Final answer:
Using the RC time constant formula t = RC ln(2), it takes approximately 17.0 seconds for a 7.9-mF capacitor to lose half of its initial stored energy when discharged through a 3.1-kΩ resistor. option B is correct.
Step-by-step explanation:
To calculate how long it will take for a 7.9-mF capacitor to lose half its initial stored energy when discharged through a 3.1-kΩ resistor, we use the formula t = RC ln(2) where t is the time, R is the resistance, C is the capacitance, and ln(2) is the natural logarithm of 2, which accounts for the time it takes to lose half the energy.
By plugging in the values, we get:
C = 7.9 mF = 7.9 × 10−6 F
R = 3.1 kΩ = 3.1 × 103 Ω
So the time t can be calculated as:
t = (7.9 × 10−6 F) × (3.1 × 103 Ω) × ln(2)
= 7.9 × 3.1 × ln(2) × 10−3 s
= 24.49 × ln(2) s
= 24.49 × 0.693 s
= 16.9557 s
Hence, it will take approximately 17.0 seconds for the capacitor to lose half its initial stored energy, which corresponds to answer choice (b).