For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:
Q(t) = Cℰ(1-e^{-t/(RC)})
Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.
The time constant of the circuit τ is the product of the resistance and capacitance:
τ = RC
Q(t) can be rewritten as:
Q(t) = Cℰ(1-e^{-t/τ})
We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Plug in these values and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
The charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
The charge stored on one of the capacitor's plates can be calculated by,
[tex]\bold{Q(t) = C\epsilon(1-e^{-t/(RC)})}[/tex]
Where,
Q(t) - the charge,
t - time,
ℰ- the battery's terminal voltage,
R - the resistor's resistance,
C - the capacitor's capacitance.
The time constant of the circuit τ is equal to the product of the resistance R and capacitance C :
τ = RC
The amount of charge Q(t) when t = τ. put the values,
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Put the values in the formula and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
Therefore, the charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
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