5.2 RC Time Constants & Energy

Why This Matters for the Exam

The behavior of capacitors over time during charging and discharging is fundamental for understanding aircraft timing circuits, delay relays, filtering networks, and transient voltage suppression. Energy storage calculations are crucial for maintenance safety (residual charge hazards) and key aerospace applications like aircraft strobe lights, emergency power packs, and fuel quantity systems.

The RC Time Constant ($\tau$)

When a DC voltage is applied to a circuit containing resistance ($R$) and capacitance ($C$), the capacitor does not charge instantly. Instead, the voltage across the capacitor and the current flowing through the circuit change exponentially over time. This transitional period is known as the transient state. The rate at which the capacitor charges or discharges is governed by the RC time constant (represented by the Greek letter tau, $\tau$).

The time constant of a series RC circuit is defined as the product of the circuit resistance and the capacitance: $\tau = R \cdot C$

Where:

• $\tau$ is the time constant in seconds (s).
• $R$ is the resistance in Ohms ($\Omega$).
• $C$ is the capacitance in Farads (F).

One time constant is the time required for:

1. The voltage across a charging capacitor to reach 63.2% of the applied voltage source.
2. The voltage across a discharging capacitor to decay to 36.8% of its initial charge.

Because the curves are exponential, a capacitor theoretically takes an infinite amount of time to become completely charged or discharged. However, for all practical engineering and aircraft maintenance purposes, the transient phase is complete after five time constants ($5\tau$). At $5\tau$, the capacitor is 99.3% charged or discharged, and the circuit is considered to have reached a stable steady state.

Charging Characteristics

When a switch is closed to connect a discharged capacitor to a DC source ($V_S$) through a resistor ($R$):

• At the instant the switch is closed ($t = 0$): The capacitor has no charge, so the voltage across it is zero ($v_C = 0$). Because there is no voltage counteracting the source, the current is at its absolute maximum, governed purely by Ohm's Law: $I_{max} = V_S / R$. At this instant, the capacitor acts as a short circuit.
• As time progresses ($t > 0$): Charge accumulates on the plates, creating a back-EMF (electromotive force) that opposes the source. The voltage across the capacitor increases exponentially, while the current flowing in the circuit decays exponentially.
• At steady state ($t \ge 5\tau$): The capacitor voltage equals the source voltage ($v_C = V_S$). The current in the circuit drops to zero ($i_C = 0$). The capacitor now acts as an open circuit, blocking further DC current flow.

The mathematical equations describing the charging phase are:

• Capacitor Voltage: $v_C(t) = V_S \cdot (1 - e^{-t/\tau})$
• Circuit Current: $i_C(t) = \frac{V_S}{R} \cdot e^{-t/\tau}$

Discharging Characteristics

When a charged capacitor ($V_{initial}$) is disconnected from the power source and connected across a discharge resistor ($R$):

• At $t = 0$: The voltage across the capacitor is $V_{initial}$, and the initial current is at its maximum ($I_{initial} = V_{initial} / R$), flowing in the opposite direction of the charging current.
• As time progresses ($t > 0$): The electrostatic field collapses as electrons flow from the negative plate back to the positive plate. Both the voltage and current decay exponentially.
• At steady state ($t \ge 5\tau$): The capacitor is completely discharged ($v_C = 0$) and current ceases ($i_C = 0$).

The mathematical equations describing the discharging phase are:

• Capacitor Voltage: $v_C(t) = V_{initial} \cdot e^{-t/\tau}$
• Circuit Current: $i_C(t) = -\frac{V_{initial}}{R} \cdot e^{-t/\tau}$

Energy Stored in a Capacitor

A capacitor stores electrical energy in the electrostatic field established between its plates. The amount of stored energy depends on both the capacitance and the square of the voltage. The formula for the stored electrostatic energy ($E$, measured in Joules) is: $E = \frac{1}{2} \cdot C \cdot V^2$

Where:

• $E$ is the stored energy in Joules (J).
• $C$ is the capacitance in Farads (F).
• $V$ is the voltage across the capacitor in Volts (V).

Using $Q = C \cdot V$, the energy formula can also be expressed as: $E = \frac{1}{2} \cdot Q \cdot V = \frac{1}{2} \cdot \frac{Q^2}{C}$

Key Physical Relationship: Stored energy is proportional to the square of the voltage. Doubling the voltage across a capacitor quadruples the amount of stored energy. This is a critical factor in high-power aircraft systems, such as engine ignition units and strobe lights.

