You have a 81.9 81.9 mF capacitor initially charged to a potential difference of 11.1 11.1 V. You discharge the capacitor throug h a 3.19 3.19 Ω resistor. What is the time constant?
2 answers:
Answer:
The time constant is 0.26 s
Explanation:
From ohm's law,
Current (I) = potential difference ÷ resistance = 11.1 ÷ 3.19 = 3.48 A
Quantity of electricity (Q) = capacitance × potential difference = 81.9×10×^-3 × 11.1 = 0.90909 C
time constant (t) = Q/I = 0.90909/3.48 = 0.26 s
<h2>
Answer: </h2>
0.261s
<h2>
Explanation: </h2>
For an R-C discharging circuit, the time constant τ, is the product of the capacitance (C) of the capacitor and the resistance (R) of the resistor. i.e
τ = R x C ------------------(i)
From the question;
R = 3.19Ω
C = 81.9mF = 81.9 x 10⁻³F
Substitute these values into equation (i) as follows;
τ = 81.9 x 10⁻³ x 3.19
τ = 261.261 x 10⁻³
τ = 0.261s
Therefore, the time constant is 0.261s
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