First-Order RC Circuit Analysis: Voltage, Current, and Responses

Welcome back. Last lesson introduced Thévenin and Norton equivalents: a source network can be replaced at a port by a voltage $V_{\mathrm{Th}}$ in series with a resistance $R_{\mathrm{Th}}$. That viewpoint now becomes dynamic. When the load at that port is a capacitor, the Thévenin resistance controls not only the current but also how quickly the capacitor voltage changes.

In this lesson, you will calculate capacitor voltage and current, derive the natural response of a discharging capacitor, and solve step responses in both simple and multi-resistor RC circuits. These ideas will later become essential for MOS amplifier bandwidth, slew-rate limits, compensation capacitors, and switching transients.

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The capacitor laws and the crucial continuity rule

A capacitor stores separated charge. Its defining voltage-charge relationship is

$$q_C=C{v}_C$$

where $C$ is capacitance in farads and $v_C$ is the voltage across the capacitor, measured using an assigned polarity.

Differentiating with respect to time gives the current-voltage law:

$$i_C=C\frac{d{v}_C}{dt}$$

This equation assumes the passive sign convention: $i_C$ is defined as entering the capacitor terminal marked positive for $v_C$.

Two consequences matter immediately:

1. A nonzero capacitor current causes the voltage to change.
2. An instantaneous change in capacitor voltage would require infinite current.

Therefore, in ordinary circuit analysis,

$$v_C(0^{+})=v_C(0^{-})$$

The notation has a precise meaning:

• $0^{-}$: the instant just before a switch changes position.
• $0^{+}$: the instant just after it changes position.

The capacitor voltage is continuous, even though resistor currents and source currents may change suddenly at the switching instant.

For a DC circuit that has remained unchanged for a long time, the capacitor reaches steady state:

$$\frac{d{v}_C}{dt}=0$$

Thus,

$$i_C=0$$

So, at DC steady state, a capacitor behaves as an open circuit. Be careful: this statement applies after a long time, not necessarily at $t=0^{+}$. Immediately after switching, the capacitor must retain its pre-switch voltage.

The energy stored in a capacitor is

$$w_C=\frac{1}{2}C{v}_C^2$$

During a natural discharge, this initially stored energy is converted to heat in the resistance connected across the capacitor.

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Natural response: a charged capacitor discharges

A natural response is the behavior produced by energy initially stored in the circuit, with no independent source actively driving the circuit after the switching instant.

Consider a capacitor initially charged to $V_0$, connected across a resistor $R$ at $t=0$. Assign $v_C$ positive at the capacitor’s upper terminal, and define $i_C$ as entering that terminal.

Applying KVL around the source-free loop gives

$$v_R+v_C=0$$

Using

$$v_R=i_{C}R$$

and

$$i_C=C\frac{d{v}_C}{dt}$$

gives

$$RC\frac{d{v}_C}{dt}+v_C=0$$

The product $RC$ has units of seconds:

$$\Omega \cdot \mathrm{F}=\mathrm{s}$$

We define the time constant as

$$\tau =RC$$

The natural-response equation becomes

$$\frac{d{v}_C}{dt}+\frac{1}{\tau }{v}_C=0$$

Its solution is an exponential decay:

$$v_C(t)=V_0{e}^{-t/\tau }$$

Since the capacitor discharges, its current is negative under the passive sign convention:

$$i_C(t)=C\frac{d{v}_C}{dt}=-\frac{V_0}{R}{e}^{-t/\tau }$$

The negative sign is physically meaningful. It says current leaves the initially positive capacitor terminal rather than entering it.

Interpreting the time constant

At one time constant,

$$t=\tau$$

the remaining fraction of the initial voltage is

$$e^{-1}\approx 0.368$$

Thus, during a discharge, $36.8\%$ of the initial voltage remains after one time constant. During a charge toward a final value, $63.2\%$ of the required change has occurred after one time constant.

Time	Remaining transient term $e^{-t/\tau }$	Fraction of voltage change completed
$0$	$1$	$0\%$
$\tau$	$0.368$	$63.2\%$
$2\tau$	$0.135$	$86.5\%$
$3\tau$	$0.050$	$95.0\%$
$5\tau$	$0.0067$	$99.3\%$

An exponential technically never reaches its final value exactly. In design calculations, however, $5\tau$ is usually treated as “settled” for a basic RC circuit.

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Example: charge, then discharge

The switching circuit below illustrates a standard natural response. Suppose the switch has been connected to position $a$ for a long time. The capacitor charges through $R_1$ from the source $V_g$. At $t=0$, the switch moves to position $b$, disconnecting the source and allowing the capacitor to discharge through $R$.

