Account for energy during charging and discharge

Use one energy boundary for each history

The charging history begins with zero stored capacitor energy and an active 6 V source. The discharge history begins with a charged capacitor and no source. Their energy accounts therefore have different starting terms:

• Charging: source work = stored energy now + resistor heat.
• Discharging: initial stored energy = stored energy now + resistor heat.

The model assumes constant ideal components and omits every loss except dissipation in R. The energy table evaluates each complete history from its own time zero. It does not add the two rows together as if they described one simultaneous circuit.

Calculate the long-time totals first

For C = 0.001 F and final voltage 6 V, final stored energy is C V²/2 = 0.018 J = 18 mJ. The source transfers final charge Q = CV = 0.006 C through a constant 6 V potential difference, so its total work is VQ = 0.036 J = 36 mJ. The remaining 18 mJ is dissipated in the resistor during charging.

During a complete ideal discharge from 6 V, the initial 18 mJ is ultimately dissipated in the resistor. The source is absent; no additional source work is required. These are infinite-time limits. The table’s finite-time values approach them without being replaced by them.

This charging result differs from the previous quasistatic geometry-change experiment. Here capacitance is fixed and a sudden source connection creates a current through resistance. Omitting resistor heating would make the accounting incomplete.

Check an intermediate time

At R = 2 kΩ and t = 2 s, one time constant has elapsed. The charging row shows approximately 22.7563 mJ of source work, 7.1924 mJ stored, and 15.5640 mJ dissipated. Add the last two using full precision before rounding to recover source work.

The discharge row starts with 18 mJ, retains approximately 2.4360 mJ, and has dissipated 15.5640 mJ. The equal heat in the two histories at equal time follows from their equal current magnitudes with these chosen initial conditions, because heating depends on I². Their stored energies are different.

The two instantaneous resistor powers are also equal, but power and accumulated heat have different units. Power is measured in milliwatts (mW); accumulated heat is millijoules (mJ). At the initial instant, resistor power is nonzero while accumulated heat is zero. Multiplying the final instantaneous power by elapsed time would underestimate the heat because current was larger earlier.

Follow the changing rate without needing calculus first

Imagine dividing the history into short intervals. In each interval, approximate energy dissipated as the local power I²R times interval duration, then add the contributions. Smaller intervals improve the approximation for a smooth history. This is the idea behind integrating power over time.

The browser uses analytic formulas. If x = t/τ and U final = C V source²/2, accumulated resistor heat in either chosen history is U final [1 − exp(−2x)]. Charging source work is C V source²[1 − exp(−x)]. These give zero accumulated work and heat at t = 0; stored discharge energy starts at U final instead of zero.

Resistance changes the rate and the observation window

For the same ideal source and capacitance, increasing R slows charging and discharge and lowers initial current. It does not change the infinite-time totals above for this model. At a fixed finite time, however, different resistances give different accumulated heat and stored energy. Compare R = 1 kΩ and R = 5 kΩ at 10 s to see why a finite-time comparison cannot be substituted for an asymptotic claim.

The model does not allow R = 0. Removing resistance from an ideal instantaneous charging circuit produces a singular case requiring additional physics; it is not a justified way to claim loss-free sudden charging. Likewise, dissipated energy does not determine resistor temperature without thermal properties and cooling information.

Transfer to a new source and capacitor

An initially uncharged 500 µF capacitor is charged through a positive resistor by a constant 4 V ideal source. Find the limiting source work, stored energy and resistor heat. What heat would a subsequent complete resistor-only discharge release?

Check the RC energy transfer

Final stored energy is 0.5 × 0.0005 × 4² = 0.004 J = 4 mJ. Source work is 0.0005 × 4² = 0.008 J = 8 mJ, so charging dissipates 4 mJ. A later discharge from 4 V dissipates the stored 4 mJ. The charging and discharge heats belong to successive processes in this hypothetical example; the browser's paired rows have independent initial conditions.