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Circuit and electromagnetism basics

An uncharged capacitor C is fully charged from a DC voltage E through a resistor R. Which correctly gives the energy dissipated in the resistor during charging?

  1. A(1/2)CE² regardless of R—half of the energy CE² supplied by the source
  2. BIt decreases as R decreases, approaching 0 as R approaches 0
  3. CCE² regardless of R
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Answer: A. (1/2)CE² regardless of R—half of the energy CE² supplied by the source

During charging the source delivers a charge Q = CE, so it does work QE = CE². The capacitor stores (1/2)CE², and the remaining (1/2)CE² becomes heat in the resistor. Integrating the resistor's power with i = (E/R)e^(−t/RC) gives ∫i²R dt = (E²/R) × RC/2 = (1/2)CE², in which R cancels out. A smaller R just makes the current larger and briefer, leaving the total loss unchanged. The dynamic power CV²f of CMOS logic comes from this same charge–discharge loss.

  • A. Correct. Half of the source's work CE² is lost in the resistor, independent of R.
  • B. A smaller R makes the current larger, losing the same energy in a shorter time.
  • C. CE² is the total work done by the source; half of it is stored in the capacitor.
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