What value resistor will discharge a 3.0 μF capacitor to 10 % of its initial charge in 2.1 ms ?
SOLUTION :
We use the following formula :
Q = Q0 e^(- t / R C)
=> Q / Q0 = e^ ( - t / RC)
where,
Q / Q0 = 0.10 (capacitor is discharged to 10% of initial value)
t = time taken for discharge to 0.20 Q0 is 2.1 ms = 2.1^10^(- 3) sec.
R = Resistance in Ωs to be determined .
C = Capacitor capacitance = 3.0 µF= 3.0 *10^(-6) F .
Hence,
0.10 = e^(- 2.1*10^(-3) / (R * 3.0 *10^(-6))
=> 0.10 = e^(- 700 / R)
Taking natural logarithm :
=> ln(0.10) = - 700 / R
=> R = - 700 / ln(0.10)
=> R = 304.0 Ω approx.
=> Resistor of 304.0 Ω approx. (ANSWER).
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