Physics · Condensed Matter Physics
The current IJ(t) through a Josephson junction and the voltage V(t) across it are given by: IJ(t) = IC sin φ(t) dφ(t)dt = 2eV(t)↔ where IC is the critical current of the junction and φ(t) is the phase difference across the junction. A resistor R is connected in parallel to the junction, and a constant current I > IC flows through the combination. The energy dissipated in R during the time phase φ changes by 2π is:
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1↔2e(I + IC)
- 2↔2e(I - IC)
- 3↔2eIC
- 4↔2eI✓
Correct answer
↔2eI
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