Aircraft Maintenance Hazards and Safety Protocols

Capacitors can retain a high-voltage charge for long periods after the aircraft electrical system is powered off and batteries are disconnected. This stored energy represents a major shock and burn hazard to technicians.

• Bleeder Resistors: High-voltage circuits incorporate high-value resistors connected in parallel with the capacitors. These 'bleeder' resistors provide a safe discharge path for the stored energy when the system is powered down. However, maintenance personnel must never assume the bleeder circuit is functional.
• Safe Discharge Procedure: Before touching any high-voltage capacitor (such as those in radar systems, strobe power supplies, or ignition exciters), the technician must perform a manual discharge. This is done using a specialized, insulated grounding probe containing a high-wattage power resistor to limit the discharge current.
• The Screwdriver Hazard: A common, dangerous maintenance mistake is short-circuiting capacitor terminals using a standard screwdriver. This causes a near-zero resistance discharge path, resulting in instantaneous discharge ($t \approx 0$). The resulting current spike can create an explosive arc flash, vaporizing metal, damaging component terminals, causing eye injuries, and physically destroying the capacitor's internal plate connections.

Worked Calculation Scenarios

Scenario 1: Strobe Light Energy & Power An aircraft anti-collision strobe light system uses a $220\,\mu$F capacitor charged to 350 VDC. The capacitor discharges through a xenon flash tube in $100\,\mu$s. Calculate the energy stored in the capacitor and the average power delivered during the flash.

1. Convert units to SI: $C = 220 \times 10^{-6}$ F, $V = 350$ V, $t = 100 \times 10^{-6}$ s.
2. Calculate stored energy ($E$): $E = \frac{1}{2} \cdot C \cdot V^2 = 0.5 \cdot (220 \times 10^{-6}) \cdot (350)^2$ $E = (110 \times 10^{-6}) \cdot 122,500 = 13.475\,\text{Joules}$
3. Calculate average power ($P$): Power is energy per unit time ($P = E / t$): $P = \frac{13.475\,\text{J}}{100 \times 10^{-6}\,\text{s}} = 134,750\,\text{Watts} = 134.75\,\text{kW}$ This demonstrates how a small amount of stored energy can produce massive peak power when discharged in a very short timeframe.

Scenario 2: Timing Circuit Transient Voltage An aircraft time-delay circuit contains a $100\,\text{k}\Omega$ resistor connected in series with a $4.7\,\mu$F capacitor. The circuit is connected to a 28 VDC bus. Calculate the circuit time constant, the time required to reach steady state, and the voltage across the capacitor 0.94 seconds after the switch is closed.

1. Calculate the time constant ($\tau$): $\tau = R \cdot C = (100 \times 10^3\,\Omega) \cdot (4.7 \times 10^{-6}\,\text{F}) = 0.47\,\text{seconds}$
2. Calculate steady-state time ($5\tau$): $\text{Steady-State Time} = 5 \cdot 0.47\,\text{s} = 2.35\,\text{seconds}$
3. Calculate voltage at $t = 0.94$ seconds: First, notice that $t = 0.94$ s is exactly two time constants ($2\tau$): $t / \tau = 0.94 / 0.47 = 2$ Using the charging voltage equation: $v_C(0.94) = V_S \cdot (1 - e^{-2})$ Since $e^{-2} \approx 0.1353$: $v_C(0.94) = 28 \cdot (1 - 0.1353) = 28 \cdot 0.8647 \approx 24.21\,\text{V}$ (Alternatively, from the standard percentages table, at $2\tau$ the voltage is 86.5% of the source: $28 \times 0.865 = 24.22$ V).

Common Exam Traps

• Unit Conversions: The most common math error is failing to convert microfarads ($\mu$F) or kilohms ($\text{k}\Omega$) to basic units. For example, calculating $\tau = 100 \times 4.7 = 470$ seconds instead of $0.47$ seconds because $100\,\text{k}\Omega$ and $4.7\,\mu$F were not converted.
• Energy Voltage-Square Dependency: If an exam question asks how the stored energy changes when voltage increases from 12 V to 24 V, the answer is that it increases by four times (because $24^2 / 12^2 = 4$). Do not select the option that says it doubles.
• Confusion of charging vs discharging currents: Always remember that current is maximum at $t=0$ in both cases, but flows in opposite directions. The voltage behaves opposite: it starts at zero on charge, and starts at maximum on discharge.