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Asked in NEET 2023 · Capacitor on an a.c. source
The capacitive reactance is X_C=1/(ω C), so lowering ω raises X_C.
The conduction current in the circuit is I=V/(X_C), which therefore falls.
Inside the capacitor there is no conduction current; the displacement current there is exactly
equal to the conduction current in the wires, so it falls with it.
The displacement current decreases.
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