
This can be rearranged so that it can be solved
in the following manner:
Differential Equation
dq/(Cξ – Q) = dt/(RC)
Now integrate each side
Q
Q
∫Q dQ/(Cξ - Q) = ∫Qdt/(RC)
o
o
-ln (Cξ-Q) +ln(Cξ-Qo) = ln[(Cξ-Qo)/(Cξ-Q)]=t/(RC)
Qo is the charge at time t = 0
Continued…
Q = Cξ(1-e-t/RC) = Q(1-e-t/RC)
Where Q = Cξ is the maximum charge on the
capacitor
I = (ξ/R)e -t/RC
Time Constant, tau
RC is the time constant, tau
ln (q/Q) = -t/(RC)
q = Qoe(-t/RC)
I = -dq/dt = Ioe(-t/RC)
Qo is the initial charge on the capacitor
Io = Q/(RC)
Problem
A 6 microfarad capacitor is charged through a 5
kΩ resistor by a 500 V power supply. How long
does it require for the capacitor to acquire 99
percent of its final charge?
Answer
Tau = RC = 5000Ω ( 6 x 10-6 F) = 0.030 sec
q = .99Q = Q (1-e-t/RC)
e -t/RC = 0.01
t/RC = -ln (0.01)
t/RC = 0.14 sec