Capacitor and displacement current.

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SUMMARY

The discussion centers on the behavior of a capacitor with capacitance 'C' being charged by a DC source, specifically addressing whether an ammeter will show a momentary deflection during the charging process. It is established that the ammeter will indeed show a momentary deflection due to the presence of both conduction and displacement current, as described by Maxwell's displacement current equation. The voltage across the capacitor, VC, builds up rapidly according to the formula VC = E(1 - e^(-t/RC)), where E is the EMF of the DC source, leading to an exponential decay of current, i = (E/R)e^(-t/RC), until the capacitor is fully charged.

PREREQUISITES
  • Understanding of RC circuits
  • Familiarity with Maxwell's displacement current equation
  • Knowledge of exponential functions and their applications in electrical circuits
  • Basic concepts of capacitance and charging processes
NEXT STEPS
  • Study the implications of Maxwell's displacement current in electromagnetic theory
  • Explore the behavior of RC circuits during transient states
  • Investigate the effects of varying resistance on capacitor charging times
  • Learn about the applications of capacitors in AC circuits
USEFUL FOR

Students of electrical engineering, physics enthusiasts, and professionals involved in circuit design and analysis will benefit from this discussion, particularly those interested in the dynamics of capacitors and current flow in circuits.

altamashghazi
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Homework Statement



a capacitor of capacitance 'C' is being charged across a dc source with an ammeter . will the ammeter show a momentary deflection during the process of charging? if so, how would it and the resulting continuity of current in the circuit be explained ?

Homework Equations



maxwell's displacement current equation

The Attempt at a Solution



continuity is bcoz conduction +displacement current is same everywhere in circuit but what about momentary deflection ?
 
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You basically have a RC circuit with very little resistance so that the voltage over the capacitor , VC,will build up very fast during the charging process according to

VC = E(1 - e-t/RC)

where E is the emf of the dc source. The voltage of the capacitor opposes that of the dc source and the current will stop when the two voltages are the same - that is the charging process stops. The charge build up on the capacitor will be

q = CVC

so that the current will be decaying exponentially with time

i = dq/dt = E/R e-t/RC
 

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