Assume

$$V_g=1.2\mathrm{V}$$ $$R=20\mathrm{k\Omega }$$ $$C=100\mathrm{pF}$$

Before switching

At position $a$, after a long time, the capacitor is an open circuit. Therefore no current flows through $R_1$, so $R_1$ has no voltage drop:

$$v_C(0^{-})=V_g=1.2\mathrm{V}$$

By continuity,

$$v_C(0^{+})=1.2\mathrm{V}$$

After switching

At position $b$, the capacitor sees only $R$, so

$$\tau =RC$$ $$\tau =(20\mathrm{k\Omega })(100\mathrm{pF})=2\mu \mathrm{s}$$

The final voltage is zero because the capacitor discharges through $R$ to the reference node:

$$v_C(\infty )=0$$

Therefore,

$$v_C(t)=1.2e^{-t/(2\mu \mathrm{s})}\mathrm{V}$$

The capacitor current is

$$i_C(t)=-\frac{1.2\mathrm{V}}{20\mathrm{k\Omega }}{e}^{-t/(2\mu \mathrm{s})}$$ $$i_C(t)=-60e^{-t/(2\mu \mathrm{s})}\mu \mathrm{A}$$

At $t=4\mu \mathrm{s}$, two time constants have passed:

$$v_C(4\mu \mathrm{s})=1.2e^{-2}$$ $$v_C(4\mu \mathrm{s})\approx 0.162\mathrm{V}$$ $$i_C(4\mu \mathrm{s})=-60e^{-2}\mu \mathrm{A}$$ $$i_C(4\mu \mathrm{s})\approx -8.12\mu \mathrm{A}$$

The initial energy stored in this capacitor was

$$w_C(0)=\frac{1}{2}(100\mathrm{pF})(1.2\mathrm{V}{)}^2=72\mathrm{pJ}$$

That energy is eventually dissipated as heat in $R$.

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The universal first-order RC response

Rather than memorizing separate charging and discharging cases, use one general form:

$$v_C(t)=v_C(\infty )+\left[v_C(0^{+})-v_C(\infty )\right]e^{-t/\tau }$$

This equation says:

• Start at the final value.
• Add the difference between the initial and final values.
• Let that difference decay exponentially.

For any first-order RC circuit with one capacitor,

$$\tau =R_{\mathrm{Th}}C$$

where $R_{\mathrm{Th}}$ is the Thévenin resistance seen by the capacitor after switching.

This directly uses the technique from the previous lesson:

1. Draw the circuit for $t>0$.
2. Remove the capacitor.
3. Deactivate independent voltage sources by replacing them with shorts.
4. Deactivate independent current sources by replacing them with opens.
5. Find the resistance seen looking into the capacitor terminals.

The initial value comes from the circuit at $t<0$:

$$v_C(0^{+})=v_C(0^{-})$$

The final value comes from the $t>0$ circuit at DC steady state, where the capacitor is open:

$$v_C(\infty )=V_{\mathrm{Th}}$$

Thus, the general capacitor-voltage response can also be written as

$$v_C(t)=V_{\mathrm{Th}}+\left[V_0-V_{\mathrm{Th}}\right]e^{-t/(R_{\mathrm{Th}}C)}$$

where

$$V_0=v_C(0^{+})$$

The corresponding capacitor current is

$$i_C(t)=\frac{V_{\mathrm{Th}}-V_0}{R_{\mathrm{Th}}}{e}^{-t/(R_{\mathrm{Th}}C)}$$

This expression uses the passive sign convention for $i_C$.

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Step response: a capacitor moves toward a new final voltage

A step response occurs when a source is connected or changes value abruptly at $t=0$. The source provides a new final voltage, but the capacitor prevents an instantaneous jump to that value.

For a Thévenin source $V_{\mathrm{Th}}$ connected to a capacitor through $R_{\mathrm{Th}}$, KVL gives

$$V_{\mathrm{Th}}=R_{\mathrm{Th}}{i}_C+v_C$$

Substitute the capacitor law:

$$V_{\mathrm{Th}}=R_{\mathrm{Th}}C\frac{d{v}_C}{dt}+v_C$$

The solution is the general expression already obtained:

$$v_C(t)=V_{\mathrm{Th}}+\left[V_0-V_{\mathrm{Th}}\right]e^{-t/(R_{\mathrm{Th}}C)}$$

Special case: an initially uncharged capacitor charges from a DC source

If

$$V_0=0$$

then

$$v_C(t)=V_{\mathrm{Th}}\left(1-e^{-t/\tau }\right)$$

and

$$i_C(t)=\frac{V_{\mathrm{Th}}}{R_{\mathrm{Th}}}{e}^{-t/\tau }$$

At the instant after switching,

$$i_C(0^{+})=\frac{V_{\mathrm{Th}}}{R_{\mathrm{Th}}}$$

Initially, the uncharged capacitor has zero voltage across it, so it draws the maximum current permitted by the resistance. As the capacitor charges, its voltage rises, leaving less voltage across the resistor and causing current to decay to zero.

