Question

What value resistor will discharge a 2.60 μFμF capacitor to 20.0%% of its initial charge in 2.20 msms ? nothing Ω

Answer 1

Answer:

The value of resistor is 525.76 ohms

Explanation:

Given;

capacitance, C = 2.6 μF = 2.6 x 10⁻⁶ F

let the initial charge on the capacitor = Q₀

time , t = 2.2 ms = 2.2 x 10⁻³ s

The discharge of the capacitor is given by;

$Q(t) = Q_oe^{-\frac{t}{RC} }$

Where;

Q(t) is the charge after 2.2ms = 20% of Q₀ = 0.2Q₀

$Q(t) = Q_oe^{-\frac{t}{RC} }\\\\0.2Q_o = Q_oe^{-\frac{t}{RC} }\\\\0.2 = e^{-\frac{t}{RC} }\\\\ln(0.2) = -\frac{t}{RC}\\\\-1.6094 = -\frac{t}{RC}\\\\R = \frac{t}{1.6094C} \\\\R = \frac{2.2*10^{-3}}{1.6094*2.6*10^{-6}}\\\\R = 525.76 \ ohms$

Therefore, the value of resistor is 525.76 ohms