Circuit Lab
v1.0.0

RC Impulse Analysis Calculator

Analyze capacitor charging through resistor: electrical and thermal parameters

Parameters

Enter RC circuit parameters

Equivalent resistance: Req = 10 Ω

If specified, affects current calculation but not power dissipation in R

Results

Calculated electrical and thermal parameters

Enter parameters to see results

Time-Domain Analysis

Current, power, and energy over time

Enter parameters to see charts

RC Impulse Physical Model

When a discharged capacitor is suddenly connected to a voltage source V through a series resistor R, the system exhibits exponential charging behavior. The time constant τ = RC characterizes the speed of this process. This analysis covers all key electrical and thermal parameters: current decay, power dissipation, energy distribution, and equivalent pulse metrics for component rating verification.

Step 1: Time Constant

The RC time constant determines the charging/discharging rate. At t = τ, the current has decayed to 36.8% (1/e) of its initial value.

τ=R×C\tau = R \times C

Step 2: Peak Current (at t = 0)

At t = 0, the uncharged capacitor acts as a short circuit. The initial current is limited only by the series resistance, following Ohm's law.

I0=VRI_0 = \frac{V}{R}

Step 3: Current as Function of Time

The current decays exponentially with time constant τ. This is the solution to the differential equation V = i·R + Q/C with Q = ∫i dt.

i(t)=I0⋅e−t/τ=VR⋅e−t/(RC)i(t) = I_0 \cdot e^{-t/\tau} = \frac{V}{R} \cdot e^{-t/(RC)}

Step 4: Instantaneous Power Dissipation

Power in the resistor follows p(t) = i²(t)·R. Since current decays exponentially, power decays as e^(-2t/τ), twice as fast as current.

p(t)=i2(t)⋅R=V2R⋅e−2t/τp(t) = i^2(t) \cdot R = \frac{V^2}{R} \cdot e^{-2t/\tau}

Step 5: Energy Distribution

Of the total energy ½CV² supplied by the source, exactly half is dissipated in the resistor and half is stored in the capacitor. This 50/50 split is independent of R and C values.

ER=12CV2E_R = \frac{1}{2} C V^2

ℹ️ This is the total energy dissipated in R as heat (from t = 0 to t = ∞)

Step 6: Joule Integral (I²t)

The I²t integral quantifies thermal stress on components. It's calculated as ∫₀^∞ i²(t) dt. This parameter appears in fuse and resistor pulse rating datasheets.

I2t=∫0∞i2(t) dt=V2C2RI^2t = \int_0^{\infty} i^2(t) \, dt = \frac{V^2 C}{2R}

Note: I²t units are squared. Example: 1 A²·s = 1,000,000 mA²·s

Step 7: Equivalent Rectangular Pulse Width

For comparison with datasheet pulse ratings, we calculate the width of a rectangular pulse at p_peak that would dissipate the same total energy E_R. This gives w_eq = τ/2.

weq=ERppeak=τ2w_{eq} = \frac{E_R}{p_{peak}} = \frac{\tau}{2}

Step 8: Average Power over Time Window

For a single pulse from t = 0 to t = T, the average power is calculated by integrating p(t) and dividing by T. For repetitive pulses at frequency f, multiply by f.

Pavg=1T∫0Tp(t) dtP_{avg} = \frac{1}{T} \int_0^T p(t) \, dt

Where:

VV= Supply voltageV
RR= Series resistanceΩ
CC= CapacitanceF
τ\tau= Time constants
I0I_0= Peak currentA
i(t)i(t)= Current at time tA
p(t)p(t)= Power at time tW
ppeakp_{peak}= Peak powerW
ERE_R= Total energy in resistorJ
I2tI^2t= Joule integralA²·s
weqw_{eq}= Equivalent pulse widths
ff= Pulse frequency (optional)[-]

📘 Usage Notes

  • • Time constant τ: At t = τ, current drops to 36.8% (1/e). At t = 5τ, it's down to 0.67% (practical "complete charge").
  • • 50% energy split: Always half in resistor (heat), half in capacitor (stored), regardless of R and C values.
  • • I²t rating: Compare calculated I²t with component datasheet pulse rating. Units are squared: 1 A²·s = 10⁶ mA²·s.
  • • Equivalent pulse: weq = τ/2 is the width of a rectangular pulse at peak power that dissipates the same energy.
  • • Average power: For single pulse over time T, Pavg = ER/T. For repetitive pulses at frequency f, Pcontinuous = f × ER.
  • • Component derating: Ensure peak current I₀ doesn't exceed resistor's pulse current rating, and I²t is within limits.
  • • Thermal analysis: Use average power calculations to verify resistor's continuous power rating isn't exceeded.
  • • Practical completion: At 5τ, the capacitor is 99.3% charged and 99.3% of energy has been transferred.