
By the end of this lesson, students should be able to:
Define an RC circuit.
Explain the charging and discharging behavior of a capacitor.
Derive the characteristic equations of RC circuits.
Understand the concept of the time constant.
Analyze current, charge, and voltage as functions of time.
Solve AP Physics C problems involving RC circuits.
An RC circuit is a circuit containing:
A resistor ((R))
A capacitor ((C))
connected in series or in a more complex arrangement.
RC circuits are important because they introduce time-dependent behavior into electric circuits.
Unlike simple resistor circuits, the current in an RC circuit changes with time.
RC circuits are used in:
Camera flashes
Timing circuits
Electronic filters
Signal processing systems
Computer memory circuits
Understanding RC circuits is an important AP Physics C topic because they combine concepts from:
Capacitance
Current
Energy conservation
Differential equations
The resistor opposes the flow of current.
According to Ohm’s Law:
$$
V_R = IR
$$
where:
(V_R) = voltage across the resistor
(I) = current
(R) = resistance
A capacitor stores charge and electrical energy.
The charge stored on a capacitor is:
$$
Q = CV
$$
where:
(Q) = charge
(C) = capacitance
(V) = voltage across the capacitor
Consider:
A battery of emf (\mathcal{E})
A resistor (R)
A capacitor (C)
Initially:
$$
Q=0
$$
The capacitor is uncharged.
When the switch is closed:
Current begins flowing.
Charge accumulates on the capacitor.
Current gradually decreases.
Eventually the capacitor becomes fully charged.
For the charging circuit:
$$
\mathcal{E}-IR-\frac{Q}{C}=0
$$
where:
$$
V_C=\frac{Q}{C}
$$
is the voltage across the capacitor.
Since:
$$
I=\frac{dQ}{dt}
$$
we obtain:
$$
\mathcal{E}=R\frac{dQ}{dt}=\frac{Q}{C}0
$$
Solving this differential equation produces the charging equations.
The charge on the capacitor as a function of time is:
$$
Q(t)=C\mathcal{E}\left(1-e^{-t/RC}\right)
$$
As:
$$
t\rightarrow\infty
$$
the exponential term approaches zero.
Therefore:
$$
Q_{max}=C\mathcal{E}
$$
The capacitor becomes fully charged.
The current in the circuit is:
$$
I(t)=\frac{\mathcal{E}}{R}e^{-t/RC}
$$
Initially:
$$
t=0
$$
so:
$$
I_0=\frac{\mathcal{E}}{R}
$$
which is the maximum current.
As time increases:
$$
I\rightarrow0
$$
The current gradually disappears because the capacitor opposes further charge flow.
The voltage across the capacitor is:
$$
V_C(t)=\mathcal{E}\left(1-e^{-t/RC}\right)
$$
Initially:
$$
V_C=0
$$
The capacitor behaves like a wire.
Eventually:
$$
V_C=\mathcal{E}
$$
The capacitor behaves like an open circuit.
The quantity:
$$
\tau=RC
$$
is called the time constant.
The time constant determines how quickly the capacitor charges or discharges.
After one time constant:
$$
t=\tau
$$
the capacitor reaches approximately:
$$
63%
$$
of its final charge.
At:
$$
t=\tau
$$
$$
Q=0.632Q_{max}
$$
At:
$$
t=2\tau
$$
$$
Q=0.865Q_{max}
$$
At:
$$
t=3\tau
$$
$$
Q=0.950Q_{max}
$$
At:
$$
t=5\tau
$$
the capacitor is considered essentially fully charged.
Suppose a charged capacitor is disconnected from the battery and connected across a resistor.
Initially:
$$
Q=Q_0
$$
Current begins flowing through the resistor.
Applying Kirchhoff’s Loop Rule:
$$
IR+\frac{Q}{C}=0
$$
Substituting:
$$
I=\frac{dQ}{dt}
$$
leads to the discharge equations.
The charge decreases exponentially:
$$
Q(t)=Q_0e^{-t/RC}
$$
As time increases:
$$
Q\rightarrow0
$$
The capacitor gradually loses all stored charge.
The magnitude of the discharge current is:
$$
I(t)=\frac{Q_0}{RC}e^{-t/RC}
$$
The current is largest immediately after the discharge begins.
It then decreases exponentially toward zero.
The voltage across the capacitor is:
$$
V_C(t)=V_0e^{-t/RC}
$$
where:
$$
V_0=\frac{Q_0}{C}
$$
Eventually:
$$
V_C\rightarrow0
$$
The capacitor becomes completely discharged.
Charge:
$$
Q(t)=Q_{max}\left(1-e^{-t/RC}\right)
$$
starts at zero and approaches a maximum.
Voltage:
$$
V_C(t)=\mathcal{E}\left(1-e^{-t/RC}\right)
$$
has the same shape.
Current:
$$
I(t)=\frac{\mathcal{E}}{R}e^{-t/RC}
$$
starts large and decreases toward zero.
Charge, current, and voltage all follow exponential decay:
$$
e^{-t/RC}
$$
and approach zero as time increases.
A circuit contains:
$$
R=2.0\times10^3\Omega
$$
and
$$
C=5.0\times10^{-6}F
$$
Find the time constant.
Use:
$$
\tau=RC
$$
Substitute:
$$
\tau=(2.0\times10^3)(5.0\times10^{-6})
$$
$$
\tau=0.010s
$$
$$
\tau=1.0\times10^{-2}s
$$
A capacitor is charging in an RC circuit.
The maximum charge is:
$$
Q_{max}=20\mu C
$$
Find the charge after one time constant.
At:
$$
t=\tau
$$
$$
Q=0.632Q_{max}
$$
Substitute:
$$
Q=(0.632)(20)
$$
$$
Q=12.6\mu C
$$
$$
Q=12.6\mu C
$$
Assuming current remains constant.
In RC circuits:
$$
I
$$
changes continuously with time.
Confusing charging and discharging equations.
Charging:
$$
1-e^{-t/RC}
$$
Discharging:
$$
e^{-t/RC}
$$
Always identify which process is occurring.
Forgetting the time constant.
The key parameter controlling RC behavior is:
$$
\tau=RC
$$
Most RC derivations start from:
$$
\sum\Delta V=0
$$
This often earns immediate partial credit.
Charge:
$$
Q=C\mathcal{E}
\left(
1-e^{-t/RC}
\right)
$$
Voltage:
$$
V_C=\mathcal{E}
\left(
1-e^{-t/RC}
\right)
$$
Current:
$$
I=
\frac{\mathcal{E}}{R}
e^{-t/RC}
$$
These equations appear frequently on AP exams.
RC circuits contain both a resistor and a capacitor.
The time constant is:
$$
\tau=RC
$$
During charging:
$$
Q=C\mathcal{E}
\left(
1-e^{-t/RC}
\right)
$$
$$
I=
\frac{\mathcal{E}}{R}
e^{-t/RC}
$$
$$
V_C=
\mathcal{E}
\left(
1-e^{-t/RC}
\right)
$$
During discharging:
$$
Q=Q_0e^{-t/RC}
$$
$$
I=
\frac{Q_0}{RC}
e^{-t/RC}
$$
$$
V_C=V_0e^{-t/RC}
$$
RC circuits exhibit exponential behavior.
The time constant determines the rate of charging and discharging.
RC circuits are one of the most important applications of differential equations in AP Physics C: Electricity and Magnetism.
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