3. A capacitor is connected to a battery and becomes fully charged (see Fig. 2),

Question

3. A capacitor is connected to a battery and becomes fully charged (see Fig. 2), It is then s then disconnected from the battery and connected to a 10 k resistor (Fig. 2A below). The voltage across the capacitor is measured and the half-life is found to be 10 seconds. What would the half-life be if the capacitor were discharged through two 10 kQ resistors connected in series instead of the single resistor (Fig. 2B)? what about for two 10 k resistors connected in parallel (Fig. 2C)? You should provide numerical values and explain your reasoning for the two cases.

Explanation / Answer

RC circuit

discharging of a capacitor

with R = 10 k.ohm

capacitance C

time constant is T = R*C , which is time taken to complete one cycle

given half life is T/2 = 10 s ==> T = 20 s , so the capacitance of the capacitor is  

T = R*C

20 = 10*10^3*C

C = 2 mF

we know that while discharging the capacitor the voltage across the capacitor is

V(t) = V0*e^(-t/R*C)

now the resistance is R = 2*10*10^3 ohm

v(t/2) = V0*e^(-t/(2*2*10*10^3*2*10^-3))

v(t/2) = Q/C *(e^-t/80)

now the resistors are connected in parallel so the resultant resistance is 5000 ohm

V(t) = V0*e^(-t/R*C)

resistance is R = 5*10^3 ohm

v(t/2) = V0*e^(-t/(5000*2*10^-3))

v(t/2) = V0*e^-t/10

v(t/2) = Q/C *(e^-t/10)

from the above two expression

compraring it with standard equation

v(t) = V0*e^(-t/RC)

the quantity RC is time constant  

first case it is RC = 80 second case RC = 10  

so the resistance is decreasing in the second case so that the time period is large so as the half life or half cycle

if we know the value of v(t ) in terms of V0 we get the value of t half

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