Part A
What value resistor will discharge a 2.60 μF capacitor to 20.0% of its initial charge in 2.50 ms?
SOLUTION :
We use the following formula :
Q = Q_{0} e^(- t / R C)
=> Q / Q_{0} = e^ ( - t / RC)
where,
Q / Q_{0} = 0.20
t = time taken in secs for discharge to 0.20 Q_{0} = 2.50 ms = 2.50^10^(- 3) sec.
R = Resistance in Ωs to be determined .
C = Capacitor capacitance = 2.60 µF= 2.60*10^(-6) F .
Hence,
0.20 = e^(- 2.50*10^(-3) / (R * 2.60*10^(-6))
=> 0.20 = e^(- 961.53846 / R)
Taking natural logarithm :
=> ln(0.2) = - 961.53846 / R
=> R = - 961.53846 / ln(0.2)
=> R = 597.44 Ω approx.
=> Resistor of 597.44 Ω approx. (ANSWER).
What value resistor will discharge a 2.60 μF capacitor to 20.0% of its initial charge in 2.50 ms?
Part AWhat value resistor will discharge a 2.60 μF capacitor to 20.0 % of its initial charge in 2.50 ms?
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