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Example: a nonzero initial voltage and a new DC level

Suppose that, after a switch changes position, the network connected to a capacitor has this Thévenin equivalent:

$$V_{\mathrm{Th}}=1.8\mathrm{V}$$ $$R_{\mathrm{Th}}=15\mathrm{k\Omega }$$ $$C=2\mathrm{nF}$$

The capacitor was previously charged to

$$v_C(0^{-})=0.3\mathrm{V}$$

First apply voltage continuity:

$$v_C(0^{+})=0.3\mathrm{V}$$

Next, calculate the time constant:

$$\tau =R_{\mathrm{Th}}C$$ $$\tau =(15\mathrm{k\Omega })(2\mathrm{nF})=30\mu \mathrm{s}$$

At DC steady state the capacitor is open, so its final voltage equals the Thévenin voltage:

$$v_C(\infty )=1.8\mathrm{V}$$

Now apply the universal formula:

$$v_C(t)=1.8+(0.3-1.8)e^{-t/(30\mu \mathrm{s})}$$ $$v_C(t)=1.8-1.5e^{-t/(30\mu \mathrm{s})}\mathrm{V}$$

At one time constant,

$$v_C(30\mu \mathrm{s})=1.8-1.5e^{-1}$$ $$v_C(30\mu \mathrm{s})\approx 1.25\mathrm{V}$$

The initial current is

$$i_C(0^{+})=\frac{1.8-0.3}{15\mathrm{k\Omega }}=100\mu \mathrm{A}$$

At one time constant,

$$i_C(30\mu \mathrm{s})=100e^{-1}\mu \mathrm{A}$$ $$i_C(30\mu \mathrm{s})\approx 36.8\mu \mathrm{A}$$

The positive current means that current enters the capacitor’s positive terminal, as expected because the capacitor voltage is rising from $0.3\mathrm{V}$ toward $1.8\mathrm{V}$.

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A reliable workflow for every first-order RC problem

For exam problems and later transistor-level circuits, use the same structured method.

1. Separate the circuit into time intervals.
   If a switch moves at $t=0$, analyze the circuit before and after that instant separately.

2. Find the initial capacitor voltage.
   Analyze the $t<0$ circuit. If it has been in place for a long time with DC sources, replace the capacitor with an open circuit.

   $$v_C(0^{+})=v_C(0^{-})$$

3. Find the final capacitor voltage.
   Analyze the $t>0$ circuit at DC steady state, again replacing the capacitor with an open circuit.

   $$v_C(\infty )$$

4. Find the resistance seen by the capacitor.
   Use the $t>0$ topology. Deactivate independent sources and calculate

   $$R_{\mathrm{Th}}$$

5. Calculate the time constant.

   $$\tau =R_{\mathrm{Th}}C$$

6. Write the capacitor-voltage response.

   $$v_C(t)=v_C(\infty )+\left[v_C(0^{+})-v_C(\infty )\right]e^{-t/\tau }$$

7. Find current by differentiation, if needed.

   $$i_C(t)=C\frac{d{v}_C}{dt}$$

A useful physical check is to inspect the endpoints:

$$v_C(0)=v_C(0^{+})$$

and

$$\underset{t\to \infty }{\lim }{v}_C(t)=v_C(\infty )$$

If either endpoint is wrong, the exponential expression is wrong even if the algebra appears tidy.

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Common errors to avoid

• Treating a capacitor as an open circuit at $t=0^{+}$.
  It is open only in DC steady state. At $t=0^{+}$, its voltage is constrained by continuity.

• Using the pre-switch resistance for the time constant.
  The time constant depends on the resistance seen by the capacitor in the post-switch circuit.

• Forgetting to deactivate sources when finding $R_{\mathrm{Th}}$.
  Independent voltage sources become shorts; independent current sources become opens.

• Using $RC$ automatically in a multi-resistor circuit.
  The correct expression is

  $$\tau =R_{\mathrm{Th}}C$$

  Only in a simple single-resistor circuit does $R_{\mathrm{Th}}=R$.

• Ignoring the sign of capacitor current.
  With current defined into the positive capacitor terminal, charging current is positive and discharge current is negative.

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Key takeaways

A capacitor follows

$$i_C=C\frac{d{v}_C}{dt}$$

and its voltage cannot change instantaneously:

$$v_C(0^{+})=v_C(0^{-})$$

For a natural discharge from $V_0$,

$$v_C(t)=V_0{e}^{-t/(RC)}$$ $$i_C(t)=-\frac{V_0}{R}{e}^{-t/(RC)}$$

For any first-order RC circuit, use the universal form:

$$v_C(t)=v_C(\infty )+\left[v_C(0^{+})-v_C(\infty )\right]e^{-t/\tau }$$

with

$$\tau =R_{\mathrm{Th}}C$$

The transition is largely complete after roughly five time constants. This is the first appearance of the resistance-capacitance time constants that will later determine the bandwidth and settling behavior of analog amplifiers.

Next, you will analyze first-order RC circuits in the sinusoidal domain using complex impedance, determine cutoff frequency, and express attenuation using decibels.