(a) Compute the charge on the capacitor just before the switch is thrown from po

Question

(a) Compute the charge on the capacitor just before the switch is thrown from position 2 back to position 1.
C

(b) Compute the voltage drops across the resistor and across the capacitor at the instant described in part (a).
Across the resistor.
V
Across the capacitor.
V

(c) Compute the voltage drops across the resistor and across the capacitor just after the switch is thrown from position 2 back to position 1.
Across the resistor.
V
Across the capacitor.
V

(d) Compute the charge on the capacitor 85.0 ms after the switch is thrown from position 2 back to position 1.
C

Explanation / Answer

a) for 85 ms capacitor will be charged.

SO potential across capacitor will be

V = E[ 1 - e^(-t/RC)]

V = 21 [ 1 - e^(-0.085 / (900x 4x 10^-15))]

V = 21 Volt

SO charge Q = CV = 4 x 10^-15 x 21 =8.4 x 10^-14 C

b) Voltage across resistor = E - V(across C)
= 21 - 21 = 0

Across capacitor = 21 Volt ( As calculated in above part)

c) potential does not change instantly (In case of RC circuit)
V_R = 0

V_C =21 V

d) 85 ms is very very large compared to RC (time constant).

so charge across capacitor will become zero again.

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