Thankyou
SOLUTION :
We use the following formula :
Q = Q0 e^(- t / R C)
=> Q / Q0 = e^ ( - t / RC)
where,
Q / Q0 = 0.20 (capacitor is dichrged to 20% of initial value)
t = time taken in secs for discharge to 0.20 Q0 = 2.60 ms = 2.60^10^(- 3) sec.
R = Resistance in Ωs to be determined .
C = Capacitor capacitance = 2.50 µF= 2.50*10^(-6) F .
Hence,
0.20 = e^(- 2.60*10^(-3) / (R * 2.50*10^(-6))
=> 0.20 = e^(- 1040 / R)
Taking natural logarithm :
=> ln(0.2) = - 1040 / R
=> R = - 1040 / ln(0.2)
=> R = 646.19 Ω approx.
=> Resistor of 646.19 Ω approx. (ANSWER).
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