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 through a 3.19 3.19 Ω resistor. What is the time constant?

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

i wanna say it’s between b and d, leaning more towards b.

he . if i may ask what i your question exactly?

explanation: